Multiplying image to coordinates and rotation
Hi,
I was wondering if anyone could help me write a Adobe Illustrator script.
The script must be able to:
1. Read a file containing several coordinates (X,Y) and angles.
2. Select an image (or load an image).
3. Copy and paste an image to the coordinates and rotate the images by the given angles.
4. Repeat pasting/rotating until all coordinates are pasted with images.
The result should be something like the image attached below.
Can this be possible with an illustrator script?
Could anyone guide me how I should approach this?
Could you recommend any online references on programming an illustrator script?
I don't always have access to Illustrator.
Many thanks in advance,
Anthony
Hi Carlos,
I have a bit of programming background. I'm familiar with VB, C, C++. I'm a novice on illustrator.
I attempted at writing quick "copy, paste & rotate" script in javascript, see below. But I need to somehow get the coordinates from another file (a text file).
Can I use something like this on the Illustrator script?
http://www.javapractices.com/topic/TopicAction.do?Id=42
var selectedObjects = app.activeDocument.selection; var topObject = app.activeDocument.selection[0];
var msgType = "";
if (app.documents.length>0)
myselection = app.activeDocument.selection;
if (myselection.length>0)
for (var i=0; i<myselection.length; i++) {
myselection[i].duplicate()
myselection[i].translate(200,0)
myselection[i].rotate(13)
myselection[i].duplicate()
myselection[i].translate(300,50)
myselection[i].rotate(50)
myselection[i].duplicate()
myselection[i].translate(600,10)
myselection[i].rotate(45)
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JS CS3 flag duplicate images with scale and rotation differnce
I've been trying to cut down on the number of images a script I have would produce. The old script would look for links that are being used multiple times and then just version the link which is fine. What I am finding this that although a document may have one image linked 9 time 6 times out of the nine the image is scaled and rotated the same amount and the other 3 times the rotation and scaling also match. In this example the image only needs 2 versions the first as the original and the second with the scale and rotation of the image used 3 time let say.
So to the code:
The following code will work through the doc and create an array of the links as duplicates:
function LinkUsage(myLink) {
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Is there a way to loop through the array and only flag the differnent scale and roatations. What I should get is that out of the multiple link uasage the image is only changed twice out of the 9 times used.
I was trying to loop through the array and then loop through it within the loop to compare the link but this does not work for obvious reasons.
for (var i=0 ; i <myNumLinks.length; i++){
var valLink = myDoc.links.itemByID( myNumLinks[i]);
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for (var j=0 ; j<myNumLinks.length; j++){
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$.write(myNumLinks[i]+"\n");
Is what I am trying to do possible and if so any suggestions would be appreciated.
Cheers, John.I think I might be one step closer to getting this to work. I've been able to just capture the file path of the duplicates using:
//count the number of times the link is used
function LinkUsage(myLink) {
var myLinkCounter = 0;
for (var myCounter = 0; myCounter < myDoc.links.length; myCounter++) {
if (myLink.filePath == myDoc.links[myCounter].filePath) {
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return myLinkCounter
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function chkVal(val){
var compCount = 0;
for(var arr = 0; arr < myNumLinks.length; arr++){
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compCount++;
if(compCount >=1){
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return true;
//check the links inthe doc
var myDoc = app.activeDocument;
var myNumLinks = [];
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A library on my hard drive became corrupted, so I over-wrote it from an external hard drive. When I now access the library on my hard drive, many of the images are vertical, and cannot be rotated other than one at a time. Ideas to correct this?
You had already done "Repair" and "Rebuild"? These usually takes scores of minutes or hours.
Does the Library on your external drive work from the external drive? Navigate to it in Finder and double-click the name.
If it does, delete the problem Library, empty the System Trash, reboot, Verify your system disk using Disk Utility (and might as well repair permissions while you're at it), reboot, test the external drive Library once more to confirm it works properly, then close it and copy the Library to your system drive. When the copying is done, double-click the Library (now on the system drive) and see what you get. -
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I have been trying to scale the image and rotate it before i pass it to the DisplayJAI but i keep running out of memory.
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private Image rotateImage(boolean rotateRight)
Image img = imageHolder.getInUseImageIcon().getImage();
Image rot = null;
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int imageHeight = imageHolder.getInUseImageIcon().getIconHeight();
int buffer[] = new int[imageWidth * imageHeight];
int rotate[] = new int[imageWidth * imageHeight];
try
MediaTracker tracker = new MediaTracker (this);
tracker.addImage (img, 0);
tracker.waitForAll();
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grabber.grabPixels();
catch(InterruptedException e) {
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for(int y = 0; y < imageHeight; y++)
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else
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for(int y = 0; y < imageHeight; y++)
for(int x = 0; x < imageWidth; x++)
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catch (Exception e)
e.printStackTrace();
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Image scaled = originalImage.getImage().getScaledInstance(
(int)(currentWidth * 1.5),
(int)(currentHeight * 1.5),
Image.SCALE_FAST);
I tried this method, it works and does what i want but the scrolling and loading is too slow.
public void paintComponent(Graphics g)
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Hello,
I would like to perform flip and +/-90� rotation on an image but after a rotation, the flip is performed in the wrong direction my code looks like this :
/*axis at the center of image*/
g2D.translate(width/2,height/2);
g2D.scale(-1,1);
g2D.rotate(Math.PI);
/*image orientation correction*/
g2D.scale(1,-1);
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g2D.rotate(flipHorizontal*flipVectical*Math.toRadians(angle));
g2D.scale(flipHorizontal,flipVectical); //1 or -1
/* drawing the image */
g2D.drawImage(imageToDraw, centreToX,centreToY, null);
// than positioning the axis back to origine I've tried to say that if the image as already been rotated then invert the flips, but in that case, after one flip the rotation is performed in the wrong direction...
Does anybody know how to solve this problem???
Thank you in advance.i have to rotate, flip & scale an image , but it get misplaced when rotated on 90 and -90 please help me public void putImage_buffer(String s,int x,int y,int width,int height,double rotation,int xScale,int yScale,boolean hFlip,boolean vFlip) {
int Rwidth=width;
int Rheight=height;
try{
ImageInputStream imageInput = ImageIO.createImageInputStream(new FileInputStream(filename));
it = ImageIO.getImageReaders(imageInput);
reader = (ImageReader) it.next();
reader.setInput(imageInput);
objBufimage=reader.read(0);
}catch(Exception e){System.out.println(e);}
System.out.println("width : height! ratation"+width+" : "+height+" : "+rotation);
if(width==0&&height==0){
offscreen.drawImage(objBufimage,0,0, null);
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Rheight=width;
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rotation=270;
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Rwidth=height;
Rheight=width;
if(vFlip )
if(rotation==90)
hFlip=!hFlip;
vFlip=!vFlip;
//Rwidth=height;
//Rheight=width;
System.out.println("width : height 90 vFlip");
if(rotation==-90)
hFlip=!hFlip;
vFlip=!vFlip;
//Rwidth=height;
//Rheight=width;
rotation = 270;
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if(rotation==180)
System.out.println("width : height 180 vFlip");
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vFlip=!vFlip;
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//Rheight=width;
if(rotation==-90)
{ rotation=270;
hFlip=!hFlip;
vFlip=!vFlip;
Rwidth=height;
Rheight=width;
if(rotation==180)
System.out.println("width : height 180 hFlip");
if(vFlip && hFlip)
if(rotation==90)
System.out.println("width : height 180 hFlip&&vFlip 90");
if(rotation==-90)
rotation=270;
System.out.println("width : height 180 hFlip&&vFlip -90");
AffineTransform tx = new AffineTransform();
AffineTransform origTransform = offscreen.getTransform();
System.out.println("hFlip:vFlip Math.PI "+hFlip+" : "+vFlip);
rotation=rotation*(hFlip?-1:1)*(vFlip?-1:1);
rotation = Math.toRadians( rotation );
AffineTransform rotate=AffineTransform.getRotateInstance(rotation, Rwidth/2d, Rheight/2d);
AffineTransform moveToImageCenter=AffineTransform.getTranslateInstance(Rwidth/2d, Rheight/2d);
moveToImageCenter.concatenate(AffineTransform.getScaleInstance(hFlip?-1:1,vFlip?-1:1));
moveToImageCenter.concatenate(AffineTransform.getTranslateInstance(Rwidth/-2d, Rheight/-2d));
AffineTransform flip=moveToImageCenter;
AffineTransform transformer = AffineTransform.getTranslateInstance(x,y);
offscreen.setTransform(transformer);
offscreen.transform(rotate);
offscreen.transform(flip);
offscreen.drawImage(objBufimage,0,0,Rheight,Rwidth,null);
//offscreen.drawImage(objBufimage.getScaledInstance(width,height,objBufimage.SCALE_DEFAULT),0,0,width,height, null);
offscreen.setTransform( origTransform );
}i get vFlip if hFlip+90or-90 and i get hFlip if vflip +90-90 means if actual flipping is vflip+90/-90 my method parameter gets hFlip +/-90 thats by i'm changing values -
How to find out the coordinates of rotated textframe
Hi All,
With the help of (idml) itemtransform horizontal / vertical distances and path point arrays, I can find out the coordinates of text frame (x1,y1), (x2,y2), (x3,y3), (x4,y4). If the textframe is rotated then item transform values are getting changed, but the path point array values are same. I can find rotation angle by the matrix [cos(θ) sin(θ) -sin(θ) cos(θ) 0 0], But I could not get the exact coordinates of rotated textframe. The textframes are given below.
Normal Text frame
<TextFrame Self="u136" ParentStory="u124" ItemTransform="1 0 0 1 101.72727272727272 -349.41818181818184">
<Properties>
<PathGeometry>
<GeometryPathType PathOpen="false">
<PathPointArray>
<PathPointType Anchor="-101.72727272727272 -46.581818181818164" LeftDirection="-101.72727272727272 -46.581818181818164" RightDirection="-101.72727272727272 -46.581818181818164"/>
<PathPointType Anchor="-101.72727272727272 -0.3272727272727103" LeftDirection="-101.72727272727272 -0.3272727272727103" RightDirection="-101.72727272727272 -0.3272727272727103"/>
<PathPointType Anchor="115.9090909090909 -0.3272727272727103" LeftDirection="115.9090909090909 -0.3272727272727103" RightDirection="115.9090909090909 -0.3272727272727103"/>
<PathPointType Anchor="115.9090909090909 -46.581818181818164" LeftDirection="115.9090909090909 -46.581818181818164" RightDirection="115.9090909090909 -46.581818181818164"/>
</PathPointArray>
</GeometryPathType>
</PathGeometry>
</Properties>
</TextFrame>
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<TextFrame Self="u136" ParentStory="u124" ItemTransform="0 1 -1 0 320.3805483338268 -125.07900895050204">
<Properties>
<PathGeometry>
<GeometryPathType PathOpen="false">
<PathPointArray>
<PathPointType Anchor="-101.72727272727272 -46.581818181818164" LeftDirection="-101.72727272727272 -46.581818181818164" RightDirection="-101.72727272727272 -46.581818181818164"/>
<PathPointType Anchor="-101.72727272727272 -0.3272727272727103" LeftDirection="-101.72727272727272 -0.3272727272727103" RightDirection="-101.72727272727272 -0.3272727272727103"/>
<PathPointType Anchor="115.9090909090909 -0.3272727272727103" LeftDirection="115.9090909090909 -0.3272727272727103" RightDirection="115.9090909090909 -0.3272727272727103"/>
<PathPointType Anchor="115.9090909090909 -46.581818181818164" LeftDirection="115.9090909090909 -46.581818181818164" RightDirection="115.9090909090909 -46.581818181818164"/>
</PathPointArray>
</GeometryPathType>
</PathGeometry>
</Properties>
</TextFrame>
When I converted the values of rotated textframe in to coordinates and drawn in a screen then I am not getting the exact position where it drawn in the original.
Can anyone help me to find out the cordinates of rotated textframe.
Thanks in advance.It seems pretty straightforward to me.
Your center point is (42.375 mm, 27.458 mm) and your box is 58.417 mm x 28.417mm.
IDML defines the rectangle by its corners, not its center, so let's find the upper-left corner. Divide the width by two and subtract from the x, same for the height and the y. You get (13.665 mm, 13.2495 mm).
That is in page coordinates relative to the upper-left corner of the page, we can see from your rulers.
But IDML coordinates are spread-relative from the center of the spread. Your spread is a single page that is 85 mm x 55 mm.
So if we translate your upper-left corner, it is (13.665, 13.2495) mm - (85/2, 55/2) => (29.3335 mm, 14.2505mm) in spread-relative coordinates.
But IDML coordinates are in points, not in mm. So we convert (multiply by 2.835). And we get (83.150 pt, 40.395 pt).
But that doesn't match up with your IDML file, which has (-82.650 pt, -39.890pt).
But look at the difference: (0.500 pt, 0.505 pt). That's really 1/2 point, because your box has a 1-pt border and that gets split evenly across all 4 sides. And there's 0.005 of round-off error, bceause floating point math sucks.
Any questions? -
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Here is the image that I'm working with...
I want to make this object rotate a tight 90 degrees in After Effects and then separate. If you're wondering why I put this up under Photoshop it is because I need to have this model in addition to 2 others, one of the top half and one of the bottom half and button, both turned 90 degrees [preferably clockwise] in order to create the GIF. I need these 3 pieces before I can even consider creating the GIF in After Effects... so if somebody either wants to try and do it for me or to try and help me to do it on my own, either one would be extremely helpfulDon't worry guys, I actually found the answer all on my own what I needed was to separate the image into 3 files, something I was probably going to do anyways to make exporting and importing the images easier and rotating the two top and bottom half images here's what I have right now (per background removal)Thanks for trying to help... but I think I'm ready to move on to After effects with this one The reason I'm moving it to after effects is simply because I understand simple movement/animation there already... if you're curious as to what I plan on doing with these images, the complete ball will sit in a hole in my border, turn a sharp 90 degrees right and then be replaced by the two half images, the bottom one will fade slowly out while the top is lifted off screen to the right... it's a fancy reveal for my upcoming project thanks again
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http://www.belightsoft.com/products/imagetricks/overview.php
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When I drag an image into IBA, it rotates 90 degrees.
I'm having trouble dragging a particular image into iBooks Author. I took a picture and imported from my camera to iPhoto. The image had to be rotated 90 degrees, which I did in iPhoto. I'm using the "Contemporary" template in iBooks Author and I'm trying to drag my image from iPhoto to replace the red, space themed image at the top of a section page. When I do so, the image reverts to its original instead of the rotated form. I can't figure out how to rotate the image without rotating the mask.
Hello Abram1,
Not sure of the reasons but I would imaging it might have to do with iPhoto and its non destruction approach to editing.
Have you tried exporting the image out as a new image that would then have the new rotation that you wish applied to it.
I hope this helps,
Regards,
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Hi there,
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// JUST TO TEST
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Thanks,
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kglad wrote:
try:
var BMP:Bitmap = Bitmap (LOADERINFO.content);
BMP.pixelSnapping = PixelSnapping.ALWAYS;
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BMPD.draw(BMP,matrix);
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// JUST TO TEST
addChild (BMP);
addChild (BMP2); -
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Hi
I have a Canon EOS 350D camera and now have PSE 9. I have taken some RAW images & imported them into PSE 9 OK. I know the original RAW image is retained after editing. However, after cropping a RAW image in the Editor and then "Opening" and "Saving as" a PSD or JPEG, I then close the image in Editor and the cropped image appears in the Organiser. That is as it should be, I think. If I then wish to start again from scratch on the original RAW image, it seems to me that I have to manually delete the edited version and the .xmp file in Windows Explorer, and then double click the original RAW file in Windows Explorer, which then opens correctly in PSE 9 Editor. Can this be done from PSE 9?When you open the original camera raw image, go to the flyout menu
and choose Camera Raw Defaults, that resets everything but the crop, straighten,
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To reset the Image rotation settings, you have manually rotate back using the image rotate buttons.
MTSTUNER -
Curious issue with extrude and rotate
I've found a curious issue with extrude and rotate in CS6. It's not a bug, but it is a nonintuitive behavior that is confusing, and sometime troublesome. I think this is something that ought to go on the to-do list for the devs. Anybody else agree? Disagree?
Example:
I began with an arrow-shaped path, to which I applied the "extrude" effect. I then put an icon on top of the arrow, and grouped the icon and arrow. Now, I'm using this icon-arrow in several different parts of my design, in different orientations. As I rotate the arrow, the relative position of the icon and the arrow appears to shift. I believe this is happening because, as you rotate the icon-arrow, Illustrator rotates both the icon and the arrow about the center of the bottom face of the arrow. This is because the top face of the arrow is essentially fictitious, just an artifact of the extrude effect. So, even though the icon appears as if it is drawn on the top face of the arrow, it behaves as if it is attached to the bottom face of the arrow. When you apply a transform, that transform affects the bottom face of the arrow, causing the top face to shift relative to the icon.
The solution would be to add an extra check box to the extrude effect, asking whether you wanted to extrude upwards (the way things are currently done) or downwards. If you could extrude downwards, then the top face of the arrow would be the "real" face, the one affected by transformations. You could then put an image on the top face, group the arrow and the image, and when you applied a transform, the icon would remain fixed relative to the top face, which would meet your intuitive expectations.
If anyone is curious to experiment with this particular example, I've posted this file to Google docs here:
https://docs.google.com/file/d/0B0Hoxcl5zTEtTXo5QnNsMGwwOGc/edit?usp=sharingbut it does work because i downloaded your file and i did jsut that. it isn't exact, but your suggestion on how to fix it wouldn't fix it anyway. the reason the acorn doesn't match is because of the distortion from the perspective not the direction of the extrusion. even if you could dictate the extrusion to go downward, the arrow would still rotate into perspective (distort) and then extrude. to avoid that, what you want is to extrude in axonometric and to do that you make your perspective rotation angle 1 degree, then make your extrusion length a lot longer to match what you want. with 1 degree there will be minimal distortion to the arrow as you rotate it and the acron will match.
i uploaded your file with your orignal 10 degree angle, and you'll see that they stay in sync pretty well. i drew a box around the acorn to better distinguish the difference. then i recreated the same thing with a 1 degree angle and on that version it stays in sync almost perfectly.
again. it doesn't matter wich direction you extrude since in perspective both faces will be distorted. if you want to keep the distortion from the perspective, you can apply the same perspective rotation to the acorn without the extrusion and it will distort in the same way as the arrow. i added that option to your file. they are all group and you can rotate them and see that they stay together pretty well, except the first one is a little off.
i uploaded your file with my solution
https://docs.google.com/file/d/0B4ilRoSBagc3WDY3MUR1bDNpV0E/edit -
Dear all,
In Photoshop CS5, it is possible to crop an image and rotate its edges manually.
Sometimes, it is possible to crop an image and move each line of the crop, BUT cannot rotate it and the rounded arrow does not appear when hover at an angle.
Sometimes, when cropping, it is possibel to rotate, BUT without the ability of moving each line of the crop.
Sometimes, when cropping, cannot moving each line of the crop and the cropping appears with fixed ratio.
1- How can I control these situations?
2- How can I move each line of the crop and rotate it?
Best Regards.Out of curiosity (using Ps CS5 32 12.03 on XP, OGL/GPU not active) I opened what appeared to be a grayscale .bmp file....
and it did crop with rotate OK,
Checking color mode it was RGB, so changed it to grayscale only, and voila, crop with rotate still works.
So all files, so far, work the same way....correctly handling crop with rotate OK.
Also tried on Ps CS5 12.03 Win7 64 laptop with OGL active, and it all works OK too. -
Coordinates of rotated textframe
With the help of itemtransform horizontal, vertical distance and path point arrays, I can find out the coordinates of text frame (x1,y1), (x2,y2), (x3,y3), (x4,y4). If the textframe is rotated then item transform values are getting changed, but the path point array values are same. I can find rotation angle by the matrix [cos(θ) sin(θ) -sin(θ) cos(θ) 0 0], But I could not get the exact coordinates of rotated textframe. The textframes are given below.
Normal Text frame
<TextFrame Self="u136" ParentStory="u124" ItemTransform="1 0 0 1 101.72727272727272 -349.41818181818184">
<Properties>
<PathGeometry>
<GeometryPathType PathOpen="false">
<PathPointArray>
<PathPointType Anchor="-101.72727272727272 -46.581818181818164" LeftDirection="-101.72727272727272 -46.581818181818164" RightDirection="-101.72727272727272 -46.581818181818164"/>
<PathPointType Anchor="-101.72727272727272 -0.3272727272727103" LeftDirection="-101.72727272727272 -0.3272727272727103" RightDirection="-101.72727272727272 -0.3272727272727103"/>
<PathPointType Anchor="115.9090909090909 -0.3272727272727103" LeftDirection="115.9090909090909 -0.3272727272727103" RightDirection="115.9090909090909 -0.3272727272727103"/>
<PathPointType Anchor="115.9090909090909 -46.581818181818164" LeftDirection="115.9090909090909 -46.581818181818164" RightDirection="115.9090909090909 -46.581818181818164"/>
</PathPointArray>
</GeometryPathType>
</PathGeometry>
</Properties>
</TextFrame>
Rotated textframe
<TextFrame Self="u136" ParentStory="u124" ItemTransform="0 1 -1 0 320.3805483338268 -125.07900895050204">
<Properties>
<PathGeometry>
<GeometryPathType PathOpen="false">
<PathPointArray>
<PathPointType Anchor="-101.72727272727272 -46.581818181818164" LeftDirection="-101.72727272727272 -46.581818181818164" RightDirection="-101.72727272727272 -46.581818181818164"/>
<PathPointType Anchor="-101.72727272727272 -0.3272727272727103" LeftDirection="-101.72727272727272 -0.3272727272727103" RightDirection="-101.72727272727272 -0.3272727272727103"/>
<PathPointType Anchor="115.9090909090909 -0.3272727272727103" LeftDirection="115.9090909090909 -0.3272727272727103" RightDirection="115.9090909090909 -0.3272727272727103"/>
<PathPointType Anchor="115.9090909090909 -46.581818181818164" LeftDirection="115.9090909090909 -46.581818181818164" RightDirection="115.9090909090909 -46.581818181818164"/>
</PathPointArray>
</GeometryPathType>
</PathGeometry>
</Properties>
</TextFrame>
Can anyone help me to find out the cordinates of rotated textframe.
Thanks in advance.I suspect you'll do better asking here: http://forums.adobe.com/community/indesign/indesign_sdk
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