Verify if string is valid ASCII
I have a string and I want to verify if it contains only valid ASCII, for example, if the string contains characters like the right allow, then it be should reject ! can anyone help? thanks!
What's the in the db? A series of bytes from a different character set? (One which, when displayed on a terminal (which may or may not understand the character set), displays a right arrow?)
Perhaps the ultimately correct solution is to use java.io.InputStreamReader, constructing it denoting the correct character set.
Or, to put data into the database using OutputStreamWriter, again specifying the correct character set.
Then you catch CharConversionExceptions. That's how you detect bad input.
Or, do the same kind of thing using String.getBytes and String constructors.
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I'm unable to install Itunes on my new PC. I'm logged in as the administrator And have deleted my browsing history. This is the error I get, ITunes installation error- “This installation package could not be opened. Verify that the package exists and that you can access it, or contact the application vendor to verify that this a valid Windows Installer package”... Any suggestions how to fix?
Hi
Try the following:
Uninstall iTunes and Quicktime
Reboot
in the task tray right click the quicktime icon and click exit.
navigate to the folder in program files and remove the quicktime directory and all its files.
Reboot.
Now try install iTunes and quicktime and it should work
Hope this helps. Please remember to click “Mark as Answer” on the post that helps you, and to click “Unmark as Answer” if a marked post does not actually answer your question. This can be beneficial to other community members reading the thread. -
I get this following message when installing the latest itunes on mozilla or IE10 was working on this machine before crash and reinstalled for HP disc I tried at leat 15 times firewall off /on microsoft essentials on/off on windows 7 64 bit see message
This installation package could net be opened verify thatthe package exist and that you can access for contact the application vendor to verify this is a valid windows installer packageIf you are not already doing so, install from an Administrator account.
If you haven't already done so, download the iTunes Installer from:
http://www.apple.com/itunes/download/
Then right click on the installer and select "Run as Administrator." (Even though you are using an admin account).
If that doesn't work, try creating a new administrator account and installing from there. -
Received message when trying to install Itunes or Safari for Windows. " Installation package could not be opened. Verify package exists or contact application vendor to verify this is a valid Windows Installer package".
I to am having the same issue. I have a admin account and logged into it as well and still have the same message pop up. Contacted live support the woman i spoke with had to send it up higher. We went through every thing including her trying (taking over as a host on my laptop) still have not bee contacted this June will be a year that i have heard nothing from them sine She di everything she could and pushed it up higher.
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Verifying a String. Is there a better way to do this?
Hi!!!
Maybe for some of you this is obvious...
I need to verify a String object. What I have to do is just a verification if there is a character different from 0 (zero) in a String. Then, the code would be something like this:
String test = "00001";
boolean ok = true;
for (int i = 0 ; i < test.length ; i++) {
if (test.charAt(i) != '0') {
ok = false;
break;
}I want to know if there is another different solution, more appropriate, without using that for loop. Is there a magic method in Java that can do that verification?
Thank you! ;-)(cue someone posting some regex foo that will dothis
in one line of something that looks liketransmission
line noise)Yeah, I've thought that. But, since I don't know
anything about regex, well, I'll do some homeworks
now...thank you!!!Your for loop is fine and much more efficient than any regex would be. -
Function module to verify if date is valid.
Anyone know a function module to verify if date is valid?thx
hiiii
This function module will be useful.check.
FM CALL FUNCTION 'DATE_CHECK_PLAUSIBILITY'
EXPORTING
date = pre_date
EXCEPTIONS
plausibility_check_failed = 1
OTHERS = 2.
IF sy-subrc 0.
MESSAGE e002(0s) WITH pre_date.
ENDIF.
reward if useful
thx
twinkal -
How to test if a day string is valid
A day in given month of a given year is valid if it meets the following conditions:
1)dayString >0
2)dayString < days in a given month of a given year
3)dayString.matches("^[0-9]{1,2}$")
i.e., the day must be a positive, number like, and must not be greater than a the number of days in a given month.
How to calculate the days in a given month of a given year?
Thanks to help.
Scott
P.S.:
In javascript it can be calculated by:
32 - new Date(year, mon-1, 32).getDate()
An option could be using SimpleDateFormat function, but unfortunately, it can only tell if a date is valid, not the specific day string. What I want to know is if the day string is valid for a given month and year.scottjsn wrote:
How to calculate the days in a given month of a given year?The java.util.Calendar class has a getActualMaximum() method.
An option could be using SimpleDateFormat function, but unfortunately, it can only tell if a date is valid, not the specific day string. That doesn't make any sense to me. If the "day string" isn't valid, the date isn't valid.
~ -
String Literal for ASCII Encoding
I need to know what is the STRING LITERAL for ASCII CHAR. ENCODING in Java (as ISO_LATIN-1 has "8859_1").
Thanks in Advance!import java.nio.charset.Charset;
import java.util.Set;
import java.util.Iterator;
class Test {
void m(String name) {
Charset s = Charset.forName(name);
System.out.println("display name= " + s.displayName());
Set aliases = s.aliases();
Iterator it = aliases.iterator();
while (it.hasNext()) {
String x = (String)it.next();
System.out.println("alias= " + x);
public static void main(String[] args) {
if (args.length > 0) new Test().m(args[0]);
}display name= US-ASCII
alias= us
alias= ISO_646.irv:1991
alias= ANSI_X3.4-1968
alias= iso-ir-6
alias= 646
alias= ISO646-US
alias= cp367
alias= ANSI_X3.4-1986
alias= csASCII
alias= ASCII
alias= iso_646.irv:1983
alias= IBM367 -
I have had to take my pc back to factory settins, so lost my itunes, tried to install again getting message saying
the installation package could not be opened verify the package exists and that you can access it or contact the vendor to verify that this is a valid windows package,Hi
Try the following:
Uninstall iTunes and Quicktime
Reboot
in the task tray right click the quicktime icon and click exit.
navigate to the folder in program files and remove the quicktime directory and all its files.
Reboot.
Now try install iTunes and quicktime and it should work
Hope this helps. Please remember to click “Mark as Answer” on the post that helps you, and to click “Unmark as Answer” if a marked post does not actually answer your question. This can be beneficial to other community members reading the thread. -
Find out if string is valid file name?
Is there a platform-independent method in the standard API to determine if a string contains only valid filename characters, and if not, convert it?
How about something like
File f = new File(String pathname)
then
if(f.isDirectory())
And
if(f.isFile())
And
if(f.exists())This would only work for existing files. You could try File.exists() as a first effort, and if this returned false, try creating a new directory with File.mkdirs(). Of course, this only works for local code (i.e. it wouldn't work in an Applet), and wouldn't work (for example) if the path contained a drive letter that didn't exist on your machine.
RObin -
Hex string conversion to ASCII Character string
I have a Hex String 494A4B4C and want this string to get converted in ASCII Character String. IJKL. How to do in Labview 8.5.
Here is a screenshot of the described code:
Ton
Message Edited by TCPlomp on 30-09-2009 01:35 PM
Free Code Capture Tool! Version 2.1.3 with comments, web-upload, back-save and snippets!
Nederlandse LabVIEW user groep www.lvug.nl
My LabVIEW Ideas
LabVIEW, programming like it should be!
Attachments:
Example_VI_BD.png 3 KB -
Hi,
I am creating a call logging system with a Java app that will collect the incoming/outgoing calls from a meridian telephone system. My question is this...........
The data from the calls comes in through the COM port, if i collect this and write it to the screen its a jumbled mess, i know the problem but dont know how to implement it, i need to force the text to convert to 7 bit ascii or something like that...
Can anyone point me in the right direction?If ur sure that the data that ur getting as ascii characters, then u can store these ascii charters in an array of bytes. then u create a string passing this array to constructor, then string will contain the characters for these ascii codes.
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Pleace have mercy! String to valid Date?
Anyone please have mercy and help me with this topic:
How can i check wether a string is of a valid date format:
YYYYMMDD? I have been trying for hours but am unable to find the right function.... :-(
THank thee a thousand times and may the source be with you!sorry will not do it again! i was just so desperate that i posted twice....
and thank thee for thy help! -
This may appear to be a trivial question. I want to ascertain a String is a String. I thought of using a string method on a variable passed into a function and catch the exception. I have done quite a bit of searching around Java Platform SE 7. This seems like a strange thing to do and I have thought of testing everything else and if they fail it has to be a string but I would like something specific.
832844 wrote:
Edit: ok, it looks like you are trying to write some kind of parser here. You might be asking if the String looks like a Java String constant (i.e. it starts and ends with a double-quote). Is that what you want?I am asking if I have been provided a String that looks like a Java String constant. So I test if I have an int through parseInt and catch exception, same with float, and finally check if it is a String. Otherwise, return false.Ok, for the rest of this post I'll assume that variable is of type String and has been verified to not be null.
This means that variable is-a String. There's no way around it. The question is if the information contained in that String can be interpreted as a floating-point literal, an integer literal or a String literal.
Since floating-point literals and integer literals are everyday things, Java provides simple method to interpret them and you already use them.
String literals as such are not so common in everyday use. Therefore Java doesn't provide a simple method to interpret them.
You could use a regex to check if a given String "looks like" a String literal:
String variable = nextToken();
boolean isStringLiteral=variable.matches("\\\"([^\"]|\\\\\\\")*\\\");
{code}
If I didn't mis-type this, it should check if your variable starts and ends with a double-quote and contains no un-quoted double-quotes (there are ugly details that this regex doesn't handle, such as an un-quoted double-quote after a quoted backslash character, but it's just a demonstration). -
Is the email asking to verify your apple id valid
i Have received an email asking to verify my apple/icloud id, is this valid as it is asking me to re enter credit card details
Hi: I have also been receiving emails lately to verify my ID (got the below today). I have done nothing and assume they are Phishing and should do nothing. Agreed?
iCloud
ID
This is the final email to notify you as of 28 - April - 2015 that you have not
yet reviewed your iCloud ID details. Under "Know your Customer (KYC)"
legislation Apple is required by law to carry out a one time a verification of
your information, failure to do so will result in deletion of your Apple ID
within the next day.
Please proceed to »
Verify you Apple ID
To cancel the deletion of your iCloud ID proceed to your Apple ID information
before the deadline.
Resolution Validation Request: #D18YMN5147413179
With Regards,
iSupport Cloud Team
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