Match Certain word in a String

Hi all,
I'm actually doing a project regrading e-learning. And i'm suppose to create a set of questions for each topics using labview. So Is there any way that i can match certain word in the string to make sure that answer is correct? Cause i'm sure that every user that input answer will be different. Thus, I want to pick out main point as an answer. Is there anyway i can do it??
Really appreciate your help!!!
Thank you!!
Solved!
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Match Strings.vi ‏8 KB

Here's another option (building on Jeff's code). Turn on the Conditional Terminal on the FOR loop and change it to "Continue if TRUE". This way, the loop will exit as soon as a failure is found. Just pass the result straight out of the loop. If none fail, then the FOR loop will exit on its own (from the auto-indexing) and a pass is passed out.
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Match first word in a String

I'm writing a program to reorganize some music. And I have a special case where bands that begin with 'The' need to be organized differently. So I'm writing a method to rename the music directories, but I am having trouble figuring out how to determine if the artist name begins with 'The'. Can someone point me in the right direction as to which java tool would help me in determining this case. The artist names are being passed in as Strings. TIA.
Edited by: prem1ers on Dec 21, 2009 7:27 PM

prem1ers wrote:
Ok, thank you. I am not so comfortable with regex, so I will look for the easier way. Thanks again.Hint: if you are looking for "begins with", you're on the right track but not using quite the correct words to find the most suitable String method.

Match ANY word

Is there a way to match ANY word in a string using CONTAINS?
- e.g. Given a set of columns holding variations of text borrowed from above
'you can optionally edit your post'
'you can optionally preview your post'
'you can optionally delete your post'
and then match all 3 with something syntactically like
and CONTAINS(<column>, 'optionally % your post') >0
I predict using % is a bad idea as expansion will kill it as the index grows, but is there an alternative that will match a single word gap (so not using NEAR)?

Although not intended for the purpose, using a stopword has the effect of requiring that it match any word. You could use an existing stopword like "the" or you could create your own stoplist and your own stopword for the purpose like "anyword", as demonstrated below. The stopword is not indexed, but its location is recorded, so a word must be present in that spot. I suspect it might be faster than % or NEAR or other methods.
SCOTT@orcl_11gR2> CREATE TABLE test_tab
2    (test_col VARCHAR2(60))
3 /
Table created.
SCOTT@orcl_11gR2> INSERT ALL
2 INTO test_tab VALUES ('you can optionally edit your post')
3 INTO test_tab VALUES ('you can optionally preview your post')
4 INTO test_tab VALUES ('you can optionally delete your post')
5 INTO test_tab VALUES ('some other data')
6 SELECT * FROM DUAL
7 /
4 rows created.
SCOTT@orcl_11gR2> BEGIN
2    CTX_DDL.CREATE_STOPLIST ('test_stoplist', 'BASIC_STOPLIST');
3    CTX_DDL.ADD_STOPWORD ('test_stoplist', 'ANYWORD');
4 END;
5 /
PL/SQL procedure successfully completed.
SCOTT@orcl_11gR2> CREATE INDEX test_idx
2 ON test_tab (test_col)
3 INDEXTYPE IS CTXSYS.CONTEXT
4 PARAMETERS ('STOPLIST test_stoplist')
5 /
Index created.
SCOTT@orcl_11gR2> SELECT token_text FROM dr$test_idx$i
2 /
TOKEN_TEXT
CAN
DATA
DELETE
EDIT
OPTIONALLY
OTHER
POST
PREVIEW
SOME
YOU
YOUR
11 rows selected.
SCOTT@orcl_11gR2> SELECT * FROM test_tab
2 WHERE CONTAINS (test_col, 'optionally ANYWORD your post') > 0
3 /
TEST_COL
you can optionally edit your post
you can optionally preview your post
you can optionally delete your post
3 rows selected.

How to search for words in a string?

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Quickest and easiest way to do it is to search through an
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Delete the first word in a string?

Hi, i have a code in which i can extract the first three words in a string. after it's found the three first words, i would like it to remove the first one, what could i add?
this is what the code looks like:
import java.util.regex.*;
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Take the length of the first word, plus one for the length of the space, that's how many characters to remove. Take the substring of the original string from the end of the removed part to the end of the whole string. Look up the String methods length(), and substring().

Even more about deleting the first word in a string

hi, i have this code that removes the first word in a string, returns the shorter string, removes the first word from the shorter string aso... what i would like it to do is to stop when it hits a non-character, but i can't get it to do that. does anyone know why \b won't work?
import java.io.*;
class Testar {
public static void main(String[] args) {
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} thanx in advance!

while (partDesc.equals("\b") == false) That compares partDesc to a string consisting of one backspace character. I suspect you're trying to use the regex word-boundary anchor, but that's a dead end. If you want to stop beheading the string when the regex stops matching, you can write the code exactly that way: class Testar {
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} If performance is a concern, you can use a pre-compiled Pattern object for greater efficiency. Thanks to Matcher's lookingAt() method, you can use the same regex for the test and the replacement: class Testar {
public static void main(String[] args) {
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      m.reset(partDesc);
} The ^ anchor isn't really necessary in this version, since lookingAt() implicitly anchors the match to the beginning of the string, but you might as well leave it in.

Can I sort a column based upon a "find" word? i.e. I want to group all the rows that contain a certain word. I'm using Numbers '09 on a Macbook Pro. Thanks for your help!

A second question: I am creating a database of products for my point of sale system. I am using a product list sent to me by a vendor. Can I pull out certain words in their description and add them to a new column. i.e. The vendors' description in one cell is "Blueridge Guitar Dreadnought." Can I automatically add the word "Dreadnought" to a new column in the same row?

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Regards,
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Search for a word in a string.

Hello everybody,
Could anybody help me how to search for a word in a string ?
Example: I have a string as below:
String[] weather = {"snow", "rain", "sunny", "temperature", "storm", "freezing"};
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}Here's the output I get:
end function
method test
functioning test
test method
test method end
end    function
ending method
rend   method
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How can I create a text pop up window in my Pages document? I want text to pop up when the reader hovers his/her cursor over a certain word.

How can I create a text pop up window in my Pages document? I want text to pop up when the reader hovers his/her cursor over a certain word in the document. I am teacher. So for example when a student came to word he/she did not know, if he/she hovered the cursor over the word, a defintion or other information would appear. You can do this in Word using bookmarks/hyperlinks but I can't figure this out in Pages. I can link it to another point in my Pages document but I just need the text to pop up - not take the reader to another location. THANK YOU!!!!!!

Have you tried Word for Mac?
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Peter

I just downloaded Pages. When typing on a blank document, I am unable to highlight certain words. My daughter usesbitbfor homework and needs to highlight her spelling words. I've read disucssions and it says gonto the a and enter fill and cg

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At the top of the screen in Page there is an "I" next to the Media Icon. If you tap that "I" icon - you bring up a menu that has Style as one of the options. Tap Style and you can change the color of the text and use all of the preinstalled Styles available in the app.
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Here's how to find the right word in a string

I needed to find the rightmost word in a string. I didn't find a simple formula in these forums, but I now have one, so I wanted to share it. Hope you find it useful.
Assuming that the string is in cell A1, the following will return the rightmost word in the string:
RIGHT(A1,LEN(A1)-FIND("*",SUBSTITUTE(A1," ","*",LEN(A1)-LEN(SUBSTITUTE(A1," ","")))))

I found the problem. Whatever character was being used in the substitution was parsed out by the forum parser. I replaced it with "œ" (option q on my keyboard).
=RIGHT(A1,LEN(A1)-FIND("œ",SUBSTITUTE(A1," ","œ",LEN(A1)-LEN(SUBSTITUTE(A1," ","")))))
Still needs an error check for a single-word "sentence" and to remove the trailing period but it does seem to work. Pretty slick.
Message was edited by: Badunit
I see below that the problem was fixed by the OP.
Message was edited by: Badunit

Getting one character at a certain position from a string array

Hi, i'm new at learning java, but have strong experience at C++ and asm. How would i get one character at a certain positon from a string array.
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ok well that didn't work, but i found getChars function can do what i want. One problem, how do i check the contents of char array defined as char Inchar[];

How to read each and every word from a string.

Hi all,
I have a string which is having many label numbers. if the string is lv_str, its value is like, 11111111111111##22222222222222##3333333333333.
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POINTS PROMISED.
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Hi,
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Use split at '#' so that you will get the individual values.
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Reward iof helpful.

How to search for a particular word in a string?

How to search for a particular word in a string?
thanks for your help....

This works fine.
public class Foo {
    public static void main(String[] args) {
    String s = "Now is the time for all good...";
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    System.out.println(s.contains(s2));
    System.out.println(s.contains("for al"));
    System.out.println(s.contains("not here"));
}output:true
true
falseYou must have something else wrong in your code.
JJ

Match Certain word in a String

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