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How to express this pattern with java regex?

I'm working on a java program that tries extracts a string with the format:
|something|=
from different string inputs, where |something| has the following constraints:
|something| can not start with "*"
|something| can not be equal to "name"
|something| can have any length
Can such pattern be represented with a java regular expression? Or should I consider a different approach. Any help would be greatly appreciated!

This is a regular expression you can use:
\|(?!name\|)[^*].*\|=
You can get a nice explanation here.

regular expression for specific URL

How could i use a regular-expression in java to extract URLs in the form
/p/{any set of characters}/bugs/{any set of numbers start from 0 to 999}
From text file. I tried one as the following
final String regex = "(\\/p\\/.*\\/bugs\\/(\\d{0,3}))";
But i didn't work fine for me.

Not sure why you have those backslashes in your regex. Additionally, use a negated character class to match any set of characters in the pattern. The following might work for you:
/p/[^/]*/bugs/[0-9]{1,3}

Try this regex:
final String regex = "/p/.+/bugs/[0-9]{1,3}";

Java regex - replace substrings between delimiters with same substrings without delimiters

I'm trying to figure out how to replace with Java 1.6 in strings like
hello ${world }! ${txt + '_t'}<br/> ${do_not_replace
any substring identified between '${' and '}' with the same substring without these delimiters.
So the output for the string above should be
hello world ! txt + '_t'<br/> ${do_not_replace
I identified a working pattern that allows me to replace the substrings with a fixed string
str.replaceAll('[${](.*?)}', '_')
and i know that i cannot use named groups with this version of Java.
Any suggestion for a simple solution to this problem are highly appreciated! Many thanks

try
s = s.replaceAll("\\$\\{(.+?)}", "$1");

why does this regex not work?

I am trying to match a string with a java regex and I cannot succeed. I'm pretty new to java and with most of my experience being linux based regex, I've had no success. Can someone help me?
Below are the codes that Im using.
The regex is-
//vod//final\_\d{0,99}.\d{0,99}\\-Frag\d{0,99}
The line that I'm trying to match is
/vod/final_1.3Seg1-Frag1
where I want 1.3, 1 and 1 to be wildcarded.
Someone please help me out... :(

You are missing the Seg1 part. Also you are escaping characters that need not to be escaped. Try out this regexp: /vod/final_\\d+\\.\\d+Seg1-Frag\\d+

This should work:
Pattern p = Pattern.compile( "/vod/final_\\d+\\.\\d+Seg\\d+-Frag\\d+" );
Notes: To protect special characters, you can use Pattern.quote()
When running into problems like this, start with a simple text and pattern and build from there. I.e. first try to match /, then /vod/, then /vod/final_1, etc.

You're escaping too much. Don't escape /, _, -.
Something like:
/vod/final_\d{0,99}.\d{0,99}-Frag\d{0,99}

Does this work?
/\/vod\/final\_\d{0,99}.\d{0,99}Seg\d-Frag\d{0,99}
Also, here's what I used to edit the regex you provided above: http://rubular.com/
It says it's for ruby, but it also mentions that it works for java too.

How do I write a regular expression to find the following pattern?

I am trying to write a regular expression to do a find and replace operation. Assume Java regex syntax. Below are examples of what I am trying to find:
12341+1
12241+1R1
100001+1R2
So, I am searching for a string beginning with one or more digits, followed by a "1+1" substring, followed by 0 or more characters. I have the following regex:
^(\d+)(1\\+1).*
This regex will successfully find the examples above, however, my goal is to replace the strings with everything before "1+1". So, 12341+1 would become 1234, and 12241+1R1 would become 1224. If I use the first grouped expression $1 to replace the pattern, I get the wrong result as follows:
12341+1 becomes 12341
12241+1R1 becomes 12241
100001+1R2 becomes 100001
Any ideas?

Your existing regex works fine, just that you are missing a \ before \d
String str = "100001+1R2";
str = str.replaceAll("^(\\d+)(1\\+1).*","$1");
Working link

IMHO, the regex is correct.
Perhaps you wrote it wrong in the code. If you want to code the regex ^(\d+)(1\+1).* in a string, you have to write something like String regex = "^(\\d+)(1\\+1).*".
Your output is the result of ^(\d+)(1+1).* replacement, as you miss some backslash in the string (e.g. "^(\\d+)(1\+1).*").

Your regex looks fine to me - I don't have access to java but in JavaScript the code..
"12341+1".replace(/(\d+)(1\+1)/g, "$1");
Returns 1234 as you'd expect. This works on a string with many 'codes' in too e.g.
"12341+1 54321+1".replace(/(\d+)(1\+1)/g, "$1");
gives 1234 5432.

Personally, I wouldn't use a Regex at all (it'd be like using a hammer on a thumbtack), I'd just create a substring from (Pseudocode)
stringName.substring(0, stringName.indexOf("1+1"))
But it looks like other posters have already mentioned the non-greedy operator.
In most Regex Syntaxes you can add a '?' after a '+' or '*' to indicate that you want it to match as little as possible before moving on in the pattern. (Thus: ^(\d+?)(1+1) matches any number of digits until it finds "1+1" and then, NOT INCLUDING the "1+1" it continues matching, whereas your original would see the 1 and match it as well).