In the following String:
Events('1234', '123456', '', 'QW233Cdse');
I need to extract "QW233Cdse"
Any suggestion?
When we are working with regular expressions then its very important that we should look for the static text in the test string that can help to create a strong regular expression.
As in your case, "Events()" seems to be a static text containing dynamic value in the round parenthesis so in order to generate the regular expression you need to keep 'Events()' text and add the expression in the round parenthesis as mentioned below:
Test String: Events('1234', '123456', '', 'QW233Cdse');
Regular Expression can be:
Events(.'(.)');
Events(.* '(.+?)');
Note: The backslash before round parentheses would avoid interpreting the round braces as unescaped character. For example, a parenthesis "(" begins the definition of a quantifier, but the leading backslash of parenthesis "(" indicates that the regular expression should match the parenthesis.
Regular expression is most important item to learn when you are working with load testing tools and you can refer to below blog post to get more information on regular expression:
https://www.redline13.com/blog/2016/01/jmeter-extract-and-re-use-as-variable/
Let me know if you have any further question
The relevant regular expression would be something like:
Events\(.* '(.+?)'\);
Demo:
References:
JMeter: Regular Expressions
Using Regular Expressions in JMeter
Perl 5 Regex Cheat sheet
Try using this regex:
\w+(?='\))
Regex would be:
, '([^']+?)');
Configuration would be:
Related
I have a requirement to parse part of url with multiple expressions using regex, expressions are like /abc/def (or) /z/a (or) /v/g. I have a regex that satisfies single expression, but not sure how to do for multiple expression.
(?<volga>.+?\/abc\/def.+?)
volga is a named capturing group.
Above regular expression satisfies anything before /abc and anything after /def, the same way url also has /z/a and /v/g. This is kind of either /abc/def/ or /z/a or /v/g can exist in url. How to write a regular expression that checks for all of above expressions.
Examples of url:
/api/com/abc/def/710660716/847170/
/api/com/z/a/
Maybe this:
(?<volga>.+?\/((abc\/def)|(z\/a)|(v\/g)).*)
Matches any string containing
abc/def
or z/a
or v/g
Regex101
Can you please help with the regular expression for the below problem pattern. I am unable to perform replace operation using regex.
Source string:
/search/zip-12345/district-324
Replacement expression would be /search.html?$1
The following regex /search/((?:"[^"]*"|[^:,])*) provides output as
/search.html?zip-12345/district-324
But the expected output is
/search.html?zip=12345&district=324
Use the following regex pattern:
(\/search)\/([^\/-]+)-([^\/]+)\/([^-]+)-(.*)
With substitution:
$1.html?$2=$3&$4=$5
https://regex101.com/r/m5lkw6/4
I am using a sphinx search module on a site I am developing and there is the option to enter regular expressions to be replaced with specified characters.
The available options are Match Expression,Replace Expression and Replace Char (these are input fields in a CMS admin panel so I'm unsure of the actual code function used behind the scenes unfortunately). My understanding is the search checks for any expressions which match Match Expression and replaces the expressions specified in Replace Expression with those specified in Replace Char. So it's a sort of find and replace on matched terms.
Some examples that work:
Example 1
Match Expression: /[a-zA-Z0-9]*-[a-zA-Z0-9]*/
Replace Expression: /-/
Replace Char: empty
Matched text: SX500-123, GLX-11A, GLZX-VXV, GLZ/123, GLZV 123, CNC-PWR1
Result text: SX500123, GLX11A, GLZXVXV, GLZ/123, GLZV-123-123, CNCPWR1
More examples here: http://mirasvit.com/doc/ssp/2.3.2/ssp/global/long_tail
What I want to do is strip any single or double quotes or apostrophes from a search query.
Example inputs: "examination papers",'examination papers,'examination' "papers",pa"pers,pa'pers
Desired outputs: examination papers,examination papers,papers,papers,papers
I have tried just replacing the - with a " in the examples listed above for now but even this hasn't worked.
Any help would be greatly appreciated! Thank you
You can use these expressions:
Match Expression - /["'][\w\s]+["']|\w+["']\w+/
This will match the following text:
"examination papers",'examination papers','examination' "papers",pa"pers,pa'pers
Then you can use this regex to replace your quotes:
Replace Expression - /["']/
Replace Char - empty
So, your output will be:
examination papers,examination papers,examination papers,papers,papers
As a context for this answer. I understand from the tool you are using that your match expression gathers a resultset where you can apply another regex expression (Replace expression) that will replace the content matched with replace char
the latex sample is as follows:
$F=K$,balalalala,balablal Bi$_x$Sb$_{1-x}$，balabla $abcd$ balabala
What I want to match is inline math expressions like $F=K$, $abcd$, while not those expressions with "_" after "$", like $_x$ and $_{1-x}$
So I write regex expression like this
\$[^_][^\$]+\$(?!_)
I add (?!_) because $Sb$ in the middle of Bi$_x$Sb$_{1-x}$ should not be considered as an math expression.
But the code is not working properly. It returns two expression
$F=K$ and $，balabla $.
What is the right regex expression for this problem?
your desired match needs a Lookbehind regex, somthing like:
\$[^$]+\$(?<!\$_[^$]+)
but we know that regex inside could not use + or * (must be fix-length), so the above regex is invalid.
I suggest to process the text in 2-pass. in first pass remove any $_xxx$ pattern :
perl -ne 's/(\$_[^\$]+\$)//g;print;'
and then match your desired pattern:
grep -oP '\$[^$]+\$'
I'm trying to match cases with regular expression to search and replace some text of given pattern. I can match the pattern, but I'd like to keep some of the literals when replacing.
For example, from the string "abcd123," I'd like to keep abcd but remove 123. I can match the pattern using a simple regular expression like [a-zA-Z0-9]+, but when I want to replace it, I don't know what to use for the replacement. Is this even possible with just regular expressions?
Thanks a lot.
The answer depends on what language/regex engine you are using. You typically use parentheses to save sections matched and either $1, $2, ... or \1, \2, ... in the replacement string to refer to those sections.
For example, from JavaScript:
var x = "Hello World";
x.replace( /([A-Z])\w+/g, '$1xx' );
// "Hxx Wxx"
What language or text editor are you using?