I have a regular expression that captures 17 matches. But these matches also include the spaces (leading or trailing).
Tried using [\s*]. But, some spaces still remain.
Screenshots are attached for your reference.
Please help.
Regards,
Ajith
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This link may help: Regex - Trimming whitespace from start and end of line
\S(.*\S)? should trim leading and trailing whitespace
It's hard to come up with a comprehensive answer without seeing your response data (at least partial)
Try the following regular expression:
edison-system?\s*(.+?)?\s*?/
Where:
?\s means optional whitespace
+ - zero or more times
More information:
JMeter Regular Expressions
Using RegEx (Regular Expression Extractor) with JMeter
Java Regular Expressions Quantifiers
Related
I need to capture below values
(1589913628397qzhQpW, 1589913628397qzhQr5, 1589883028581e8w3u6,
1589915918676qzhQkS, 1589915918676qzhQkS)
through regular expression post processor in jmeter. please let me know the regular expression for the same.
<div class="tspacer" onclick="treeClickDocument(
'1589913628397qzhQpW','1589913628397qzhQr5',
'1589883028581e8w3u6','1589915918676qzhQkS',
'1589915918676qzhQkS',false,
'Perf3575','10050')">
If the output looks like exactly as in your question the relevant regular expression would be:
<div class="tspacer" onclick="treeClickDocument\(\n\s+'(\w+)','(\w+)',\n\s+'(\w+)','(\w+)',\n\s+'(\w+)'
Demo:
Explanation of potentially not obvious stuff:
\ - escape symbol for meta-characters
\n - matches new line
\s+ - matches arbitrary amount of white spaces
\w+ - matches a "word" - sequence of alphanumeric characters
More information:
Using RegEx (Regular Expression Extractor) with JMeter
JMeter - Regular Expressions
Perl 5 Regex Cheat sheet
I have a link in my Request -
example.com/people/3176972
and my regular expression extractor is-
Regular expression: example.com/people/(.+?)
Template: $1$
Match no: 1
but it is only extracting only 3. I want to extract 3176972 number.
What am I doing wrong?
If you want to retrieve anything after the last slash, then just remove question mark from your expression:
example.com/people/(.+)
(question mark tells it to be non-greedy, hence it's taking 1 character).
If the last portion is always numeric, use
example.com/people/([0-9]+)
You should try this regular expression:
[0-9]+
Amend your regular expression to look like example.com/people/(.*) or example.com/people/(\d+) as your regex stops after first match.
See Regular Expressions chapter of JMeter's User Manual for more information on JMeter Regular Expressions.
Convenient way of testing regular expressions is using "RegExp Tester" mode of the View Results Tree listener.
Check out How to debug your Apache JMeter script guide for more information on different debugging techniques for JMeter tests.
I want to scrape data from some text and dump it into an array. Consider the following text as example data:
| Example Data
| Title: This is a sample title
| Content: This is sample content
| Date: 12/21/2012
I am currently using the following regex to scrape the data that is specified after the 'colon' character:
/((?=:).+)/
Unfortunately this regex also grabs the colon and the space after the colon. How do I only grab the data?
Also, I'm not sure if I'm doing this right.. but it appears as though the outside parens causes a match to return an array. Is this the function of the parens?
EDIT: I'm using Rubular to test out my regex expressions
You could change it to:
/: (.+)/
and grab the contents of group 1. A lookbehind works too, though, and does just what you're asking:
/(?<=: ).+/
In addition to #minitech's answer, you can also make a 3rd variation:
/(?<=: ?)(.+)/
The difference here being, you create/grab the group using a look-behind.
If you still prefer the look-ahead rather than look-behind concept. . .
/(?=: ?(.+))/
This will place a grouping around your existing regex where it will catch it within a group.
And yes, the outside parenthesis in your code will make a match. Compare that to the latter example I gave where the entire look-ahead is 'grouped' rather than needlessly using a /( ... )/ without the /(?= ... )/, since the first result in most regular expression engines return the entire matched string.
I know you are asking for regex but I just saw the regex solution and found that it is rather hard to read for those unfamiliar with regex.
I'm also using Ruby and I decided to do it with:
line_as_string.split(": ")[-1]
This does what you require and IMHO it's far more readable.
For a very long string it might be inefficient. But not for this purpose.
In Ruby, as in PCRE and Boost, you may make use of the \K match reset operator:
\K keeps the text matched so far out of the overall regex match. h\Kd matches only the second d in adhd.
So, you may use
/:[[:blank:]]*\K.+/ # To only match horizontal whitespaces with `[[:blank:]]`
/:\s*\K.+/ # To match any whitespace with `\s`
Seee the Rubular demo #1 and the Rubular demo #2 and
Details
: - a colon
[[:blank:]]* - 0 or more horizontal whitespace chars
\K - match reset operator discarding the text matched so far from the overall match memory buffer
.+ - matches and consumes any 1 or more chars other than line break chars (use /m modifier to match any chars including line break chars).
I'm opening a file and finding the line I need, but then I have trouble creating a variable from the found string
70c 08:04:04.014 rexx TRACE 2203 8=4.4|9=892|35=J|49=ICE_SM_S|56=SM|34=280|70=0241608914160889|71=0|626=2|793=16|72=|466=1164266784|857=0|73=1|11=|37=1156426784|526=1156426674|38=1|198=1310883PTM|54=1|6=117.2100000000|336=R|625=P|55=B|461=FXXXXX|200=20120901|207=IFEU|53=1|30=ICE|453=2|448=SLM|447=C|452=7|448=FFC|447=C|452=12|75=20120210|60=20120310-09:04:04|77=O|58=CYU795|232=14|233=GL_TRADEJOBOUT|234=N|233=GL_ORDERJOBOUT|234=N|233=GL_TAKEN|234=0|233=GL_TRADETYPE|234=E|
This is the string and I want to assign it to a variable of tag198, so it would be
tag198 = '1310883PTMS'
Anything after | is not needed.
tag198 = line.match(/198=(.*)/)[1]
puts tag198
but that keeps all after 198; I need just the string prior to the |.
Change your regular expression to:
/198=(.+?)\|/
That makes it non-greedy and stop at the vertical bar. You have to escape the vertical bar because it normally would mean "OR" in a regular expression.
Your regular expression's * is greedy, and will consume all characters it can without stopping the rest of the expression from matching. There is nothing in the expression that tells ruby when to stop collecting characters.
Look at regular-expressions.info. A partial fix for your problem would be to put a '|' after your capture:
tag198=line.match(/198=(.*)\|/)[1] puts tag198
The '|' is escaped as it has special meaning in regexes otherwise. This doesn't yet work though, because the * can still consume '|' characters, so long as it leaves one behind to match the '|' in our expression. To fix completely, prevent the * from capturing any pipes:
tag198 = line.match(/198=([^|]*)\|/)[1] puts tag198
See results of this change here.
If it is only letters and numbers you could use
/198=([A-Za-z0-9]*)/
Also, in case you didn't know, you can test regular expressions on rubular.com, it also provides some information about special charters in regular expressions, it is a great site for all your regular expressions needs even if it isn't for ruby.
My question is how to match the first three characters of certain lines within a string using regular expressions the regex i have should work however when i run the program it only matches the first three characters of the first line the string is
.V/RTEE/EW\n.N/ERER/JAN/21
my regex is ^(.[VN]/)* so it needs to match .V/ and .N/ any help I will be very grateful
You need to suppress the special meaning of the . and /
Use \ in-front of them.