Problems with Regex

[rf222]

I have executed the following regex:

- (PNMIS_)(\S*\S*\S*\S*\S*\S*\S*\S*\S*\S*\S*\S*\S*\S*\S*\S*\S*)\S*_REPORT

Replacement:

- pr1.tss.mis.$2.c


Sample data:

/* ----------------- PNMIS_test6732323323_REPORT ----------------- */
/* ----------------- PNMIS_test23_REPORT ----------------- */

The systems hangs! I need 17 characters on position $2 only. Can not have more.

Please help. Thanks

 

[tsuji]

> I need 17 characters on position $2 only.

Your description is ambiguous and your pattern does not satisfy any of the sample?! Try this?
[tt]
var rx=new RegExp("(PNMIS_)(\\S{17})\\S_REPORT")
var s="/* ----------------- PNMIS_test6732323323ssst_REPORT ----------------- */";
alert(RegExp.$2);
[/tt]

 

[tsuji]

I missed out a line! Here is a re-take.
[tt]
var rx=new RegExp("(PNMIS_)(\\S{17})\\S_REPORT")
var s="/* ----------------- PNMIS_test6732323323ssst_REPORT ----------------- */";
var cm=rx.exec(s);
alert(RegExp.$2);
[/tt]

 

[rf222]

Thanks. I will try it today. Why is there "\\S". I thought "\S" was enough. please explain. Thanks. "\S" means one no space character. I did not know about {n} syntax... Great help

 

[tsuji]

The format used here to establish regexp will not recognize "\" as literal, it will take it as escape character. If you use another format, say,
[tt]/PNMIS_)(\S{17})\S_REPORT/[/tt]
then you don't need to. You didn't reveal what approach you took.

 

[tsuji]

My choice of words may not be perfect, upon re-reading it, you get the meaning...

 

[rf222]

Shoul not this be:
/PNMIS_)(\S{17})\S*_REPORT/

What {17} means? First 17 characters? What if there are only 5 characters after "PNMIS_" ? Example: "PNMIS_test1_REPORT.

By the way, the other regex finished sucessfully, yet it took 10 hours to run!

 

[kaht]

{x} - matches exactly x instances of the preceding character

If that does not work for you, instead you could use this terminology:
{x,y} - matches at least x and at most y instances of the preceding character

So if you wanted to match anywhere between 0 and 17 \S, then you would do it like this:
(take note that there is no space between the 0, the comma, and the 17 - that is very important as it will not work if you have any spaces included)

/PNMIS_)(\S{0,17})\S*_REPORT/

-kaht

[small] <P> <B> <P> <B> <P> <B> <P> <B> <P> <B> <P> <B> <P> <B> <P> <B> <P> <B> <P> <B> <P> <B> <P> <B> <P> <B> <P> <.</B>[/small]
[small](He's back)[/small]

 

[tsuji]

The /.../ expression, I had missed out an open parenthesis during cut and paste operation. I guess it would be obvious. Hence, it should be read /[highlight]([/highlight]PNMIS_).../.

The {17} is what I understand from what you said I need 17 characters on position $2 only, I quoted already.

>it took 10 hours to run
No idea what you have been doing...

 

[rf222]

It works fast now! Great, thank you!

I wonder, maybe you can help me extend this, I have a feeling I need to do lookbehind, but not sure how.

Sample Data:

----------START OF DATA ---------------

/* ----------------- PNMIS_MSTANDARD_REPORT ----------------- */

insert_job: PNMIS_MSTANDARD_REPORT job_type: c
condition: success(PNMIS_NYMONTHLY2_REPORT)

/* ----------------- PNMIS_NYMONTHLY_REPORT ----------------- */

insert_job: PNMIS_NYMONTHLY_REPORT job_type: c
condition: success(PNMIS_MSTANDARD_REPORT)

/* ----------------- PNMIS_NYMONTHLY2_REPORT ----------------- */

insert_job: PNMIS_NYMONTHLY_REPORT job_type: c
condition: success(PNMIS_MSTANDARD55_REPORT) and success(PNMIS_THIS_ONE_IS_NOT_IN_THIS_FILE_REPORT)

blah, blah, blah...
----------END OF DATA ---------------


1) So this regex will replace the header:
- PNMIS_(\S{0,17})\S*_REPORT

However I would like to also:

2) Replace also what is after "insert_job", I guess I need to add some or statement at the begining...

3) Replace what is after "success(", however only for entries, which are in the file (header).

So step 3 would be more tricky, I am thinking that I may need to do this in 2 itteriations, by executing step 3 first with look behind feature, than step 1 and 2. I am not sure what the syntax would be... I also wonder if it is possible to do this all in one itteriation only.

Thanks

 

[tsuji]

Lookbehind is not yet supported in regexp engine of javascript. You have to use submatch, just like your using .$1 to do that.