Re: LPEX RegEx Replace

Yea, that's not how it works...

I think all you needed in replace was
SetFail()..

Unless you've got
SetFail(parm1, parm2;
SetFail(parm1;


In that case, you'd need a replacement group
Something like so
find: (SetFail\(.+[^\)])(;)

replace: \1\)\2

The extra parens define a capture group, you'd need two one for the
"SetFail(..." and another for the ending ";"
The \1\2 are where the captured text is used in replacement...

Then it's just a matter of inserting the ")" between them.

I think :) Regex is not my strong suit...

Personally...I'd have selected SetFail and filter or find all.

Then just manually added the closing ) to the lines that needed it.

Charles




On Fri, Jun 28, 2019 at 8:39 AM Justin Taylor <JUSTIN@xxxxxxxxxxxxx> wrote:

More info:

I have a SetFail() method that is missing the closing parenthesis. This
finds it:
SetFail\(.+[^\)];

I put this in the replace:
SetFail\(.+[^\)]\);

Rather than actually evaluate the regular expression in the replace, it
literally overwrote every SetFail() method with:
SetFail\(.+[^\)]\);

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