Find and replace in Atom - atom-editor

I'm attempting to find and replace \$GET['[^']+'] with ($0) but it literally replaces with ($0) rather than ($GET[''])
Regex mode is active.
How can I search and replace using the results of the search in the replace?

The issue was that \$GET['[^']+'] doesn't have a capture group, and the first returned string is $1 not $0.
Find (\$GET['[^']+'])
Replace ($1)
Works as expected

Related

GA Search & Replace Filter

I wonder whether someone can help me please.
I have the following URI in GA: /invite/accept-invitation/accepted/B
Which I'd like to change to: /invite/accept-invitation/accepted
I've tried a 'Search and Replace filter as follows:
Search String - /invite/accept-invitation/accepted/*
Replace String - /invite/accept-invitation/accepted
But the result I get is:
/inviteaccept-invitation/accepted/B
Could someone tell me where I've gone wrong with this please?
Many thanks and kind regards
Chris
Google Analytics "Search and replace" filter uses regular expressions. More precisely:
Replace string is either a regular string or it can refer to group
patterns in the search expression using backslash-escaped single
digits like (\0 to \9).
More details are available on the filter settings UI, which also refers to this link.
So in your case, the search string would be something like this.
\/invite\/accept-invitation\/accepted\/\w+
In this expression \ is escaped. Your last string part is captured with \w+, which
matches any word character (equal to [a-zA-Z0-9_]), between one and unlimited times, as many times as possible.
The Replace string doesn't have to be a regular expression. So in your case, your original version could be used:
/invite/accept-invitation/accepted/
Putting this together would result something like this, which gives the desired output in my test view:

How to remove up to a certain punctuation when there are same punctuations [duplicate]

I'm new at regular expressions and wonder how to phrase one that collects everything after the last /.
I'm extracting an ID used by Google's GData.
my example string is
http://spreadsheets.google.com/feeds/spreadsheets/p1f3JYcCu_cb0i0JYuCu123
Where the ID is: p1f3JYcCu_cb0i0JYuCu123
Oh and I'm using PHP.
This matches at least one of (anything not a slash) followed by end of the string:
[^/]+$
Notes:
No parens because it doesn't need any groups - result goes into group 0 (the match itself).
Uses + (instead of *) so that if the last character is a slash it fails to match (rather than matching empty string).
But, most likely a faster and simpler solution is to use your language's built-in string list processing functionality - i.e. ListLast( Text , '/' ) or equivalent function.
For PHP, the closest function is strrchr which works like this:
strrchr( Text , '/' )
This includes the slash in the results - as per Teddy's comment below, you can remove the slash with substr:
substr( strrchr( Text, '/' ), 1 );
Generally:
/([^/]*)$
The data you want would then be the match of the first group.
Edit   Since you’re using PHP, you could also use strrchr that’s returning everything from the last occurence of a character in a string up to the end. Or you could use a combination of strrpos and substr, first find the position of the last occurence and then get the substring from that position up to the end. Or explode and array_pop, split the string at the / and get just the last part.
You can also get the "filename", or the last part, with the basename function.
<?php
$url = 'http://spreadsheets.google.com/feeds/spreadsheets/p1f3JYcCu_cb0i0JYuCu123';
echo basename($url); // "p1f3JYcCu_cb0i0JYuCu123"
On my box I could just pass the full URL. It's possible you might need to strip off http:/ from the front.
Basename and dirname are great for moving through anything that looks like a unix filepath.
/^.*\/(.*)$/
^ = start of the row
.*\/ = greedy match to last occurance to / from start of the row
(.*) = group of everything that comes after the last occurance of /
you can also normal string split
$str = "http://spreadsheets.google.com/feeds/spreadsheets/p1f3JYcCu_cb0i0JYuCu123";
$s = explode("/",$str);
print end($s);
This pattern will not capture the last slash in $0, and it won't match anything if there's no characters after the last slash.
/(?<=\/)([^\/]+)$/
Edit: but it requires lookbehind, not supported by ECMAScript (Javascript, Actionscript), Ruby or a few other flavors. If you are using one of those flavors, you can use:
/\/([^\/]+)$/
But it will capture the last slash in $0.
Not a PHP programmer, but strrpos seems a more promising place to start. Find the rightmost '/', and everything past that is what you are looking for. No regex used.
Find position of last occurrence of a char in a string
based on #Mark Rushakoff's answer the best solution for different cases:
<?php
$path = "http://spreadsheets.google.com/feeds/spreadsheets/p1f3JYcCu_cb0i0JYuCu123?var1&var2#hash";
$vars =strrchr($path, "?"); // ?asd=qwe&stuff#hash
var_dump(preg_replace('/'. preg_quote($vars, '/') . '$/', '', basename($path))); // test.png
?>
Regular Expression to collect everything after the last /
How to get file name from full path with PHP?

TextPad Find Replace Commands Wild Cards

I am trying to figure out how I can put together a find and replace command with wildcards or figure out a way to find and replace the following example:
I would like to find terms that contain double quotes in front of them with a single quote at the end:
Example:
find "joe' and replace with 'joe'
Basically, I'm trying to find all terms with terms having "in front and at the end.'
Check the [x] Regular expression checkbox in textpad's replace dialog and enter the following values:
Find what:
"([^'"]*)'
Replace with:
'\1'
Explanation:
In a regular expression, square brackets are used to indicate character classes. A character class beginning with a caret will match anything not in the class.
Thus [^'"] will match any character except ' and ". The following * indicates that any number of these characters can follow. The ( and ) mark a group. And the group we're looking for starts with " and ends with '. Finally in the replace string we can refer to any group via \n where n is the nth group. In our case it is the first and only group and that is why we used \1.

Creating RegEx That Reads Entire String

My current regex is only picking up part of my string. It creates a match as soon as one if found, even though I need the longer version of that match to hit. For example, I am creating matches for both:
SSS111
and
SSS111-L
The first SSS111 matches fine with my current regex, but the SSS111-L is only getting matched to the SSS111, leaving the -L out.
How can I create a greedy regex to read the whole line before matching? I am currently using
[-A-Z0-9]{3,12}
to capture the numbers and letters, but have not had any luck outside of this.
Regex are allways greedy. This ist mostly the Problem.
Here i think you have only to escape the '-'
#"[-A-Z]{3-12}"

TextPad: Find all the lines not starting with a pattern and replace with a back space

I want to introduce a backspace character at the beginning of the line where a particular pattern is not found. Please advise.
Thanks,
Sagar
If you mean that you want to "remove the first character" then you can do this:
1) Write your regex pattern of what you want to find. For example, if you want to match Remove me at the start of the line, use:
^R\(emove me\)
Here we use ^ to assert the position to the start of the string. We also capture everything apart from the string we wish to keep in a backreference so it can be used later.
2) Replace the matches we find with whatever we grabbed in our backreference, in this case emove me, in effect backspacing the first character.
3) Make sure regular expression is checked and the cursor is at the start of the file, and hit Replace All.
Before
After:

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