I need to replace hundreds of if (ereg("avion","$keyw")) by if (preg_match("#avion#","$keyw")) {
I tried this:
1st : ereg\("(.*)"
2nd : preg_match\("#$1#"
But it replaces the first group by '$1'... Any idea please?
This is working on the UltraEdit version I have:
I'm just not sure why you have # in your replace.
You can pick "Current file" for all instead of the "Selected Text" I used for the comparison.
Another thing to note, it might be safer to use:
ereg\("(.*?)",
Replace:
preg_match("#$1#",
Using sed you can try this way:
sed -i".bak" 's/(ereg("\([a-z]*\)","\$keyw"))/\(preg_match\("#\1#","$keyw"\)\)/' filename
EDIT: Updated to match any word, not only 'avion'
Related
I am struggling with a generic regex to switch values around.
I want a quick way to rearrange
{"AAA","AA1"},
into
{"AA1","AAA"},
ideally using notepad++
Cheers.
Open Replace with Ctrl+H & tick the Regular Expression box
Find: \{(.*),(.*)\}
Replace With: {\2,\1}
(Assumes no } in the quoted values)
in notepad++ :
find :
\{\s*\"(.*?)\"s*\,\s*\"(.*?)\"\s*}
replace by :
{"\2","\1"}
Try this one:
Search: (\{\s*")([^"}]*)(",\s*")([^"}]*)("\})
Replace: \1\4\3\2\5
Pattern: \{("[^"]*")\s*,\s*("[^"]*")\}
Replacement: {$2,$1}
Explanation:
group the "..." parts
use the inverse character class [^"]* to match everything until the first "
reverse their orders with the backreferences: {$2,$1}
%s#{fileID: \(213[0-9]*\)#\='{fileID: '.(submatch(1)-1900)#
I am using this regex search and replace command in vim to subtract a constant from each matching id.
I can do the regex find in VSCode but how can I reference the submatch for maths & replace? submatch(1) does not work in VSCode?
Thanks.
Given a regular expression of (foobar) you can reference the first group using $1 and so on if you have more groups in the replace input field.
To augment Benjamin's answer with an example:
Find Carrots(With)Dip(Are)Yummy
Replace Bananas$1Mustard$2Gross
Result BananasWithMustardAreGross
Anything in the parentheses can be a regular expression.
Just to add another example:
I was replacing src attr in img html tags, but i needed to replace only the src and keep any text between the img declaration and src attribute.
I used the find+replace tool (ctrl+h) as in the image:
For beginners, the accepted answer is correct, but a little terse if you're not that familiar with either VSC or Regex.
So, in case this is your first contact with either:
To find and modify text,
In the "Find" step, you can use regex with "capturing groups," e.g. I want to find (group1) and (group2), using parentheses. This would find the same text as I want to find group1 and group2, but with the difference that you can then reference group1 and group2 in the next step:
In the "Replace" step, you can refer to the capturing groups via $1, $2 etc, so you could change the sentence to I found $1 and $2 having a picnic, which would output I found group1 and group2 having a picnic.
Notes:
Instead of just a string, anything inside or outside the () can be a regular expression.
$0 refers to the whole match
In my case $1 was not working, but $0 works fine for my purpose.
In this case I was trying to replace strings with the correct format to translate them in Laravel, I hope this could be useful to someone else because it took me a while to sort it out!
Search: (?<=<div>).*?(?=</div>)
Replace: {{ __('$0') }}
Regex Replace String for Laravel Translation
Another simple example:
Search: style="(.+?)"
Replace: css={css`$1`}
Useful for converting HTML to JSX with emotion/css!
Just another example for someone figuring out.
In this example, I've added #### to the start of the string and placed the first group $1 after that. Everything outside group (.*) is going to be deleted.
<h4.*">(.*)</h4>
#### $1
# before:
<h4 id="extract-inline-json-with-regex">Extract inline JSON data with Regex</h4>
# after:
#### Extract inline JSON data with Regex
I'm trying to select text between ></ . Example below I want "text"
>text</
but I'm unable to do so.
tried the following but it doesn't like the slash at the end of the regex
\>(.*?)\<\
I'm trying to do this in TextPad. How is this supposed to be done?
I'm ultimately wanting to delete all text between these two characters so all I'm left with is something like: <element></element>
RegEx wise, you can use 3 groupings and for the replace only use the first and 3rd group: \1\3.
Find: (>)(.*)(</)
Replace: \1\3
Try doing:
\>(.*?)\<\/
The regex that you were trying would actually have given error because you had a \ and nothing after that.
You are close.. use the following:
(>).*?(<\/)
And replace with \1\2
See DEMO
OR
You can use lookbehind and lookaheads:
(?<=>)(.*?)(?=<\/)
And replace with '' (empty string)
See DEMO
I want to use Notepadd++ replace option with regular expression to accomplish this:
From: IntegrationName
To: Integration_Name
How can do this ?
My RegEx to search is: .[A-Z]
this finds: "oN"
But I don't know what to put in the replace box so it will only add "_" between the "o" and the "N"...
Another solution using lookaround assertions would be:
(?<=[a-z])(?=[A-Z])
and replace with
_
Note: The "Match case" option needs to be active, otherwise Notepad++ will find a match between every two letters.
This regex will find every position where a lowercase is on the left and an uppercase is on the right.
You can make use of capture groups. If I have to take your current attempt and edit it as little as possible, you would get:
(.)([A-Z])
This will store the match of . into $1 and the uppercase letter in $2, thus, you can use the following in the replace entry box:
$1_$2
I know you've accepted an answer, but when I ran it, I got From: _Integration_Name
Here's my idea;
(:\s)(.{1})([a-z]*)([A-Z]{1})
And use the following replace
$1$2$3_$4
I finaly did it like this:
Find: ([a-z])([A-Z])
Replace with: $1_$2
I have a set of words like this
"text1
"text2
"text3
which i need to convert to a quoted text which I was able to do with stackoverflow help like this
Find: ^\".*$
replace: $0"
There are another set of words like this
text1
text2
text3
which i need to convert to 1 quoted text which i was able to do
Find: ^(.+)$
Replace: "$0"
But in this case i was able to do it only by clicking replace all and not replace.
Can anyone tell why it's happening like this and How can I achieve this without using replace all and by using only replace?
Take care of the second replacement, it will replace "text1" by ""text1"".
Use this instead:
find what: ^[^"\n]+$
replace: "$0"
It should be working with Replace All and not Replace in any case. The function can also be affected by the position of the cursor and the direction to find/replace.
To be certain, you can place the cursor at the beginning of the file, ensure that the find direction is 'Down' and use Replace All.
Now to the regex, yes, you can do it in a single replace. Use this find:
^(?:")?(.*)$
And this replace:
"$1"
regex101 demo.
The (?:")? will consume the first double quote if present so that it isn't placed in the replace later on.
There's no way you used "Replace" only once in the first case! You must have pressed it 3 times (as many as your matches).
In both situations you need "Replace all", since you want multiple global matches. When pressing the "Replace all" button it's like using Perl's /g & /m modifiers in your substitution, you can experiment here to see what I mean.
BTW the use of a capture group (the parenthesis) is redundant in your 2nd example, use find: ^.+$ and replace: "$0".