Hey I have a string that is kind of like this:
String:
foo | bar
|foo | bar
|foo bar
|bar foo foo
foo bar
|bar foo
|foo bar
bar
And I need a regular expression that matches everything from the first "|" at a line start to the first line that doesn't start with a "|" (as one match). So the match in the shown example would be like this:
Correct Match:
[
|foo | bar
|foo bar
|bar foo foo
,
|bar foo
|foo bar
]
I came up with this /\|([\s\S]+)(\n\|)/g but that matches the following and that incorrect:
Incorrect Match:
[
| bar
|foo | bar
|foo bar
|bar foo foo
foo bar
|bar foo
|
]
I hope you can understand what I need and thank you for your time!
In accordance with your question, to match everything from the first "|" at a line start to the first line that doesn't start with a "|", you can use this regex:
/^\|([\s\S]*?)(?:(?!^[^\|])[\s\S])*/gm
Its important to note that the m flag (multiline) is required in order to match multiline strings.
Given you example string of
foo | bar
|foo | bar
|foo bar
|bar foo foo
foo bar
|bar foo
|foo bar
bar
this will match
|foo | bar
|foo bar
|bar foo foo
and
|bar foo
|foo bar
The ^\| matches any line that starts with a "|" character. Then, the ([\s\S]*?) matches 0 or more of any character including new lines "Lazily" (meaning only capturing the minimum). Finally the (?:(?!^[^\|])[\s\S])* matches up until a new line (^) that does not begin with a "|" ([^\|]).
Here is a link to Regexr that shows more about how it works and an example of it in action.
const str =
`foo | bar
|foo | bar
|foo bar
|bar foo foo
foo bar
|bar foo
|foo bar
bar`
console.log(str.match(/^\|([\s\S]*?)(?:(?!^[^\|])[\s\S])*/gm))