正则表达式匹配特定字符前后的字符串

Question

I would like to make a grouping to select the strings before AND after a specific character, in this case it's the Colon.

Example:

First: foo Last: bar

I would like to Match First and foo in group 1 and Last bar in group 2

Group1: First foo

Group2: Last bar

---

I currently have

([^:]*)+([^:*])

Which only matches everything that's not a colon, which isn't exactly what i'm looking for. What are some ways or patterns with regex where i can match before and after a certain character?

Answer 1

As you want the colon removed, but still have the surrounding text in one group, you'll need to use some string manipulation after executing the regular expression:

 var s = "First: foo Last: bar", re = /\\s*([^:]*?)\\s*:\\s*([^:\\s]*)/g, result = []; while (match = re.exec(s)) { result.push(match[1] + ' ' + match[2]); } console.log(result); 

Note that it can be ambiguous which word belongs where, when there are more spaces, for example in First: foo hello there: bar .

Answer 2

The texts you want to get into single group is not a streak of continuous chars, thus, it is impossible to achieve. You need to grab the two parts, key and value separately, and then join them:

 var rx = /([^:]+):\\s*(.*?)(?=\\w+:|$)/g; var s = "First: foo Last: bar"; var m, res=[]; while (m=rx.exec(s)) { res.push([m[1].trim(),m[2].trim()].join(" ")); } console.log(res); 

See the regex demo at regex101.com . Details:

• ([^:]+) - Group 1: one or more chars other than :
• : - a :
• \\s* - 0+ whitespaces
• (.*?) - Group 2: any 0+ chars other than line break chars, as few as possible
• (?=\\w+:|$) - followed with 1+ word chars and : or end of string.

Answer 3

You can use the following regex:

([a-zA-Z]+:\s*[a-zA-Z]+) 

You can try it out on the link HERE .

'First: foo Last: bar'.match(/([a-zA-Z]+:\s*[a-zA-Z]+)/g) // ["First: foo", "Last: bar"]