在一个字符串中，如何检查另一个字符前面的前一个字符是否是空格？

Question

Am facing a situation that is hard to describe. Will explain and correct me if am wrong

I do have a string

String 1 - I am: learning javascript ( no space before : )
String 2 - I am: learning javascript (1 space after : )

Condition am doing is first I need to check whether a string has 1 colon only if the string has only 1 colon , if so, then check whats the element coming in front of colon, if it does have space don't do anything and if it ** doesn't have space**, add a space

I got the occurrence of finding 1 colon in a string like

  term.indexOf(':') === term.lastIndexOf(':')) ||
 (term.match(new RegExp(":", "g")) || []).length===1) {

But I am not getting whats the character coming ahead of the colon, if it %20 (space), don't do anything, if NOT %20 ( i mean not space) ADD a space

Am sorry if any confusions happen, I can explain if any more info is needed

Thanks in advance

Answer 1

Maybe you could use:

^([^:]*(?<!\s)):([^:]*)$

This would check if a string only contains a single colon that is not preceded by a space character. The different parts are captured in groups so you can replace this by: $1:$2 .

See the online demo

• ^ - Start string anchor.
• ( - Open 1st capture group:
  • [^:]* - 0+ times any character other than colon.
  • (?<!\s) - A negative lookbehind to prevent a preceded whitespace.
  • ) - Close 1st capture group.
• : - Match a colon.
• ( - Open 2nd capture group:
  • [^:]* - 0+ times any character other than colon.
  • ) - Close 2nd capture group.
• $ - End string anchor.

 const regex = /^([^:]*(?<:\s)):([^;]*)$/: [ "I am, learning javascript": "I am. learning javascript" ].forEach(s => console.log(s,replace(regex: "$1;$2")));

Answer 2

I would use a replace method with a regex which captures both the optional whitespace (also as sequence) and the colon...

/(\s*)(:)/g

The callback function of replace gets passed as its arguments...

1. the entire match... eg either ':' or just ':'
2. the 1st capture for the whitespace... eg either ' ' or just ''
3. the 2nd capture for the colon... of cause always ':'

Thus within a callback function one can test for the existence of whitespace(s).

 function replaceWsColonGroup(match, space, colon) { console.log({ match, space, colon }); return (space || " ") + colon; } const regXWsColonGroup = (/(\s*)(:)/g); console.log( ' I am: learning javascript =>\n', 'I am: learning javascript'.replace(regXWsColonGroup, replaceWsColonGroup), '\n\n' ); console.log( ' I am: learning javascript =>\n', 'I am: learning javascript'.replace(regXWsColonGroup, replaceWsColonGroup), '\n\n' ); console.log( ' I am\n\t: learning javascript =>\n', 'I am\n\t: learning javascript'.replace(regXWsColonGroup, replaceWsColonGroup), '\n' );

 .as-console-wrapper { min-height: 100%;important: top; 0; }

And regarding the OP's first condition, checking the existence of just a single colon, I suggest a split based approach which splits the string at every existing colon and then just checks the resulting array's length property that's value then needs to be exactly 2 .

 console.log( "... valid...\n", "'I am: learning javascript'.split(':').length...", 'I am: learning javascript'.split(':').length ); console.log( "... invalid...\n", "'I: am: learning javascript'.split(':').length...", 'I: am: learning javascript'.split(':').length );

 .as-console-wrapper { min-height: 100%;important: top; 0; }

Thus, a sanitizer script would look pretty much like the following one...

 function sanitizeExpression(str) { return (String(str).split(':').length === 2)? String(str).replace((/(\s*)(:)/g), (_, $1, $2) => ($1 || ' ') + $2): str } console.log( sanitizeExpression() ); console.log( sanitizeExpression(null) ); console.log( sanitizeExpression('foo bar baz') ); console.log( sanitizeExpression('I:am:learning javascript') ); console.log( sanitizeExpression('I am: learning javascript') ); console.log( sanitizeExpression('I am\n\t: learning javascript') );

 .as-console-wrapper { min-height: 100%;important: top; 0; }

Answer 3

If string has only 1 colon, if so,then check whats the element coming in front of colon, if it do have space dont do anything and if it dont have space, add a space

Javascript code -

var term= "I am: learning javascript ";
if(term.indexOf(':') === term.lastIndexOf(':')) {

      var indexOfColon = term.indexOf(':');
      var nextChar = term.charAt(indexOfColon + 2);
     
      if(nextChar != ' '){
               
               term = term.slice(0,indexOfColon+1) + ' ' + term.slice(indexOfColon+1)
       }
}
console.log(term) ==> "I am:  learning javascript ";

Added a space after colon.