Javascript Regular Expressions Use variables

Example 1,replace with variable

The Replace function can use regular expressions to match string implementation substitutions.

The problem encountered today is that the regular expression has a variable, taking today's actual situation as an example:

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/(^ ' +arr2[i]+ ' |) | (|' +arr2[i]+ ' $)/ Arr2[i] is an array element and is a variable Str=str.replace (/^ ' +arr2[i]+ ' |) | (|' +arr2[i]+ ' $)/, ');

You cannot replace success with this approach:

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var reg=eval_r ('/^ ' +arr2[i]+ ' ||) | (\|' +arr2[i]+ ' $)/'); Str=str.replace (Reg, ');

If it is a direct variable, we can replace it with the following actions

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var match_str = ' bitch '; var regex = "/," + Match_str + ",/ig"; var target_str = ", Abcde,bitch,fghij,"; var val = target_str.replace (eval (regex), ","); Alert (val);

Val should be:, Abcde,fghij,

Example 2,regexp mode with variable

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function Checkbeginwithchar (str,minlen,maxlen,msg) { Minlen = minlen-1; MaxLen = maxlen-1; var re=new RegExp ("^[a-za-z]{1}" ([a-za-z0-9]) {"+minlen+", "+maxlen+"); if (!re.exec (str)) { Alert (msg); return false; } return true; }

The above is a JavaScript function that can be invoked like this:

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If (!checkbeginwithchar ("A123DF", 6, 12, "The pin you enter must begin with a letter and the number of characters is between 6-12!") ”)) return false;

This is a case of using a variable, and the following function is not using a variable.

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function Checkbeginwithchar (str,msg) { var patrn=/^[a-za-z]{1} ([a-za-z0-9]) {5,50}$/; if (!patrn.exec (str)) { Alert (msg); return false; } return true; }

Note: When there is no variable, the regular expression can have a backslash on both sides, not as a string, if you use a variable with the new RegExp () and the inside string has no two backslashes.