str.match(/substr/g).length

“体”中的“子”发生率。 例如:

str = "22.5 98 12 22 322"
num = str.match(/22/g).length
alert(num) // 3

If you want to match integers rather than just substrings, make sure the number is preceded and followed by a space:

str = "22.5 98 12 22 322"
num = str.match(/(^|s)22($|s)/g).length
alert(num) // 1

另一种选择,没有定期表述:

str = "22.5 98 12 22 322"
num = str.split(" ").filter(function(item) {
    return Number(item) === 22;
}).length;

尝试使用:

if (numbers.indexOf("22") != -1) {
   //The number 22 is in the String
}

它并非完全是reg,而是为什么在你不需要的时候使用它呢?

如果你只想找到在同一条范围内重复的号码,那么你可以这样做:

var test_str =  22 34 55 45345 34 22 555 ;
var matches  = test_str.match(/([0-9]+)(?=.+1)/g); 
// matches contains ["22", "34"], but not "55"

var str = "22.5 98 12 22 322";
var pattern = /22/g;
if (pattern.test(str)) {
    // it matches "22", do your logic here...
}

http://www.w3schools.com/jsref/jsref_obj_regexp.asp”rel=“nofollow”

Corrolating integer/floating point numbers from text to regex is
very tricky, since regex has no such concept.

With dynamic input, all the work is constructing the regex programatically.
This example doesn t include scientific notation, but it could.

Pseudo Code:

If the input matches this parse regex  
    /^  (?=.*d)     // must be a digit in the input
        ([+-]?)      // capt (1), optional +/-
        [0]*         // suck up all leading 0 s
        (d*)        // capt (2), optional digits, will start with [1-9]
        ([.]?)       // capt (3), optional period
        (d*?)       // capt (4), non-greedy optional digits, will start with [1-9]
        [0]*         // suck up all trailing 0 s
     $/x

    sign =  [+]? ;

    If !length $2 AND !length $4
         sign =  [+-] ;
    Else
        If $1 ==  - 
            sign =  [-] ;
        Endif
    Endif

    whole =  [0]* ;

    If  length $2
         whole =  [0]*  + $2;
    Endif

    decfrac =  [.]? [0]* ;

    If  length $4
         $decfrac =  [.]   + $4 +  [0]* ;
    Endif

    regex =  /(?:^|s)(?=S*d)   +  "$sign $whole $decfrac" +   (?:s|$)/x ;

Else
    //  input is not valid
Endif

This is a Perl test case.
The dynamic regex produced has not been tested,
I asume it works but I could be wrong.

@samps = (
 00000.0 ,
  +.0 ,
  -.0 ,
  0. ,
  -0.00100 ,
  1 ,
  +.1 ,
  +43.0910 ,
  22 ,
  33.e19 ,
);


for (@samps) 
{
   print "
Input:  $_
";

   if( /^ (?=.*d) ([+-]?) [0]*(d*) ([.]?) (d*?)[0]*$/x )   # 1,2,3,4
   {

      my ($sign, $whole, $decfrac);

      $sign =  [+]? ;

      if ( !length $2  && !length $4) {
           $sign =  [+-] ;
      } elsif ($1 eq  - ) {
           $sign =  [-] ;
      }

      $whole =  [0]* ;

      if ( length $2) {
          $whole =  [0]* .$2;
      }

      $decfrac =  [.]? [0]* ;

      if ( length $4) {
          $decfrac =  [.]  .$4. [0]* ;
      }


      my $regex =  /(?:^|s)(?=S*d)  . "$sign $whole $decfrac" .   (?:s|$)/x ;

      print "Regex:  $regex
";
   }
   else {
      print "**input   $_   is not valid
";
      next;
   }
}

产出

Input:  00000.0
Regex:  /(?:^|s)(?=S*d) [+-] [0]* [.]? [0]* (?:s|$)/x

Input:  +.0
Regex:  /(?:^|s)(?=S*d) [+-] [0]* [.]? [0]* (?:s|$)/x

Input:  -.0
Regex:  /(?:^|s)(?=S*d) [+-] [0]* [.]? [0]* (?:s|$)/x

Input:  0.
Regex:  /(?:^|s)(?=S*d) [+-] [0]* [.]? [0]* (?:s|$)/x

Input:  -0.00100
Regex:  /(?:^|s)(?=S*d) [-] [0]* [.] 001[0]* (?:s|$)/x

Input:  1
Regex:  /(?:^|s)(?=S*d) [+]? [0]*1 [.]? [0]* (?:s|$)/x

Input:  +.1
Regex:  /(?:^|s)(?=S*d) [+]? [0]* [.] 1[0]* (?:s|$)/x

Input:  +43.0910
Regex:  /(?:^|s)(?=S*d) [+]? [0]*43 [.] 091[0]* (?:s|$)/x

Input:  22
Regex:  /(?:^|s)(?=S*d) [+]? [0]*22 [.]? [0]* (?:s|$)/x

Input:  33.e19
**input   33.e19   is not valid

Here s a good tool for you to explore regular expressions :) http://www.regular-expressions.info/javascriptexample.html

从上述网站通过:

function demoShowMatchClick() {
  var re = new RegExp("(?:^| )+(22(?:$| ))+");
  var m = re.exec("22.5 98 12 22 322");
  if (m == null) {
    alert("No match");
  } else {
    var s = "Match at position " + m.index + ":
";
   for (i = 0; i < m.length; i++) {
     s = s + m[i] + "
";
   }
   alert(s);
 }
}

这将使你们都能够相匹配,并且要求通过铺.或白色空间的开端或尾端打消数量。