javascript regular expression to check for IP addresses

Question

I have several ip addresses like    115 42 150 37 115 42 150 38 115 42 150 50   What type of regular expression should I write if I want to search for the all the 3 ip addresses  Eg  if I do 115 42 150    I will be able to search for all 3 ip addresses   What I can do now is something like    0-9  1-3    0-9  1-3    0-9  1-3    0-9  1-3   but it can t seems to work well    Thanks

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If you want something more readable than regex for ipv4 in modern browsers you can go with  function checkIsIPV4 entry      var blocks   entry split         if blocks length     4        return blocks every function block          return parseInt block 10   gt  0  amp  amp  parseInt block 10   lt   255                return false

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Try this one  it s a shorter version       0            1  d  d 25 0-5  2 0-4  d         4     Explained      start of string      0          Assume IP cannot start with 0                 Make sure string does not end with a dot               1  d  d    A single digit  two digits  or 100-199     25 0-5     The numbers 250-255     2 0-4  d   The numbers 200-249              the number must be followed by either a dot or end-of-string - to match the last number     4          Expect exactly four of these   end of string   Unit test for a browser s console   var rx      0            1  d  d 25 0-5  2 0-4  d         4     var valid   1 2 3 4   11 11 11 11   123 123 123 123   255 250 249 0   1 12 123 255   127 0 0 1   1 0 0 0    var invalid   0 1 1 1   01 1 1 1   012 1 1 1   1 2 3 4    1 2 3 n4   1 2 3 4 n   259 0 0 1   123    1 2 3 4 5    1 2 3 4   1 2 3 4   1 2 333 4   1 299 3 4    valid forEach function s  if   rx test s  console log  bad valid    s      invalid forEach function s  if  rx test s   console log  bad invalid    s

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Simple Method  const invalidIp   ipAddress     split          map ip   gt  Number ip   gt   0  amp  amp  Number ip   lt   255      includes false    if invalidIp       IP address is invalid     throw error here

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Try this one   Source from here     b 25 0-5  2 0-4  0-9   01   0-9  0-9      25 0-5  2 0-4  0-9   01   0-9  0-9      25 0-5  2 0-4  0-9   01   0-9  0-9      25 0-5  2 0-4  0-9   01   0-9  0-9    b

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This is what I did and it s fast and works perfectly   function isIPv4Address inputString        let regex   new RegExp      0-9  1 3     3  0-9  1 3           if regex test inputString            let arInput   inputString split              for let i of arInput               if i length  gt  1  amp  amp  i charAt 0       0                   return false              else                  if parseInt i   lt  0    parseInt i   gt  256                     return false                                    else         return false      return true      Explanation  First  with the regex check that the IP format is correct  Although  the regex won t check any value ranges   I mean  if you can use Javascript to manage regex  why not use it   So  instead of using a crazy regex  use Regex only for checking that the format is fine and then check that each value in the octet is in the correct value range  0 to 255   Hope this helps anybody else  Peace

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it is maybe better   function checkIP ip        var x   ip split       x1  x2  x3  x4       if  x length    4            x1   parseInt x 0   10           x2   parseInt x 1   10           x3   parseInt x 2   10           x4   parseInt x 3   10            if  isNaN x1     isNaN x2     isNaN x3     isNaN x4                 return false                     if   x1  gt   0  amp  amp  x1  lt   255   amp  amp   x2  gt   0  amp  amp  x2  lt   255   amp  amp   x3  gt   0  amp  amp  x3  lt   255   amp  amp   x4  gt   0  amp  amp  x4  lt   255                 return true                      return false

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Don t write your own regex or copy paste  You probably won t cover all edge ceses  IPv6  but also octal IPs  etc   Use the is-ip package from npm    var isIp   require  is-ip     isIp  192 168 0 1     isIp  1 2 3 4 5 6 7 8      Will return a Boolean   Downvoters  care to explain why using an actively maintained library is better than copy pasting from a website

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Below Solution doesn t accept Padding Zeros  Here is the cleanest way to validate an IP Address  Let s break it down   Fact  a valid IP Address is has 4 octets  each octets can be a number between 0 - 255  Breakdown of Regex that matches any value between 0 - 255   25 0-5  matches 250 - 255 2 0-4  0-9  matches 200 - 249 1 0-9  0-9  matches 100 - 199  1-9  0-9   matches 1 - 99 0 matches 0   const octet     25 0-5  2 0-4  0-9  1 0-9  0-9   1-9  0-9   0      Notes  When using new RegExp you should use     instead of    since string will get escaped twice    function isValidIP str      const octet     25 0-5  2 0-4  0-9  1 0-9  0-9   1-9  0-9   0      const regex   new RegExp     octet      octet      octet      octet        return regex test str

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Always looking for variations  seemed to be a repetitive task so how about using forEach    function checkIP ip        assume IP is valid to start  once false is found  always false   var test   true       uses forEach method to test each block of IPv4 address   ip split      forEach validateIP4      if   test       alert  Invalid IP4 format n  ip     else      alert  IP4 format correct n  ip      function validateIP4 num  index  arr          returns NaN if not an Int     item   parseInt num  10         test validates Int  0-255 range and 4 bytes of address         amp  amp  test  at end required because this function called for each block     test    isNaN item   amp  amp   isNaN num   amp  amp  item  gt  0  amp  amp  item  lt  256  amp  amp  arr length  4  amp  amp  test

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May be late but  someone could try   Example of VALID IP address  115 42 150 37 192 168 0 1 110 234 52 124   Example of INVALID IP address  210 110     must have 4 octets 255     must have 4 octets y y y y     only digits are allowed 255 0 0 y     only digits are allowed 666 10 10 20     octet number must be between  0-255  4444 11 11 11     octet number must be between  0-255  33 3333 33 3     octet number must be between  0-255    JavaScript code to validate an IP address  function ValidateIPaddress ipaddress        if     25 0-5  2 0-4  0-9   01   0-9  0-9      25 0-5  2 0-4  0-9   01   0-9  0-9      25 0-5  2 0-4  0-9   01   0-9  0-9      25 0-5  2 0-4  0-9   01   0-9  0-9      test ipaddress           return  true            alert  You have entered an invalid IP address        return  false

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Regular expression for the IP address format       d d    1 d d   0 d d   2 0-4  d   2 0-5      d d    1 d d   0 d d   2 0-4  d   2 0-5      d d    1 d d   0 d d   2 0-4  d   2 0-5

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A less stringent when testing the type not the validity  For example when sorting columns use this check to see which sort to use   export const isIpAddress    ipAddress    gt            d  1 3     3   d  1 3    test ipAddress    When checking for validity use this test  An even more stringent test checking that the IP 8-bit numbers are in the range 0-255   export const isValidIpAddress    ipAddress    gt           25 0-5  2 0-4  0-9   01   0-9  0-9       3  25 0-5  2 0-4  0-9   01   0-9  0-9      test ipAddress

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b    0-9  1 3     3  0-9  1 3  b   matches 0 0 0 0 through 999 999 999 999 use if you know the seachdata does not contain invalid IP addresses   b 25 0-5  2 0-4  0-9   01   0-9  0-9      25 0-5  2 0-4  0-9   01   0-9  0-9      25 0-5  2 0-4  0-9   01   0-9  0-9      25 0-5  2 0-4  0-9   01   0-9  0-9    b   use to match IP numbers with accurracy - each of the 4 numbers is stored into it s own capturing group  so you can access them later

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The answers over allow leading zeros in Ip address  and that it is not correct  For example   quot 123 045 067 089 quot should return false   The correct way to do it like that   function isValidIP ipaddress      if     25 0-5  2 0-4  0-9   1   1-9  1-9      25 0-5  2 0-4  0-9   1   1-9  1-9      25 0-5  2 0-4  0-9   1   1-9  1-9      25 0-5  2 0-4  0-9   1   1-9  1-9      test ipaddress         return  true         return  false      This function will not allow zero to lead IP addresses

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If you are using nodejs try   require  net   isIP  10 0 0 1    doc net isIP

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A short RegEx                 2   5 0-5   0-4  d  1  d  d   4    Example   x000D   x000D  const isValidIp   value   gt                   2   5 0-5   0-4  d  1  d  d   4    test value    true   false   x000D   x000D   x000D     valid x000D  console log  isValidIp  0 0 0 0        isValidIp  0 0 0 0     x000D  console log  isValidIp  115 42 150 37        isValidIp  115 42 150 37     x000D  console log  isValidIp  192 168 0 1        isValidIp  192 168 0 1     x000D  console log  isValidIp  110 234 52 124       isValidIp  110 234 52 124     x000D  console log  isValidIp  115 42 150 37        isValidIp  115 42 150 37     x000D  console log  isValidIp  115 42 150 38        isValidIp  115 42 150 38     x000D  console log  isValidIp  115 42 150 50        isValidIp  115 42 150 50     x000D   x000D     Invalid x000D  console log  isValidIp  210 110        isValidIp  210 110     x000D  console log  isValidIp  255        isValidIp  255     x000D  console log  isValidIp  y y y y       isValidIp  y y y y     x000D  console log   isValidIp  255 0 0 y        isValidIp  255 0 0 y     x000D  console log  isValidIp  666 10 10 20        isValidIp  666 10 10 20     x000D  console log  isValidIp  4444 11 11 11        isValidIp  4444 11 11 11     x000D  console log  isValidIp  33 3333 33 3        isValidIp  33 3333 33 3     x000D   x000D   x000D

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0 d  1  d  d 25 0-5  2 0-4  d         4      Full credit to oriadam  I would have commented below his her answer to suggest the double zero change I made  but I do not have enough reputation here yet     change    -   0     Because IPv4 addresses starting with zeros   0 248 42 223   are valid  but not usable      0 d   Because IPv4 addresses with leading zeros   63 14 209 00  and  011 012 013 014   can sometimes be interpreted as octal

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The regex you ve got already has several problems   Firstly  it contains dots  In regex  a dot means  match any character   where you need to match just an actual dot  For this  you need to escape it  so put a back-slash in front of the dots   Secondly  but you re matching any three digits in each section  This means you ll match any number between 0 and 999  which obviously contains a lot of invalid IP address numbers   This can be solved by making the number matching more complex  there are other answers on this site which explain how to do that  but frankly it s not worth the effort -- in my opinion  you d be much better off splitting the string by the dots  and then just validating the four blocks as numeric integer ranges -- ie   if block  gt   0  amp  amp  block  lt   255           Hope that helps

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Throwing in a late contribution                   1-9   d 1 d d 2 5 0-5   0-4  d    4     Of the answers I checked  they re either longer or incomplete in their verification  Longer  in my experience  means harder to overlook and therefore more prone to be erroneous  And I like to avoid repeating similar patters  for the same reason   The main part is  of course  the test for a number - 0 to 255  but also making sure it doesn t allow initial zeroes  except for when it s a single one     1-9   d 1 d d 2 5 0-5   0-4  d    Three alternations - one for sub 100   1-9   d  one for 100-199  1 d d and finally 200-255  2 5 0-5   0-4  d    This is preceded by a test for start of line or a dot    and this whole expression is tested for 4 times by the appended  4    This complete test for four byte representations is started by testing for start of line followed by a negative look ahead to avoid addresses starting with a             and ended with a test for end of line       See some samples here at regex101

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If you wrtie the proper code you need only this very simple regular expression    d 1 3    function isIP ip        let arrIp   ip split           if  arrIp length     4  return  Invalid IP       let re     d 1 3        for  let oct of arrIp            if  oct match re      null  return  Invalid IP          if  Number oct   lt  0    Number oct   gt  255              return  Invalid IP         return  Valid IP       But actually you get even simpler code by not using any regular expression at all   function isIp ip        var arrIp   ip split           if  arrIp length     4  return  Invalid IP       for  let oct of arrIp            if   isNaN oct     Number oct   lt  0    Number oct   gt  255              return  Invalid IP         return  Valid IP

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And instead of   1-3    you should put   1 3

[javascript] javascript regular expression to check for IP addresses

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