javascript if statement if url contains substring

 

This question already has an answer here:

  • How to check whether a string contains a substring in JavaScript? 49 answers

    • You can use the indexOf method 

    // Function is used to determine whether a string contains another string
function contains(search, find) {
    ///<summary>Sees if a string contains another string</summary>
    ///<param type="string" name="search">The string to search in</param>
    ///<param type="string" name="find">The string to find</param>
    ///<returns type="bool">A boolean value indicating whether the search string contained the find string</returns>
    return search.indexOf(find) !== -1;
}

    Here's some sample usage: 

    var url = document.URL;
var substring = 'foo';
if (contains(url.toLowerCase(), substring.toLowerCase()) { 
// Contains string
}

    The contains function is case sentitive , however; as demonstrated in my example, you can make it incasesensitive by calling the StringPrototype.toLowerCase Method

    例如，您可以使用indexOf： 

    if (url.indexOf(substring)>=0) {

    This is a forward-looking answer, and won't work in current implementations.

    ECMAScript 6 is currently defining a String.prototype.contains method. This will allow you to do: 

    if (url.contains(substring)) {

    Again, this is a future addition. Currently ECMAScript 6 (Harmony) is being drafted, and this could technically be removed, though it doesn't seem likely.

    Current draft:

    15.5.4.24 String.prototype.contains (searchString, position = 0 )

    The contains method takes two arguments, searchString and position, and performs the following steps:

  • Let O be CheckObjectCoercible(this value) .
  • Let S be ToString(O) .
  • ReturnIfAbrupt(S) .
  • Let searchStr be ToString(searchString) .
  • ReturnIfAbrupt(searchStr) .
  • Let pos be ToInteger(position) . (If position is undefined , this step produces the value 0 ).
  • ReturnIfAbrupt(pos) .
  • Let len be the number of elements in S .
  • Let start be min(max(pos, 0), len) .
  • Let searchLen be the number of characters in searchStr .
  • If there exists any integer k not smaller than start such that k + searchLen is not greater than len , and for all nonnegative integers j less than searchLen , the character at position k+j of S is the same as the character at position j of searchStr , return true ; but if there is no such integer k , return false .
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