Jquery.Inarray(), How to Use It Right

jQuery.inArray(), how to use it right?

inArray returns the index of the element in the array, not a boolean indicating if the item exists in the array. If the element was not found, -1 will be returned.

So, to check if an item is in the array, use:

if(jQuery.inArray("test", myarray) !== -1)

jquery inArray usage

You could check for the keyword with JavaScript indexOf method. This is check for each element of the array:

$("#get").click(function(){
    $(".name").each(function(){
        var fisier = $(this).text();       
        if(fisier.indexOf("migr") !=-1){
           alert('Keyword found');
        }else{
           alert('Keyword not found');
        } 
    });
});

Should I use jQuery.inArray()?

Well internally inArray makes a simple loop, I would recommend you to check if there is a native Array.prototype.indexOf implementation and use it instead of inArray if available:

function findPoint(point, list) {
  var l = list.map(function anonMapToId(p) { return p.id });
  var found = ('indexOf' in Array.prototype) ? l.indexOf(point.id)
                                             : jQuery.inArray(point.id, l);
  return found;
}

The Array.prototype.indexOf method has been introduced in browsers that implement JavaScript 1.6, and it will be part of the ECMAScript 5 standard.

Native implementations are way faster than non native ones.

JS jQuery - check if value is in array

You are comparing a jQuery object (jQuery('input:first')) to strings (the elements of the array).

Change the code in order to compare the input's value (wich is a string) to the array elements:

if (jQuery.inArray(jQuery("input:first").val(), ar) != -1)

The inArray method returns -1 if the element wasn't found in the array, so as your bonus answer to how to determine if an element is not in an array, use this :

if(jQuery.inArray(el,arr) == -1){
    // the element is not in the array
};

jQuery inArray returning false when needle exists in haystack

Your code seems to be working just as expected.

Note that $.inArray() will perform a type comparison; I expect you are passing a string. If you are testing for a key that is a string, your isValidCode function will return false, as your array only contains integer values. You can cast the code to an integer with parseInt to avoid this issue:

function isValidCode(code){
    var arr = [1, 2, 3];

    return ($.inArray( parseInt( code, 10 ), arr ) !== -1);
}

---

Related Topics