Need help regarding jQuery $.inArray()

问题

Could anybody tell me what is wrong with this code ??

jQuery(document).ready(function(){

    var val = ["Hourly", "Daily", "Weekly", "Monthly", "Yearly"];
    var myArr = ["Weekly", "something"];

    $( myArr ).each(function( j ){
        if ( $.inArray( myArr[j] == val ) ) {
            alert( 'yes, Matched !!' );
            console.log(  myArr[j] );
        } else {
            alert( 'Nops ' );
        }
    });

    //console.log( val );
});

I need to match the array elements, i used $.inArray(), but it never goes to ELSE condition even it doesn't exist in the array. Any help would be appreciatd.

回答1:

$.inArray() takes two arguments, the value and the array, and returns > -1 if it finds a match, so it should be like this:

jQuery(document).ready(function(){
   var val = ["Hourly", "Daily", "Weekly", "Monthly", "Yearly"];
   var myArr = ["Weekly", "something"];
   $.each(myArr, function(i, v) {
     if ($.inArray(v, val) != -1) {
        alert( 'yes, Matched !!' );
        console.log(v);
     } else {
        alert( 'Nops ' );
     }
   });
});

You can test it here. Also note the use of $.each() for non-element sets, no reason creating an invalid jQuery object to run the loop.

回答2:

jQuery(document).ready(function() {

    var val = ["Hourly", "Daily", "Weekly", "Monthly", "Yearly"];
    var myArr = ["Weekly", "something"];

    $(myArr).each(function(j) {
        if ($.inArray(myArr[j], val) != -1) {
            alert('yes, Matched !!');
        } else {
            alert('Nops ');
        }
    });

});

$.inArray returns -1 when it doesn't find the value in the array, otherwise it returns the position in the array, which could be 0.

来源：https://stackoverflow.com/questions/4169106/need-help-regarding-jquery-inarray