Introduction to the $.inarray () method of jquery traversal array

The $.inarray () function searches the array for the specified value and returns its index value. Returns 1 if the value does not exist in the array;

The $.inarray (Value,array)--value is the value to look for, and the array is the one being looked up.

Use the $.inarray () method must pay attention to the point, otherwise it will fall out of the pit

(1) Look at the following code:

        $ (function  () {              var arr=[{"name": "Zhang San"},{"name": "John Doe"},{"name": "Harry"}];              Alert ($.inarray ({"name": "Zhang San"}, arr);        });

The above code does not carefully analyze how to see it is not wrong, but the return value is-1. Below to analyze the cause of the discovery:

The cause of the error is {"name": "Zhang San"} and {"name": "Zhang San"} is two different references, so {"name": "Zhang San"} is not found in arr array; If you change that, you can.

        $ (function  () {            var obj = {"Name": "Zhang San" };             var arr = [{"Name": "John Doe"}, obj, {"name": "Harry"}];            Alert ($.inarray (obj, arr);        });

This code returns 1; The return value is normal;

(2) It is known that JavaScript is a weakly typed language, and for numeric types and character types, you can switch freely (for example: 1+ "" = "1"), so there is the following code:

        $ (function  () {  var arr = [1, 2, 3, 4, 5];
var a = 2;
Alert ($.inarray (A, arr);        });

This code returns a normal value of 1;

        $ (function  () {            var arr = [1, 2, 3, 4, 5];             var a = "2";            Alert ($.inarray (A, arr);        });

When you change the value of a to a string 2 The return value is 1; So when you use this $.inarray () method, make sure that the data type is the same, although JavaScript is a weakly typed language;

Description of the $.inarray () method of the jquery traversal array