javascript根据关联数组键值从数组中查找值

Question

I have an associative array:

var array1 = {
    a1: {
        id: "box1",
        title: "box 1"
    },
    a2: {
        id: "box2",
        title: "box 2"
    },
    a3: {
        id: "box3",
        title: "box 3"
    }
};

I then have another array which has references to the first array:

var array_order = ["a3:positionA", "a2:postitionB", "a1:positionC"];

I want to loop through the first list and then use the second list to find the position text

I am using jQuery so I have

$.each(array1, function(i,o) {
  something in here where I can use i to find out what position. e.g. if a1 I would get positionC
}

Answer 1

Easier and faster iterate over array_order instead of array1 :

for( var i = 0, len = array_order.length; i < len; i++ ) {
    var ref = array_order[i].split(':');

    if( ref.length === 2 && ({}).hasOwnProperty.call( array1, ref[0] ) ) {
        var array1Property = array1[ ref[0] ];
        var array1PropertyPosition = ref[1];
    }
} 

Iteration over array1 properties can be realized also, but it's significantly slower:

for( var prop in array1 )
    if( ({}).hasOwnProperty.call( array1, prop ) )
        for( var i = 0, len = array_order.length; i < len; i++ ) {
            var ref = array_order[i].split(':');
            if( ref.length === 2 && prop === ref[1] ) {
                var array1Property = array1[ prop ];
                var array1PropertyPosition = ref[1];
            }
        }

And there is no need in jQuery.

Answer 2

Ok first things first JS calls associative arrays as objects only.

next to iterate over each key in object

for(var prop in array1) {
    if(array1.hasOwnProperty(prop))
        for(key in array_order){
           if( key === prop){
           doSomethingwith(array_order[key]);

           }

        }
}

Answer 3

var temp_array_order = {};
$.each(array_order, function(key, val) { 
    temp_array_order[val.substr(0,2)] = val.substr(3); 
};
$.each(array1, function(key, val) {
    var category = temp_array_order[key];
    ...
});