访问对象数组中的对象属性

I have this set

我有这套

var data = [
    {"outlet_name":"Easy Lane Supermart","20130102_20130108":"0"},
    {"outlet_name":"Eunilaine Foodmart Kalayaan","20130102_20130108":"0"},
    {"outlet_name":"PUREGOLD PRICE CLUB, INC - VISAYAS","20130102_20130108":"0"}
];

$.each(data, function (i, item) {
    $.each(item, function (k,v) {
        $('#result').append(k,v);
    });
});

How can I make it only view all the values of outlet_name without using the item.outlet_name?

如何在不使用item.outlet_name的情况下仅查看outlet_name的所有值?

3 个解决方案

#1

$.each(data, function (i, item) {
    console.log(item.outlet_name);
});

Or without jQuery:

或者没有jQuery:

for (var i=0;i<data.length;i+=1) {
    console.log(data[i].outlet_name);
}

Ok, if you want to iterate over each object you can write:

好的,如果你想迭代每个对象你可以写:

$.each(data, function (i, item) {
    console.log("Values in object " + i + ":");
    $.each(item, function(key, value) {
        console.log(key + " = " + value);
    });
});

#2

This will provide exact answer...

这将提供准确答案......

var data = [
    {"outlet_name":"Easy Lane Supermart","20130102_20130108":"0"},
    {"outlet_name":"Eunilaine Foodmart Kalayaan","20130102_20130108":"0"},
    {"outlet_name":"PUREGOLD PRICE CLUB, INC - VISAYAS","20130102_20130108":"0"}
];
for(i=0;i<data.length;i++){
	for(var x in data[i]){
	    console.log(x + " => " + data[i][x]);
	}
}

#3

If anyone is needing to do this from a JSON string for example

例如,如果有人需要从JSON字符串执行此操作

var myJson = [{"Code":"slide_1.png","Description":"slide_1"},{"Code":"slide_2.png","Description":"slide_2"},{"Code":"slide_3.png","Description":"slide_3"}]

You can use var newJsonArray = JSON.Parse(myJson) and you will get Array[3] 0 : Object 1 : Object 2 : Object

你可以使用var newJsonArray = JSON.Parse(myJson),你将得到数组[3] 0:对象1:对象2:对象

At which point you can access it by simply saying newJsonArray[i].Code or whatever property inside the array you want to use. Hope this helps!

此时您可以通过简单地说newJsonArray [i] .Code或您想要使用的数组中的任何属性来访问它。希望这可以帮助!

#1

$.each(data, function (i, item) {
    console.log(item.outlet_name);
});