是否始终需要在整个链中（或依次）始终调用对象？

Question

I'm trying to look through reddit's API. I have seen a fiddle where they grab the image url from the site. However, I'm confused as to how they are getting the objects themselves. In their each() statement, they used data.data.children and item.data.url , however I cannot find a data object right after another data object in the json here http://www.reddit.com/r/pics.json . I also do not see item.data.url . if I look from the very beginning (the two first objects are "kind" and "data"). What does this mean?

Answer 1

Given the code in the fiddle:

$.getJSON("http://www.reddit.com/r/pics/.json?jsonp=?", function(data) {
    $.each(data.data.children, function(i,item){
        $("<img/>").attr("src", item.data.url).appendTo("#images");
    });
});

data and item are both function parameters, and their names have no relation to any named key within the JSON.

The parameter data represents the entire JSON structure, but as it's just a parameter whose name is not significant I shall call it foo instead to disambiguate it from the contents of that structure, hence foo.data refers to the child element named data within the first level of the JSON.

The item parameter represents each element in the array foo.data.children , so item.data.url for the first item would be equivalent to data.children[0].data.url within the JSON.

Answer 2

The structure of the JSON returned is:

{
    data: {
        after: "",
        before: null,
        children: [
            {
                data: {
                    url: ""
                }
            },
            ...
        ],
        modhash: ""
    },
    kind: "Listing"
}

The name of the result variable is called data in the callback function parameters. So to get to the children you need to type:

data.data.children;

Next, in the $.each loop, each entry from the children array is assigned to an item variable in that callback function. It is from there where you drill down to the url . property ( item.data.url ).

You could name these whatever you want if data.data is confusing.