从API记录响应

Question

So, I've made this code that is supposed to scan an API for Wikipedia, and then use console.log to write all of the information from the API. However, it's not returning anything.

function display() {
  $.get('https://en.wikipedia.org/w/api.php?format=json&action=query&prop=extracts&exintro=&explaintext=&titles=Stack%20Overflow', function(Data){
    var elements = document.querySelectorAll( 'body *' );
    console.log(elements[0].innerHTML);
    console.log('ya');
  });
}

display();

The variable elements is supposed to get all of the elements inside of the body of the API that it just fetched, and then the console.log statement after that is supposed to log the innerHTML of the first element. However, it's not logging anything.

Answer 1

So there were a number of issues with the code:
1. You are querying the document, but your data is held in the Data variable. Trying logging out the Data variable.
2. The GET was not formed correctly to allow for a Cross Browser request, when I tried it. The following code worked for me:

function display(){
        $.ajax({
            type: "GET",
            url: "https://en.wikipedia.org/w/api.php?format=json&action=query&prop=extracts&exintro=&explaintext=&titles=Stack%20Overflow&callback=?",
            contentType: "application/json; charset=utf-8",
            async: true,
            dataType: "json",
            success: function (data, textStatus, jqXHR) {
                console.log(data);
            },
            error: function (errorMessage) {
            }
        });
    }

    display();

Note: I had to add a callback parameter to the original search URL.

I used the following blog as a reference: http://www.9bitstudios.com/2014/03/getting-data-from-the-wikipedia-api-using-jquery/

The query shown here returns a data object, not an HTML string. So, using query selectors would not be appropriate.

Answer 2

You are not setting elements to the right thing, try:

function display(){
    $.get('https://URL', function(data){
        var parser = new DOMParser()
        , doc = parser.parseFromString(data, "text/xml");
        console.log(doc);
        var elements = doc.querySelectorAll( 'body *' );
        console.log(elements[0].innerHTML);
    })
}

display()