Javascript 从json_encode访问数据
下面是json_encode在ajax页面生成的代码。现在,我想检索数据标题ISBN,作者。。。。谁能帮我一把吗?我已经看过jQuery.parseJSON函数,但感到困惑Javascript 从json_encode访问数据,javascript,ajax,json,Javascript,Ajax,Json,下面是json_encode在ajax页面生成的代码。现在,我想检索数据标题ISBN,作者。。。。谁能帮我一把吗?我已经看过jQuery.parseJSON函数,但感到困惑 { "9780077225957": { "Items_Data": { "Title": "Developing Management Skills: What Great Managers Know and Do", "Data_Source": "
{
"9780077225957": {
"Items_Data": {
"Title": "Developing Management Skills: What Great Managers Know and Do",
"Data_Source": "Amazon",
"Item_ID": "1329",
"ISBN": "9780077225957",
"Authors": "Timothy Baldwin",
"Edition": "1",
"Year": "2007",
"Publisher": "McGraw-Hill/Irwin",
"Amazon_Thumb_URL": "http://ecx.images-amazon.com/images/I/41pVf7GKujL._SL160_.jpg"
}
}
}
虽然我认为您首先需要将9780077225957更改为键,因为您将无法使用它来访问它,但只需分配一个JavaScript变量就可以了。操作人员我已经把它改成了9780077225957本书,工作很好。这是我使用的代码
var book = { "Book9780077225957" : { "Items_Data": { "Title": "Developing Management Skills: What Great Managers Know and Do", "Data_Source": "Amazon", "Item_ID": "1329", "ISBN": "9780077225957", "Authors": "Timothy Baldwin", "Edition": "1", "Year": "2007", "Publisher": "McGraw-Hill/Irwin", "Amazon_Thumb_URL": "http://ecx.images-amazon.com/images/I/41pVf7GKujL.SL160.jpg" } } };
alert(book.Book9780077225957.Items_Data.Title);
我希望做一些类似的事情:
var Books = {
toElement: function(data) {
if (data) {
$book = $('<div></div>')
.addClass('book');
$('<span></span>').className("title").text(data.Title)appendTo($book);
$('<span></span>').className("authors").text(data.Authors)appendTo($book);
$('<span></span>').className("isbn").text(data.ISBN)appendTo($book);
$book.appendTo($('#books'));
}
}
};
$(function() {
$("#Button1").click(function() {
var params = [];
$.getJSON("book_url.js", params, function(returnedJSON) {
Books.toElement(returnedJSON.Items_Data);
});
});
});
我不确定JSON中的第一个元素是什么。它可能不会返回json.Items\u数据。出于某种原因,我认为第一个层次被忽略了。但我可能在想别的事情
希望这能有所帮助。您是从$.getJSON中检索到这个吗?你想用这些数据做什么?创建元素?Objects?@natedavisolds它目前来自纯ajax,但我认为我可以在我的ajax上添加dateType:json。我将它们用于纯HTML。那么您将基于返回的JSON在DOM中创建一个元素?是吗?尽管我认为您需要使用$.getJSON通过Ajax检索数据: