解析与访问 JSON 数据

发布于 2024-10-29 08:16:53 字数 1098 浏览 6 评论 0原文

我有 JSON 数据（添加了换行符以提高可读性）返回为

[{"pk": 7, "model": "pycourt_login.orders", "fields": {"status": 0, "delivered": false, "order_id": "6count1%2", "student_id": 5, "counterid": "counter1", "datetime": "2011-04-05 01:44:01", "dish": 6, "transaction_id": "2008010661301935441", "quantity": 2}},
{"pk": 8, "model": "pycourt_login.orders", "fields": {"status": 0, "delivered": false, "order_id": "9count1%1", "student_id": 5, "counterid": "counter1", "datetime": "2011-04-05 01:44:14", "dish": 9, "transaction_id": "2008010661301935454", "quantity": 1}}]

How do I access every item in jQuery.

我尝试做类似的事情（假设 /pendingorders 返回 JSON）

$(document).ready(function (){ 
jQuery.ajax({
    cache: false,
    url: "/pendingorders",
    type: "GET",
    success: function(json) {
        $.each(json.results, function(i,dish){
            alert(dish.pk); 
            },'json')
        },  
    }); 
});

这是我在 Firebug 中遇到的错误：

object is undefined
[Break On This Error] length = object.length,
jquery.js (line 605)

原文

I have JSON data (added a newline for readability) returned as

[{"pk": 7, "model": "pycourt_login.orders", "fields": {"status": 0, "delivered": false, "order_id": "6count1%2", "student_id": 5, "counterid": "counter1", "datetime": "2011-04-05 01:44:01", "dish": 6, "transaction_id": "2008010661301935441", "quantity": 2}},
{"pk": 8, "model": "pycourt_login.orders", "fields": {"status": 0, "delivered": false, "order_id": "9count1%1", "student_id": 5, "counterid": "counter1", "datetime": "2011-04-05 01:44:14", "dish": 9, "transaction_id": "2008010661301935454", "quantity": 1}}]

How do I access each item in jQuery.

I tried doing something like (assume that /pendingorders returns JSON)

$(document).ready(function (){ 
jQuery.ajax({
    cache: false,
    url: "/pendingorders",
    type: "GET",
    success: function(json) {
        $.each(json.results, function(i,dish){
            alert(dish.pk); 
            },'json')
        },  
    }); 
});

This is the error I get in Firebug:

object is undefined
[Break On This Error] length = object.length,
jquery.js (line 605)

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凉世弥音 2024-11-05 08:16:55

使用当前的结构，我认为您只想要：

jQuery.ajax({
    cache: false,
    url: "/pendingorders",
    type: "GET",
    success: function(json) {
        $.each(json, function(i,dish){
              alert(dish.pk); 
        });
    }); 
});

您有一个杂散的 , 'json' 作为 each 的第三个参数，以及成功值后面的逗号（即好的，但可能是无意的）

但是，由于 JSON 劫持。

如果结构是：

{"results":
[{"pk": 7, "model": "pycourt_login.orders", "fields": {"status": 0, "delivered": false, "order_id": "6count1%2", "student_id": 5, "counterid": "counter1", "datetime": "2011-04-05 01:44:01", "dish": 6, "transaction_id": "2008010661301935441", "quantity": 2}},
{"pk": 8, "model": "pycourt_login.orders", "fields": {"status": 0, "delivered": false, "order_id": "9count1%1", "student_id": 5, "counterid": "counter1", "datetime": "2011-04-05 01:44:14", "dish": 9, "transaction_id": "2008010661301935454", "quantity": 1}}]}

你原来的每一行：

$.each(json.results, function(i,dish){

将是正确的。

With the current structure, I think you just want:

jQuery.ajax({
    cache: false,
    url: "/pendingorders",
    type: "GET",
    success: function(json) {
        $.each(json, function(i,dish){
              alert(dish.pk); 
        });
    }); 
});

You had a stray , 'json' as the third parameter to each, and a comma after the success value (which is okay but probably unintended)

However, you may want to make the top level an object instead of an array, due to JSON hijacking.

If the structure were:

{"results":
[{"pk": 7, "model": "pycourt_login.orders", "fields": {"status": 0, "delivered": false, "order_id": "6count1%2", "student_id": 5, "counterid": "counter1", "datetime": "2011-04-05 01:44:01", "dish": 6, "transaction_id": "2008010661301935441", "quantity": 2}},
{"pk": 8, "model": "pycourt_login.orders", "fields": {"status": 0, "delivered": false, "order_id": "9count1%1", "student_id": 5, "counterid": "counter1", "datetime": "2011-04-05 01:44:14", "dish": 9, "transaction_id": "2008010661301935454", "quantity": 1}}]}

your original each line:

$.each(json.results, function(i,dish){

would be correct.

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三生池水覆流年 2024-11-05 08:16:55

您将“json”作为参数传递给 each() （注意缩进）。您通常可以在 get() 上执行此操作，但 ajax() 的工作方式不同 - 设置 dataType 选项就足够了：

jQuery.ajax({
    cache: false,
    url: "/pendingorders",
    type: "GET",
    dataType: 'json',
    success: function(json) {
        $.each(json.results, function(i,dish){
            alert(dish.pk); 
        });
    }
});

请参阅 ajax方法文档供参考。

You're passing 'json' as a parameter to each() (watch the indentation). You can usually do that on get(), but ajax() works differently - setting the dataType option should suffice:

jQuery.ajax({
    cache: false,
    url: "/pendingorders",
    type: "GET",
    dataType: 'json',
    success: function(json) {
        $.each(json.results, function(i,dish){
            alert(dish.pk); 
        });
    }
});

See the ajax method documentation for reference.

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