循环 JSON 时出现问题
我想将数据附加到选择框。但是，我收到了这个错误...
“错误：$("").attr("value", guruMapelId).html 不是函数
源文件： http://localhost:8084/controller?aksi=kurikulum
线：5"
这是我的 js 代码：
function dataGuruMapelSelect(dataSelect){
$.getJSON("controller", "aksi=dataGuruMapel", function(json){
$.each(json.guruMapelData, function(k,v){
var guruMapelId = v.guruMapelId;
var guruNama = v.guruNama ;
$('<option />').attr('value',guruMapelId).html(guruNama).appendTo(dataSelect);
})
});
}
这是 JSON 数据
{
"guruMapelData": [
{
"guruMapelId ": "1",
"guruNip ": "1331/001",
"guruNama ": "HARI BUDIYONO DRS.",
"mapelNama ": "PPKn",
"tahunAjarNama ": "2010/2011",
"mapelKategoriNama ": "Normatif",
"mapelId ": "1"
},
{
"guruMapelId ": "2",
"guruNip ": "1331/002",
"guruNama ": "PENI WARDAYANI DRA",
"mapelNama ": "Kewirausahaan",
"tahunAjarNama ": "2010/2011",
"mapelKategoriNama ": "Produktif",
"mapelId ": "2"
}
]
}
我的错是什么？ 先谢谢了...
I want to append data to selectbox. But, I got this error...
"Error: $("").attr("value", guruMapelId).html is not a function
Source File: http://localhost:8084/controller?aksi=kurikulum
Line: 5"
this is my js code:
function dataGuruMapelSelect(dataSelect){
$.getJSON("controller", "aksi=dataGuruMapel", function(json){
$.each(json.guruMapelData, function(k,v){
var guruMapelId = v.guruMapelId;
var guruNama = v.guruNama ;
$('<option />').attr('value',guruMapelId).html(guruNama).appendTo(dataSelect);
})
});
}
And this is the JSON data
{
"guruMapelData": [
{
"guruMapelId ": "1",
"guruNip ": "1331/001",
"guruNama ": "HARI BUDIYONO DRS.",
"mapelNama ": "PPKn",
"tahunAjarNama ": "2010/2011",
"mapelKategoriNama ": "Normatif",
"mapelId ": "1"
},
{
"guruMapelId ": "2",
"guruNip ": "1331/002",
"guruNama ": "PENI WARDAYANI DRA",
"mapelNama ": "Kewirausahaan",
"tahunAjarNama ": "2010/2011",
"mapelKategoriNama ": "Produktif",
"mapelId ": "2"
}
]
}
What is my fault?
Thanks before...
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评论（3）
您的 JSON 键的名称后面包含空格，因此该键是
"guruMapelId "，而不是"guruMapelId"。您可以删除 JSON 键中的空格，也可以使用var guruMapelId = v["guruMapelId "];和var guruNama = v["guruNama "];代替。Your JSON keys contain spaces after their names, so the key is
"guruMapelId ", not"guruMapelId". You can either remove the space from your JSON keys or usevar guruMapelId = v["guruMapelId "];andvar guruNama = v["guruNama "];instead.这是一个棘手的问题。如果将
undefined传递给attr(attr('value', undefined)) 或html(html(undefined))，它实际上与调用attr('value')或html()相同，它们都返回一个字符串而不是一个 jQuery 对象。但是等等，你会说，我没有传递
undefined，我传递的是guruMapelId和guruNama 的值！让我们仔细看看：JSON 中的键都有一个尾随空格。这意味着，
v.guruMapelId不存在（因此是undefined）。您必须使用Same for
guruNama访问数据。但更好的方法是创建正确的 JSON。演示
This was a tricky one. If you pass
undefinedtoattr(attr('value', undefined)) orhtml(html(undefined)), it will actually be the same as callingattr('value')orhtml(), which both return a string and not a jQuery object.But wait, you say, I'm not passing
undefined, I'm passing the values ofguruMapelIdandguruNama!Let's have a closer look: The keys in the JSON all have a trailing space. That means,
v.guruMapelIddoes not exist (hence isundefined). You would have to access the data withSame for
guruNama. But better would be to create proper JSON.DEMO
检查您在 $.getJSON （只需使用alert(guruMapelId) 和 guruNama ）调用中得到的内容，然后您可以按如下方式进行选择：
Check what you get in your $.getJSON (just use alert(guruMapelId) and guruNama ) call then you can just make your option as follow: