javascript 未捕获的类型错误：无法读取未定义的属性 x
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Uncaught TypeError: Cannot read property x of undefined
提问by Nick Germi
Getting the above error in Chrome console while the actual script works and generates the right output, wonder how I can get rid of this error and what is causing it.
在实际脚本运行并生成正确的输出时，在 Chrome 控制台中出现上述错误，想知道如何摆脱此错误以及导致它的原因。
JSFiddle: http://jsfiddle.net/wJUeP/
JSFiddle：http: //jsfiddle.net/wJUeP/
HTML Code:
HTML代码：
<ul id="menu"></ul>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
JS Code:
JS代码：
$(function(){
var data = [{"weekending":"09\/10\/2013","jobs":[{"jobnumber":"1001","jobaddress":"Test1001","employees":[{"employeenumber":"1","name":"James Blabla","class":"FHM","notes":"xx","nt-wkmon":"2","t12-wkmon":"5","dt-wkmon":"4","status-wkmon":"Public Holiday","startTime-wkmon":"4","finishTime-wkmon":"6","nt-wktue":"7"}]},{"jobnumber":"1002","jobaddress":"Test1002","employees":[{"employeenumber":"1","name":"Cameron Le","class":"FHQ","notes":"xx","nt-wkmon":"2","t12-wkmon":"5","dt-wkmon":"4","status-wkmon":"Public Holiday","startTime-wkmon":"4","finishTime-wkmon":"6","nt-wktue":"7"},{"employeenumber":"2","name":"David Le","class":"FHQ","notes":"xx","nt-wkmon":"2","t12-wkmon":"5","dt-wkmon":"4","status-wkmon":"Public Holiday","startTime-wkmon":"4","finishTime-wkmon":"6","nt-wktue":"7"}]},{"jobnumber":"1003","jobaddress":"Test1003","employees":[{"employeenumber":"1","name":"Nick G","class":"sdf","notes":"sdf","nt-wkmon":"2","t12-wkmon":"5","dt-wkmon":"4","status-wkmon":"Public Holiday","startTime-wkmon":"4","finishTime-wkmon":"6","nt-wktue":"7"}]}]}];
for(var i = 0, j = data[0].weekending.length; i<j; i++) {
rootMenu = data[0].jobs[i];
$("#menu").append("<li id='job_" + rootMenu.jobnumber + "'>" + rootMenu.jobnumber);
if(rootMenu.hasOwnProperty("employees")) {
$("#menu").append("<ul id='employees_job_" + rootMenu.jobnumber + "'>");
for(var n = 0, m = rootMenu.employees.length; n<m; n++) {
var subMenu = rootMenu.employees[n];
if(subMenu.hasOwnProperty("name")) {
$("#employees_job_" + rootMenu.jobnumber).append("<li>" + subMenu.name + "</li>");
}
}
$("#menu").append("</ul>");
} else {
$("#menu").append("</li>");
}
}
});
Note: I'm still in the development stage of my application and I have the flexibility to change and manipulate the data structure, if embedded JSON data looks bad I can change it, actual data is stored in a XML file and then read by PHP and outputted as JSON.
注意：我仍处于应用程序的开发阶段，我可以灵活地更改和操作数据结构，如果嵌入的 JSON 数据看起来不好我可以更改它，实际数据存储在 XML 文件中，然后由 PHP 读取并输出为 JSON。
回答by Abhidev
the for loop condition is incorrect.
for 循环条件不正确。
you have used j = data[0].weekending.lengthwhich is equal to 10 and you are iterating over the data[0].jobsobject which has only 3 jobs. You are iterating more than 3 times over the jobs and hence you are getting the error.
您已经使用j = data[0].weekending.lengthwhich 等于 10 并且您正在迭代data[0].jobs只有 3 个作业的对象。您对作业进行了 3 次以上的迭代，因此出现错误。
checkout the fiddle http://jsfiddle.net/wJUeP/7/
回答by jasonscript
The error is here:
错误在这里：
j = data[0].weekending.length
This is returning 10, which is the length of the string in the weekending property == 10 ("09/10/2013") I think you need this instead:
这将返回 10，这是周末属性 == 10 ("09/10/2013") 中字符串的长度，我认为您需要这个：
j = data[0].jobs.length
