jQuery 未捕获的类型错误：无法读取未定义的属性“长度”
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Uncaught TypeError: Cannot read property 'length' of undefined
提问by elksie5000
How do I avoid the following error when trying to take data from an AJAX call and plug it into another function when I click the submit button?
在单击提交按钮时尝试从 AJAX 调用中获取数据并将其插入另一个函数时，如何避免以下错误？
The console.log call in the ajax function shows that the data is grabbed and I'd hoped then stored in json_data.
ajax 函数中的 console.log 调用显示数据已被抓取，我希望然后将其存储在 json_data 中。
The aim was then to use this data to alter a string submitted through an HTML form.
目的是使用这些数据来改变通过 HTML 表单提交的字符串。
The error is then returned at the line in the 'click' function:
然后在“click”函数中的行返回错误：
console.log(json_data.length);
<title>Test Form</title>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.7/jquery.min.js"></script>
<style>
#results_box {
border: red 5px solid;
}
#place {
border: #cccccc 1px solid;
}
</style>
<script type="text/javascript">
$(document).ready(function() {
var json_source = "https://spreadsheets.google.com/feeds/list/0ApL1zT2P00q5dG1wOUMzSlNVV3VRV2pwQ2Fnbmt3M0E/od7/public/basic?alt=json";
var string_data ="";
var json_data = $.ajax({
dataType: 'jsonp',
url: json_source,
success: function(data){
var data_obj = [];
for (i=0; i<data.feed.entry.length; i++){
var el = {'key': data.feed.entry[i].title['$t'], 'value': '<p><a href="'+data.feed.entry[i].content['$t']+'>'+data.feed.entry[i].title['$t']+'</a></p>'};
data_obj.push(el)};
console.log("data grabbed");
return data_obj;
},
error: function(jqXHR, textStatus, errorThrown){
$('#results_box').html('<h2>Something went wrong!</h2><p><b>' + textStatus + '</b> ' + errorThrown + '</p>');
}
});
$(':submit').click(function(event, json_data){
event.preventDefault();
console.log(json_data.length);
//function
if ($('#place').val() !=''){
var copy_string = $('#place').val();
var converted_string = copy_string;
for (i=0; i<json_data.length; i++){
//console_log(data.feed.entry[i].title['$t']);
converted_string = converted_string.replace(json_data.feed.entry[i].title['$t'],
'<a href="'+json_data.feed.entry[i].content['$t']+'>'+json_data.feed.entry[i].title['$t']+'</a>');
}
$('#results_box').text(converted_string).html();
}
});
});//document ready end
</script>
</head>
<body>
<div id="wrapper">
<div id="query_box" class="panel">
<form id="form_submit"><h4>Copy to process:</h4>
<textarea id="place"></textarea>
<input type="submit" value="Go" />
</form>
</div>
<div id="results_box" >Results will appear here</div>
</div>
采纳答案by Miquel Las Heras
You are not passing the variable correctly. One fast solution is to make a global variable like this:
您没有正确传递变量。一个快速的解决方案是创建一个像这样的全局变量：
var global_json_data;
$(document).ready(function() {
var json_source = "https://spreadsheets.google.com/feeds/list/0ApL1zT2P00q5dG1wOUMzSlNVV3VRV2pwQ2Fnbmt3M0E/od7/public/basic?alt=json";
var string_data ="";
var json_data = $.ajax({
dataType: 'json', // Return JSON
url: json_source,
success: function(data){
var data_obj = [];
for (i=0; i<data.feed.entry.length; i++){
var el = {'key': data.feed.entry[i].title['$t'], 'value': '<p><a href="'+data.feed.entry[i].content['$t']+'>'+data.feed.entry[i].title['$t']+'</a></p>'};
data_obj.push(el)};
console.log("data grabbed");
global_json_data = data_obj;
return data_obj;
},
error: function(jqXHR, textStatus, errorThrown){
$('#results_box').html('<h2>Something went wrong!</h2><p><b>' + textStatus + '</b> ' + errorThrown + '</p>');
}
});
$(':submit').click(function(event){
var json_data = global_json_data;
event.preventDefault();
console.log(json_data.length);
//function
if ($('#place').val() !=''){
var copy_string = $('#place').val();
var converted_string = copy_string;
for (i=0; i<json_data.length; i++){
//console_log(data.feed.entry[i].title['$t']);
converted_string = converted_string.replace(json_data.feed.entry[i].title['$t'],
'<a href="'+json_data.feed.entry[i].content['$t']+'>'+json_data.feed.entry[i].title['$t']+'</a>');
}
$('#results_box').text(converted_string).html();
}
});
});//document ready end
回答by danludwig
console.log(typeof json_data !== 'undefined'
? json_data.length : 'There is no spoon.');
...or more simply...
……或者更简单的……
console.log(json_data ? json_data.length : 'json_data is null or undefined');
回答by Manthesh M R
You are accessing an object that is not defined.
您正在访问未定义的对象。
The solution is check for null or undefined (to see whether the object exists) and only then iterate.
解决方案是检查 null 或 undefined（以查看对象是否存在），然后才进行迭代。
回答by Do?uhan Elma
"ProjectID" JSON data format problem Remove "ProjectID": This value collection objeckt key value
“ProjectID”JSON数据格式问题 去掉“ProjectID”：这个值集合objeckt键值
{ * * "ProjectID" * * : {
"name": "ProjectID",
"value": "16,36,8,7",
"group": "Genel",
"editor": {
"type": "combobox",
"options": {
"url": "..\/jsonEntityVarServices\/?id=6&task=7",
"valueField": "value",
"textField": "text",
"multiple": "true"
}
},
"id": "14",
"entityVarID": "16",
"EVarMemID": "47"
}
}
