使用jQuery从JSON访问数据
[英]accessing data from JSON using jQuery
问题描述
Need help on code to provide iteration of this JSON response: 
需要有关代码的帮助以提供此JSON响应的迭代：
JSON Response: JSON响应：
[{"id":"1","FK_country":"USA","FK_state":"Arizona","FK_city":"Phoenix","zip":"85001","update_by":"SYSTEM","update_when":"0000-00-00 00:00:00"},
{"id":"2","FK_country":"USA","FK_state":"Arizona","FK_city":"Phoenix","zip":"85002","update_by":"SYSTEM","update_when":"0000-00-00 00:00:00"},
{"id":"3","FK_country":"USA","FK_state":"Arizona","FK_city":"Phoenix","zip":"85003","update_by":"SYSTEM","update_when":"0000-00-00 00:00:00"}]
The code I have UPDATED with assistance from suggestions below is 我在以下建议的帮助下已更新的代码是
function test3 ()
{
var myCriteria = "";
var key = "mykey";
myCriteria = $( "#city" ).val();
$('#myTestDiv').empty().append(myCriteria);
var myDataRequest = $.ajax({
url: 'ajx_zip.php',
type: 'POST',
dataType: 'json',
data: {city:myCriteria, api_key:key},
success: function(myData)
{
alert( "Data Request Success!" );
$('#zip')
.find('option')
.remove()
.end();
$( "#myTestDiv" ).append( "<p>" + myData + "</p>" );
var myNewData = $.parseJSON(myData);
$( "#myTestDiv" ).append( "<p>" + myNewData + "</p>" );
$.each(myNewData, function(i, value)
{
$('#zip').append($('<option></option>').val(value.FK_city).html(value.FK_city));
});
}
});
myDataRequest.fail(function(jqXHR, textStatus)
{
if (jqXHR.status === 0)
{
alert('Not connect.n Verify Network.');
}
else if (jqXHR.status == 404)
{
alert('Requested page not found. [404]');
}
else if (jqXHR.status == 500)
{
alert('Internal Server Error [500].');
}
else if (exception === 'parsererror')
{
alert('Requested JSON parse failed.');
}
else if (exception === 'timeout')
{
alert('Time out error.');
}
else if (exception === 'abort')
{
alert('Ajax request aborted.');
}
else
{
alert('Uncaught Error.n' + jqXHR.responseText);
}
});
}
The HTML code I have UPDATED from suggestions for this is: 我从建议中得到更新的HTML代码是：
<!doctype html public "-//w3c//dtd html 3.2//en">
<html>
<head>
<title>Test</title>
<script src="//ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script language="javascript" src="select.js"></script>
</head>
<body>
<h3>Test Address for Javascript</h3>
<FORM name="address" action="testresult.php" method="POST" >
<SELECT ID="country" NAME="country" >
<Option value="">Select Country</option>
<Option value="USA">United States</option>
<Option value="CAN">Canada</option>
</SELECT>
<br><br>
<SELECT id="state" NAME="state">
<Option value="Arizona">Arizona</option>
<Option value="California">California</option>
</SELECT>
<br><br>
<SELECT id="city" NAME="city" onchange="test3();">
<Option value="Phoenix">Phoenix</option>
<Option value="Glendale">Glendale</option>
<Option value="Chandler">Chandler</option>
<Option value="California">California</option>
</SELECT>
<br><br>
<SELECT id="zip" NAME="zip">
<Option value="Select Zip">Select Zip</option>
</SELECT>
</form>
<div id="myTestDiv"></div>
</body>
</html>
In addition, I have created a simple html-form POST test from same server to PHP processing page and I produce now a limited data-set speeding client-side performance created to test the API - all works well in producing expected results as demonstrated at the top of this post (an echo of the 'results' page). 此外，我已经创建了一个从同一台服务器到PHP处理页面的简单html形式的POST测试，现在我生成了一个有限的数据集，以加快为测试API而创建的客户端性能-所有这些都很好地产生了预期的结果，如下所示这篇文章的顶部（“结果”页面的回声）。 However the error now occurs with processing of the JSON object within the Javascript function at the $.parseJSON level. 但是，错误现在在$ .parseJSON级别在Javascript函数中处理JSON对象时发生。 In the test output DIV my appended output is as follows... 在测试输出DIV中，我附加的输出如下...
Chandler
[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]
3 个解决方案
解决方案1
3 已采纳 2014-10-18 22:00:26
Replace this: 替换为：
<SELECT id="city" NAME="city" onselect="AddZipOptions();">
with this: 有了这个：
<SELECT id="city" NAME="city" onchange="AddZipOptions();">
onSelect is an event fired when you select some text with the mouse! onSelect是当您用鼠标选择一些文本时onSelect的事件！
You should also clean the zip select before appending the results: 您还应该在添加结果之前清理zip select：
$("#zip").children(':not(:first)').remove()
$("#zip").append($('<option>', {text: result.zip, value: result.zip}));
解决方案2
0 2014-10-18 22:45:56
Try adding this to the ajax parameters 尝试将其添加到ajax参数
$.ajax({
datatype: "json",
});
and try calling this in your success function before the each. 并尝试在每个成功函数中调用此函数。
var newMyData = $.parseJSON(myData);
and then use the new variable. 然后使用新变量。
解决方案3
0 2014-10-20 16:25:23
The answer to this question was solved by both @lazel and @dancer and a previous post found here... Ajax success function returning [object, object] ...the JSON response was already parsed as one commentator suggested, but the data returned had already processed as objects within an object. @lazel和@dancer都解决了这个问题的答案，并且在这里找到了以前的文章... Ajax成功函数返回[object，object] ...已经将JSON响应解析为一个注释者的建议，但是返回了数据已经作为对象在对象中处理。
Solved and working with this updated code. 解决并使用此更新的代码。
Hopefully this will help other JQuery newbies. 希望这对其他JQuery新手有所帮助。 Thank you contributors! 谢谢贡献者！
function slct_zip ()
{
var myCriteria = "";
var key = "yourAPIkeyhere";
myCriteria = $( "#city" ).val();
var myDataRequest = $.ajax({
url: 'yourphpqueryscriptinJSON.php',
type: 'POST',
dataType: 'json',
data: {city:myCriteria, api_key:key},
success: function(myData)
{
alert( "Please Select Zip" );
$('#zip')
.find('option')
.remove()
.end();
for (var i=0;i<myData.length;i++)
{
$( "#zip" ).append($('<option></option>').val(myData[i].zip).html(myData[i].zip));
}
}
});
myDataRequest.fail(function(jqXHR, textStatus)
{
if (jqXHR.status === 0)
{
alert('Not connect.n Verify Network.');
}
else if (jqXHR.status == 404)
{
alert('Requested page not found. [404]');
}
else if (jqXHR.status == 500)
{
alert('Internal Server Error [500].');
}
else if (exception === 'parsererror')
{
alert('Requested JSON parse failed.');
}
else if (exception === 'timeout')
{
alert('Time out error.');
}
else if (exception === 'abort')
{
alert('Ajax request aborted.');
}
else
{
alert('Uncaught Error.n' + jqXHR.responseText);
}
});
}
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