在 AJAX 调用中从 PHP 返回 JSON 对象？
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Returning a JSON object from PHP in AJAX call?
提问by Kumar Kush
Here's my PHP code called during jQuery AJAX call:
这是我在 jQuery AJAX 调用期间调用的 PHP 代码：
<?php
include '../code_files/conn.php';
$conn = new Connection();
$query = 'SELECT Address_1, Address_2, City, State, OfficePhone1, OfficePhone2, Fax1, Fax2, Email_1, Email_2
FROM clients WHERE ID = ?';
$conn->mysqli->stmt_init();
$stmt = $conn->mysqli->prepare($query);
$stmt->bind_param('s', $_POST['ID']);
$stmt->execute();
$result = $stmt->get_result();
$row = $result->fetch_assoc();
echo json_encode($row);
?>
And the client-side code is:
客户端代码是：
$.post(url, {ID:$('#ddlClients').val()},
function(Result){
// Result
}
);
The AJAX call is successfully completed. I get the value of Resultas
AJAX 调用成功完成。我得到的值，结果为
"{"Address_1":"Divisional Office 1","Address_2":"The XYZ Road",.....and so on
"{"Address_1":"Divisional Office 1","Address_2":"The XYZ Road",.....and so on
What I want is to be able to use the values returned like Result.Address_1, Result.Address_2 and so on. But I can't do it using the above code. I tried using $row = $result->fetch_object()and $row = $result->fetch_array(), but no use.
我想要的是能够使用返回的值，如 Result.Address_1、Result.Address_2 等。但是我不能使用上面的代码来做到这一点。我尝试使用$row = $result->fetch_object()and $row = $result->fetch_array()，但没有用。
And I know that this can be done by this code on the server side:
而且我知道这可以通过服务器端的这段代码来完成：
$row = $result->fetch_assoc();
$retVal = array("Address_1"=>$row['Address_1'], "Address_2"=>$row['Address_2'].......);
echo json_encode($retVal);
or
或者
$row = $result->fetch_object();
$retVal = array("Address_1"=>$row->Address_1, "Address_2"=>$row->Address_2.......);
echo json_encode($retVal);
Is there a way to send the $rowdirectly to the client side JavaScript and ready to be used as JSON object, without manually creating an array first?
有没有办法将$rowJavaScript 直接发送到客户端并准备用作 JSON 对象，而无需先手动创建数组？
回答by Jasper
The response you are getting from your PHP script is in plain text. You can however parse that string into an object using $.parseJSONin your callback function:
您从 PHP 脚本中得到的响应是纯文本格式。但是，您可以$.parseJSON在回调函数中使用该字符串将该字符串解析为一个对象：
$.ajax({
url : url,//note that this is setting the `url` property to the value of the `url` variable
data : {ID:$('#ddlClients').val()},
type : 'post',
success : function(Result){
var myObj = $.parseJSON(Result);
//you can now access data like this:
//myObj.Address_1
}
}
);
You can let jQuery do this for you by setting the dataTypeproperty for your AJAX call to json:
您可以通过将dataTypeAJAX 调用的属性设置为 ，让 jQuery 为您执行此操作json：
$.ajax({
url : url//note that this is setting the `url` property to the value of the `url` variable
data : {ID:$('#ddlClients').val()},
dataType : 'json',
type : 'post',
success : function(Result){
//you can now access data like this:
//Result.Address_1
}
}
);
The above examples expect that the response from the server to be in this format (from your question):
上面的示例期望来自服务器的响应采用这种格式（来自您的问题）：
"{"Address_1":"Divisional Office 1","Address_2":"The XYZ Road"}
回答by Tim Withers
In your $.postcall, the last argument could be the data-type: json:
在您的$.post调用中，最后一个参数可能是数据类型json：：
$.post(url, {ID:$('#ddlClients').val()},
function(Result){
alert(Result.Address_1);
},'json'
);
Everything should work then, as it looks like you are doing everything right.
一切都应该正常工作，因为看起来你做的一切都是正确的。
回答by Lightness Races in Orbit
json_encodeaccepts objects, so there's no need to do that automatic array-building.:
json_encode接受对象，因此无需进行自动数组构建。：
$row = $result->fetch_object();
echo json_encode($row);
It's as simple as that!
就这么简单！
