如何将JSON传回PHP的AJAX请求?

I've built a simple hangouts application that notifies me when someone launches a hangout from on my webpage - outside of Google plus.

我已经构建了一个简单的hangouts应用程序，当有人在我的网页上发布hangout时——在谷歌plus之外。

I'm at the stage where I've called my server side script that notifies me that a new user has initiated a hangout and I need to pass a message back to the hangout to display an updated prompt to the users. Something along the lines of - A support engineer has been alerted to this hangout and will be with you shortly.

我正在调用我的服务器端脚本，它通知我一个新用户已经启动了一个hangout，我需要将消息传回hangout以向用户显示更新的提示符。一些类似的东西——一个支持工程师已经被提醒到这个hangout，并且很快就会和你在一起。

I've read in the hangout data, including who has joined from the hangouts api and passed this to my server side code no problems. But I'm not sure how I pass information back to my hangout app via its ajax request.

我已经阅读了hangout数据，包括谁加入了hangouts api并将它传递给了我的服务器端代码，没有问题。但我不确定如何通过ajax请求将信息传回hangout应用程序。

Here's my ajax request using jQuery.

这是我使用jQuery的ajax请求。

var hangoutUrl = gapi.hangout.getHangoutUrl();
var personsInHangout = gapi.hangout.getParticipants();

var callbackUrl = 'http://styxofdynamite.bitnamiapp.com/notify.php?';

$.ajax({
    url: callbackUrl,
    dataType: 'text',
    data: {
        "personsInHangout" : personsInHangout,
        "hangoutUrl" : hangoutUrl,
        "topic" : params['gd'],
    }
}).done( function(data, status, xhr){
    //call was made process result
    $('.msg').html(data.msg);
}).fail( function(xhr, status, error){
    $('.msg').html("There was a problem contacting the server. (" + status + ")");
});

At the moment I'm not returning anything so I'm happily hitting the .fail()ideally I need to return something from notify.php back to this hangout application.

现在我没有返回任何东西，所以我很高兴地点击了。fail()，理想情况下，我需要返回一些通知。php回到这个hangout应用程序。

Sever side snippet:

服务器端代码片段:

$personsInHangout = $_GET['personsInHangout'];
$hangoutUrl = $_GET['hangoutUrl'];
$topic = $_GET['topic'];

$emailMessage = 'Hey Support Team, ';

for($i = 0, $size = count($personsInHangout); $i < $size; ++$i) {
    $emailMessage .= $personsInHangout[$i]['person']['displayName'];
    $emailMessage .= ' is currently waiting in the support ';
    $emailMessage .= '<a href="'.$hangoutUrl.'">hangout</a>';
    $emailMessage .= ' to discuss ' . $topic;
}

print $emailMessage;

$to      = 'support@team.com';
$subject = 'New Hangout Request';
$headers = 'From: hangout.monitor@team.com' . "\r\n" .
'Reply-To: hangout.monitor@team.com' . "\r\n" .
'X-Mailer: PHP/' . phpversion();

mail($to, $subject, $emailMessage, $headers);

//return something to say that a support engineer has been notified.

3 个解决方案

#1

Use php's built in json_encode docs to send back a JSON formatted response. You will also need to send some headers in addition to the data.

使用json_encode文档中内置的php返回JSON格式的响应。除了数据之外，还需要发送一些标题。

<?php
// This tells the browser the following data should be interpreted as JSON, as opposed to text or html
header('Content-Type: application/json');
// Take any $data and format it correctly as JSON.  Works with arrays, strings, numbers
echo json_encode($data);

?>

3 个解决方案

#1

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