通过PHP访问JSON对象 | 通过PHP访问JSON对象

本文介绍了通过PHP访问JSON对象的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有以下代码.

if (config.sendResultsURL !== null)
{
  console.log("Send Results");
  var collate =[];
  for (r=0;r<userAnswers.length;r++)
  {
    collate.push('{"questionNumber'+parseInt(r+1)+ '"' + ': [{"UserAnswer":"'+userAnswers[r]+'", "actualAnswer":"'+answers[r]+'"}]}');
  }
  $.ajax({
    type: 'POST',
    url: config.sendResultsURL,
    data: '[' + collate.join(",") + ']',
    complete: function()
    {
      console.log("Results sent");
    }
  });
}

使用Firebug，我可以从控制台获得此信息.

Using Firebug I get this from the console.

[{"questionNumber1": [{"UserAnswer":"3", "actualAnswer":"2"}]},{"questionNumber2": [{"UserAnswer":"3", "actualAnswer":"2"}]},{"questionNumber3": [{"UserAnswer":"3", "actualAnswer":"2"}]},{"questionNumber4": [{"UserAnswer":"3", "actualAnswer":"1"}]},{"questionNumber5": [{"UserAnswer":"3", "actualAnswer":"1"}]}]

脚本从此处将数据发送到emailData.php，其内容为...

From here the script sends data to emailData.php which reads...

$json = json_decode($_POST, TRUE);
$body = "$json";
$to = "[email protected]";
$email = 'Diesel John';

$subject = 'Results';
$headers  = "From: $email\r\n";
$headers .= "Content-type: text/html\r\n";

// Send the email:
$sendMail = mail($to, $subject, $body, $headers);

现在我收到了电子邮件，但是它为空.

Now I do get the email however it is blank.

我的问题是如何将数据传递到emailData.php并从那里访问它?

My question is how do I pass the data to emailData.php and from there access it?

推荐答案

创建要传递给PHP的对象
使用JSON.stringify()为该对象创建JSON字符串.
使用POST或GET并使用名称将其传递到PHP脚本.
根据您的请求从$_GET['name']或$_POST['name']捕获它.
在php中应用json_decode以获取JSON作为本机对象.

Create an object that you want to pass to PHP
Use JSON.stringify() to make a JSON string for that object.
Pass it to PHP script using POST or GET and with a name.
Depending on your request capture it from $_GET['name'] OR $_POST['name'].
Apply json_decode in php to get the JSON as native object.

在您的情况下，您可以只传递userAnswers [r]和答案[r].数组序列被保留.

In your case you can just pass userAnswers[r] and answers[r]. Array sequence are preserved.

在循环使用中，

collate.push({"UserAnswer":userAnswers[r], "actualAnswer":answers[r]});

在ajax请求使用中，

In ajax request use,

data: {"data" : JSON.stringify(collate)}

在PHP端，

 $json = json_decode($_POST['data'], TRUE); // the result will be an array.

这篇关于通过PHP访问JSON对象的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！