使用PHP发回json值时发生意外错误 [英] Unexpected error when posting back json value using PHP

问题描述

我正在使用Ajax将一些数据从服务器端代码PHP发送回客户端，这是完成的操作

I am posting back some data from my server side code PHP to my client side using Ajax , this is how its done

//server side
$json="{
"payout_history":"0",
"round_shares":"1816",
"workers":
   {
    "jbo.5970":
      {
        "alive":"1",
        "hashrate":"1253"
      },
    "jbo.5970cpu":
      {
        "alive":"1",
        "hashrate":"21"
      },
    "jbo.5970-2":
      {
        "alive":"1",
        "hashrate":"1062"
      }
  }
}";
echo json_encode($json);

我在萤火虫的响应页面中收到此错误，我无法弄清它是怎么了

I am getting this error in the response page of firebug and i cant figure out whats wrong with it

    <br />
<b>Parse error</b>:  syntax error, unexpected 'payout_history' (T_STRING) in         
<b>C:\xampp\htdocs\exercise5json\display.php</b> on line <b>38</b><br />

推荐答案

您没有正确嵌套引号.您需要将JSON字符串括在单引号中，而不是双引号中:

You're not nesting the quotes properly. You need to enclose your JSON string in single quotes, not double quotes:

$json = '{"myTag":"myData"}';

更好-将数组创建为PHP数组，并使用json_encode()为您生成JSON.

Or better - create the array as a PHP array and use json_encode() to produce the JSON for you.

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