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服务器之家 - 编程语言 - PHP教程 - ajax调用返回php接口返回json数据的方法(必看篇)

ajax调用返回php接口返回json数据的方法(必看篇)

2021-05-14 21:16PHP教程网 PHP教程

下面小编就为大家带来一篇ajax调用返回php接口返回json数据的方法(必看篇)。小编觉得挺不错的,现在就分享给大家,也给大家做个参考。一起跟随小编过来看看吧

php代码如下:

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<?php
 
  header('Content-Type: application/json');
  header('Content-Type: text/html;charset=utf-8');
 
  $email = $_GET['email'];
 
  $user = [];
 
  $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database");
  mysql_select_db("Test",$conn);
  mysql_query("set names 'UTF-8'");
  $query = "select * from UserInformation where email = '".$email."'";
  $result = mysql_query($query);
  if (null == ($row = mysql_fetch_array($result))) {
    echo $_GET['callback']."(no such user)";
  } else {
    $user['email'] = $email;
    $user['nickname'] = $row['nickname'];
    $user['portrait'] = $row['portrait'];
    echo $_GET['callback']."(".json_encode($user).")";
  }
 
?>

js代码如下:

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<script>
    $.ajax({
      url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@gmail.com",
      type: "GET",
      dataType: 'jsonp',
      //      crossDomain: true,
      success: function (result) {
        //        data = $.parseJSON(result);
        //        alert(data.nickname);
        alert(result.nickname);
      }
    });
  </script>

其中遇到了两个问题:

1、第一个问题:

 

Uncaught SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and getting the error.

This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

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$ret['foo'] = "bar";
finish();
 
function finish() {
  header("content-type:application/json");
  if ($_GET['callback']) {
    print $_GET['callback']."(";
  }
  print json_encode($GLOBALS['ret']);
  if ($_GET['callback']) {
    print ")";
  }
  exit;
}

Hopefully that will help someone in the future.

2、第二个问题:

解析json数据。从上面的javascript中可以看到,我没有使用jquery.parseJSON()这些方法,开始使用这些方法,但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1的错误,后来不用jquery.parseJSON()这个方法,反而一切正常。不知为何。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了,希望能给大家一个参考,也希望大家多多支持服务器之家。

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