dataType json的jQuery $ .ajax请求不会从PHP脚本中检索数据

Question

我一直在寻找解决方案,但我找不到任何有效的方法.我试图从数据库中获取一堆数据,然后通过表单中的AJAX自动完成输入字段.为此,我决定使用json,因为为什么不呢,对吧？另外,我一直在考虑发回一个分隔的字符串,然后将它标记出来,在后视中它会更容易并且让我免于头疼......因为我已经决定使用json,我想我应该坚持下去,找出问题所在!发生的事情是,当执行get_member_function()时,警报对话框中会弹出一个错误,并显示"[object Object]".我也尝试过使用GET请求,并将contentType设置为"application/json; 字符集= UTF-8".唉,没有骰子.任何人都可以建议我做错了什么？小心,彼得.

我的javascript/jQuery函数如下:

function get_member_info()
   {

   var url = "contents/php_scripts/admin_scripts.php"; 
   var id = $( "select[ name = member ] option:selected" ).val();

   $.ajax(
   {

      type: "POST",
      dataType: "json",
      url: url,
      data: { get_member: id },
      success: function( response ) 
      { 

          $( "input[ name = type ]:eq( " + response.type + " )" ).attr( "checked", "checked" );
          $( "input[ name = name ]" ).val( response.name );
          $( "input[ name = fname ]" ).val( response.fname );
          $( "input[ name = lname ]" ).val( response.lname );
          $( "input[ name = email ]" ).val( response.email );
          $( "input[ name = phone ]" ).val( response.phone );
          $( "input[ name = website ]" ).val( response.website );
          $( "#admin_member_img" ).attr( "src", "images/member_images/" + response.image );

      },
      error: function( error )
      {

         alert( error );

      }

   } );

}

"contents/php_scripts/admin_scripts.php"中的相关代码如下:

   if( isset( $_POST[ "get_member" ] ) )
   {

      $member_id = $_POST[ "get_member" ];
      $query = "select * from members where id = '$member_id'";

      $result = mysql_query( $query );

      $row = mysql_fetch_array( $result );

      $type = $row[ "type" ];
      $name = $row[ "name" ];
      $fname = $row[ "fname" ];
      $lname = $row[ "lname" ];
      $email = $row[ "email" ];
      $phone = $row[ "phone" ];
      $website = $row[ "website" ];
      $image = $row[ "image" ];

      $json_arr = array( "type" => $type, "name" => $name, "fname" => $fname, "lname" => $lname, "email" => $email, "phone" => $phone, "website" => $website, "image" => $image );

      echo json_encode( $json_arr );

   }

Answer 1

我想我知道这个......

尝试使用PHP的header()函数将JSON作为JSON发送:

/**
 * Send as JSON
 */
header("Content-Type: application/json", true);

虽然你传递的是有效的JSON,但jQuery的$ .ajax并不这么认为,因为它缺少标题.

jQuery曾经很好没有标题,但它改回了几个版本.

也

确保您的脚本返回有效的JSON.使用Firebug或Google Chrome的开发人员工具在控制台中检查请求的响应.

UPDATE

您还需要更新代码以清理$ _POST以避免sql注入攻击.以及提供一些错误捕获.

if (isset($_POST['get_member'])) {

    $member_id = mysql_real_escape_string ($_POST["get_member"]);

    $query = "SELECT * FROM `members` WHERE `id` = '" . $member_id . "';";

    if ($result = mysql_query( $query )) {

       $row = mysql_fetch_array($result);

       $type = $row['type'];
       $name = $row['name'];
       $fname = $row['fname'];
       $lname = $row['lname'];
       $email = $row['email'];
       $phone = $row['phone'];
       $website = $row['website'];
       $image = $row['image'];

       /* JSON Row */
       $json = array( "type" => $type, "name" => $name, "fname" => $fname, "lname" => $lname, "email" => $email, "phone" => $phone, "website" => $website, "image" => $image );

    } else {

        /* Your Query Failed, use mysql_error to report why */
        $json = array('error' => 'MySQL Query Error');

    }

     /* Send as JSON */
     header("Content-Type: application/json", true);

    /* Return JSON */
    echo json_encode($json);

    /* Stop Execution */
    exit;

}

Answer 2

在获取数据时尝试使用jQuery.parseJSON.

type: "POST",
dataType: "json",
url: url,
data: { get_member: id },
success: function(data) { 
    response = jQuery.parseJSON(data);
    $("input[ name = type ]:eq(" + response.type + " )")
        .attr("checked", "checked");
    $("input[ name = name ]").val( response.name);
    $("input[ name = fname ]").val( response.fname);
    $("input[ name = lname ]").val( response.lname);
    $("input[ name = email ]").val( response.email);
    $("input[ name = phone ]").val( response.phone);
    $("input[ name = website ]").val( response.website);
    $("#admin_member_img")
        .attr("src", "images/member_images/" + response.image);
},
error: function(error) {
    alert(error);
}

Answer 3

该$.ajax error函数有三个参数,而不是一个:

error: function(xhr, status, thrown)

你需要转储第二个和第三个参数来找到你的原因,而不是第一个.