parsererror：SyntaxError：JSON.parse：JSON数据的第2行第1列的意外字符200 OK

Question

我试图从名为Users的MySQL表中填充jqGrid加载数据。 JS脚本是这样的：

jQuery脚本

$(function() { 
    "use strict"; 
    jQuery("#list2").jqGrid({ 
     url:'users_grid_load_data.php?q=2', 
     datatype: "json", 
     mtype: "GET",   
     colNames:['Id','First Name', 'Last Name', 'Username','Level'], 
     colModel:[ 
      {name:'id_user',index:'id_user', width:55}, 
      {name:'firstname',index:'firstname', width:90}, 
      {name:'lastname',index:'lastname', width:90}, 
      {name:'username',index:'username', width:90}, 
      {name:'level',index:'level', width:80, align:"right"}  
     ], 
     rowNum:10, rowList:[10,20,30], 
     pager: '#pager2', 
     sortname: 'id_user', 
     viewrecords: true, 
     sortorder: "asc", 
     height:"auto", 
     width:"auto", 
     caption:"LIST OF USERS" 
    }); 
    jQuery("#list2").jqGrid('navGrid','#pager2',{edit:false,add:false,del:false}); 
}); 

现在，这是users_grid_load_data.php文件：

$page = $_GET['page']; 
$limit = $_GET['rows']; 
$sidx = $_GET['sidx']; 
$sord = $_GET['sord']; 

$result = $mysqli->query("SELECT COUNT(*) AS count FROM users"); 
$row = mysqli_fetch_array($result,MYSQLI_ASSOC); 

$count = $row['count']; 
if($count > 0 && $limit > 0) { 
     $total_pages = ceil($count/$limit); 
} else { 
     $total_pages = 0; 
} 
if ($page > $total_pages) $page=$total_pages; 
$start = $limit*$page - $limit; 
if($start <0) $start = 0; 

$SQL = "SELECT * FROM users ORDER BY $sidx $sord LIMIT $start , $limit"; 
$result = $mysqli->query($SQL); 

$i=0; 
$responce = new stdClass(); 
$responce->page = $page; 
$responce->total = $total_pages; 
$responce->records = $count; 
while($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) { 
    $responce->rows[$i]['id']=$row['id_user']; 
    $responce->rows[$i]'cell']= 
       array($row['id_user'],$row['firstname'], 
        $row['lastname'],$row['username'],$row['level']); 
    $i++; 
} 

echo json_encode($responce); 

的jqGrid的加载，但它的中间显示消息：

parsererror: SyntaxError: JSON.parse: unexpected character at line 2 column 1 of the JSON data 200 OK {"page":"1","total":1,"records":"1","rows":[{"id":"4","cell":["4","Alexandre","Araujo","alexaraujo73","2"]}]} 

我可以看到从MySQL加载的注册t有能力的用户，但我陷入了这个错误。 任何人都可以帮助我吗？我非常感谢任何帮助。 预先感谢您。

Answer 1

这可能是因为任何变量的未定义错误。从你的代码是当你在发送JSON数据来创建

$responce = new stdClass(); 
$responce->page = $page; 
$responce->total = $total_pages; 
$responce->records = $count; 
$responce->rows=array();// create an array of rows here 
while($row = mysqli_fetch_array($result,MYSQLI_ASSOC)) { 
    $arr = array('id'=>$row['id_user'],'cell'=>array($row['id_user'], 
            $row['firstname'], 
            $row['lastname'], 
            $row['username'], 
            $row['level']) // cell closing 
     ); // closing of arr 
    $responce->rows[]=$arr; // push it to $rows of $reponce 
    //$i++; // no need of it 
} 

命名为$responce和while loop您使用的是以前没有定义的属性rows，所以尽量使用像之前声明它的响应变量， jqgrid，所以你的回应必须是json而不是别的。为了防止响应添加使用error_reporting()像其他人一样的错误，

// hide all notice/warnings 
error_reporting(0);