Question

由于之前的计算机开始疯狂，我只是将我的代码从一台计算机交换到另一台计算机。现在，当我尝试使用我的网站时，它不会将任何信息提取到div中。我真的不确定这里发生了什么，它给了我一个错误

未捕获的SyntaxError：意外的令牌＆lt;在JSON的第2位JSON.parse（）

我之前没有得过的。

<script>
        //$('#agent').change(function(){
            $(document).on('change', "#agent", function(event) { 

        //jquery
         var IEX_ID=$(this).val();


        if  (IEX_ID){
            //ajax call
            $.ajax({

                type:'POST',
                url:'GetAllData.php',
                cache:false,
                data: 'Fullname='+IEX_ID,
                success: function(data){





                    var obj = JSON.parse(data);

                    console.log(obj);



                    $('#IEX_ID').html(obj.IEX_ID);
                    $('#BalancedScore').html(obj.BalancedScore);
                    $('#CustomerEx').html(obj.CEPt);
                    $('#CEM1').html(obj.CEM1);
                    $('#CEM2').html(obj.CEM2);
                    $('#CEM3').html(obj.CEM3);
                    $('#CETlabel').html(obj.CET1);
                    $('#CET2label').html(obj.CET2);
                    $('#CET3label').html(obj.CET3);

             </script>

下面是从中拉出数组的php。如果我做控制台日志，它会给我“尝试获取非对象的属性'num_rows'”这通常意味着我的查询存在问题。我没有发现任何问题？它与在另一台计算机上使用的完全相同的代码。我只是复制所有文件并传输它们以查看是否修复了任何问题。我不确定为什么一台计算机上的一切都完美无缺，但在这一台计算机上却给了我这些错误。

    <?php

    $Fullname=$_POST['Fullname'];

    if(!empty($_POST['Fullname']))


  {



            $query=$conn->query("SELECT Supervisor, Fullname, IEX_ID, BalancedScore, CEPt, CEM1, CEM2, CEM3, FCRPt, FCRM1, FCRM2, AIMPt, AIMScore, AIMCnt, ProdPt, CET1, CET2, CET3, FCRT1, FCRT2, AYHT, AYHM, M_AYHT, AYHPt, Total, skill1, skill2, skill3, skill4, skill5, skill6, skill7, skill8, skill9, skill10, skill11, skill12, skill13, skill14, skill15, skill16, skill17, skill18, skill19, IncentiveMonth FROM CurrentSC WHERE IEX_ID = '".$Fullname."' order by Fullname ASC ");


            $rowCount=$query->num_rows;

            if($rowCount>0)


            {
                            while($row = mysqli_fetch_array($query))
                            {

                                            echo json_encode;

                   $output = array ("IEX_ID" => $row['IEX_ID'], "BalancedScore" => $row['BalancedScore'], "CEPt" => $row['CEPt'], "CEM1" => $row['CEM1'] , "CEM2" => $row['CEM2'], "CEM3" => $row['CEM3'], "FCRPt" => $row['FCRPt'], "FCRM1" => $row['FCRM1'] ,"FCRM2" => $row['FCRM2'],"AIMPt" => $row['AIMPt'],"AIMScore" => $row['AIMScore'],"AIMCnt" => $row['AIMCnt'],"ProdPt" => $row['ProdPt'],"CET1" => $row['CET1'] ,"CET2" => $row['CET2'],"CET3" => $row['CET3'],"FCRT1" => $row['FCRT1'],"FCRT2" => $row['FCRT2'],"AYHT" => $row['AYHT'],"AYHPt" => $row['AYHPt'],"M_AYHT" => $row['M_AYHT'],"AYHM" => $row['AYHM'],"Total" => $row['Total'],"skill1" => $row['skill1'],"skill2" => $row['skill2'],"skill3" => $row['skill3'],"skill4" => $row['skill4'],"skill5" => $row['skill5'],"skill6" => $row['skill6'],"skill7" => $row['skill7'],"skill8" => $row['skill8'],"skill9" => $row['skill9'],"skill10" => $row['skill10'],"skill11" => $row['skill11'],"skill12" => $row['skill12'],"skill13" => $row['skill13'],"skill14" => $row['skill14'],"skill15" => $row['skill15'],"skill16" => $row['skill16'],"skill17" => $row['skill17'],"skill18" => $row['skill18'],"skill19" => $row['skill19'],"IncentiveMonth" => $row['IncentiveMonth']);  






              echo json_encode($output);



                            }

                      }
         }


        ?>

Answer 1

未捕获的SyntaxError说：意外的令牌＆lt;在JSON的第2位JSON.parse（）

这意味着网址：'GetAllData.php' 可能会返回PHP错误。这可能是数据库，PHP更新，缺少函数等......当我的PHP源格式错误时，我看到同样的错误。

QED：检查website.com/GetAllData.php的结果，看看它们是否是正确的JSON。

为什么我的JSON不能突然工作？

1 个答案: