php – 未显示JSON输出

发布时间：2020-12-13 16:50:46 所属栏目：PHP教程来源：网络整理

导读：我有一个 PHP文件,它以JSON格式提供输出.代码如下 – ?php include 'configure.php'; $qr = "SELECT * FROM student_details"; $res= mysql_query($qr); $i=0; while($row = mysql_fetch_array($res)) { $stud_arr[$i]["full_name"] = $row["full_name"]; $s

我有一个 PHP文件,它以JSON格式提供输出.代码如下 –

<?php
    include 'configure.php';
    $qr = "SELECT * FROM student_details";
    $res= mysql_query($qr);
    $i=0;
    while($row = mysql_fetch_array($res))
         {
         $stud_arr[$i]["full_name"] = $row["full_name"];
         $stud_arr[$i]["reg_no"] = $row["regno"];
         $stud_arr[$i]["address"] = $row["address"];
         $stud_arr[$i]["mark1"] = $row["mark1"];
         $stud_arr[$i]["mark2"]= $row["mark2"];
         $stud_arr[$i]["mark3"] = $row["mark3"];
    $i++;
     }
    header('Content-type: application/json'); 
    echo json_encode($stud_arr);
    ?>

这个文件在我的服务器上运行时,完美地给了我结果,即所有学生的详细信息和他们的标记在这里 –

[{"full_name":"Lohith","reg_no":"100","address":"street,lane","mark1":"90","mark2":"87","mark3":"88"},{"full_name":"Ranjeet","reg_no":"101","address":"dfkljg","mark1":"56","mark2":"45","mark3":"39"},{"full_name":"karthik","reg_no":"102","address":"askjldf","mark1":"85","mark2":"90","mark3":"100"}]

现在我试图在HTML文件上显示这个 –

function getAllDetails()
{
var myTable = '' ;
myTable += '<table id="myTable" cellspacing=0 cellpadding=2 border=1>' ;
  myTable +=   "<tr><td><b>No</b></td><td><b>Full Name</b></td><td><b>Mark1</b></td><td><b>Mark2</b></td><td><b>Mark3</b></td></tr>";var url = "json-example2.php";
  $.getJSON(url,function(json) { $.each(json,function(v) {    
                myTable +=   "<tr><td>"+v.reg_no+"</td><td>"+v.full_name+"</td><td>"+v.mark1+
                "</td><td>"+v.mark2+
                "</td><td>"+v.mark3+
                "</td></tr>";   });

                $("#stud_tbl").html(myTable);});};

上面的代码显示了一个表,但在表的每个数据单元格中都显示“undefined”.

No  Full Name     Mark1           Mark2           Mark3
undefined   undefined   undefined   undefined   undefined
undefined   undefined   undefined   undefined   undefined
undefined   undefined   undefined   undefined   undefined

请帮忙解决这个问题.

解决方法

jQuery.each()的第一个参数是值的索引,第二个参数是值.

解决方案更改$.each(json,function(v){to $.each(json,function(i v){

function getAllDetails() {
  var myTable = '';
    myTable += '<table id="myTable" cellspacing=0 cellpadding=2 border=1>';
    myTable += "<tr><td><b>No</b></td><td><b>Full Name</b></td><td><b>Mark1</b></td><td><b>Mark2</b></td><td><b>Mark3</b></td></tr>";
    var url = "data.json";
    $.getJSON(url,function(json) {
                $.each(json,function(i,v) {
                            myTable += "<tr><td>" + v.reg_no + "</td><td>"
                                    + v.full_name + "</td><td>" + v.mark1
                                    + "</td><td>" + v.mark2 + "</td><td>"
                                    + v.mark3 + "</td></tr>";
                        });

                $("#stud_tbl").html(myTable);
            });
};

演示：Plunker

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!