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,给出了完美的结果,即这里所有学生的详细信息和他们的分数 -
[{"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"}]
现在我尝试使用 - 在 HTML 文件中显示它
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>"; });
$("#stud_tbl").html(myTable);});};
上面的代码正在显示一个表格,但在表格的每个数据单元格中都显示“未定义”。
No Full Name Mark1 Mark2 Mark3
undefined undefined undefined undefined undefined
undefined undefined undefined undefined undefined
undefined undefined undefined undefined undefined
请帮助解决这个问题。
最佳答案
jQuery.each()的第一个参数是值的索引,第二个是值。
解决方案将$.each(json, function(v) {
改为$.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
关于php - 不显示 JSON 输出,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16144908/