[英]Json_encode data using JSON.parse in javascript
PHP文件
<?php
include"connection.php";
$id = $_GET['edit'];
// echo $id;
$sql = "SELECT * FROM form WHERE id=$id";
$result = $conn->query($sql);
$row=$result->fetch_assoc();
$json_res=array();
$json_res = array('user'=>($row));
echo json_encode($json_res);
$conn->close();
?>
JavaScript 文件
<script type="text/javascript">
function loadup(str){
var xmlhttp;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
// var json = '{"user":[{"id":"45","name":"dfgMohan","loc":"Cherthala","dist":"Alappuzha","phone":"456","email":"fgdfgh@gmail.c","gender":"male","proofid":"2","sslc":"SSLC","plus2":"PLUS 2","degree":"DEGREE","pg":""}]}';
var x=xmlhttp.responseText;
var json1=JSON.parse(x);
alert(json1.user[0].id);
alert(json1.user[0].name);
document.getElementById("myd").innerHTML=json1;
}
}
xmlhttp.open("GET","plain.php?edit="+str,true);
xmlhttp.send();
}
</script>
为什么我不能提醒这个?
alert(json1.user[0].id);
alert(json1.user[0].name);
当我尝试提醒它时,没有显示提醒消息。
当我提醒一些消息时,它在解析代码之前显示,然后在它不执行之后显示。 代码卡在解析代码行上。 如何解决这个问题?
<?php
include"connection.php";
$id = $_GET['edit'];
// echo $id;
$sql = "SELECT * FROM form WHERE id=$id";
$result = $conn->query($sql);
$row = $result->fetch_assoc();
$json_res = array();
array_push($json_res, $row);
echo json_encode($json_res);
$conn->close();
?>
javascript文件包含
<script type="text/javascript">
function loadup(str) {
var xmlhttp;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function ()
{
if (xmlhttp.readyState == 4 && xmlhttp.status == 200)
{
//var json = '{"user":[{"id":"45","name":"Raj Mohan","loc":"Cherthala","dist":"Alappuzha","phone":"9846320121","email":"rajmohan6535@gmail.c","gender":"male","proofid":"2","sslc":"SSLC","plus2":"PLUS 2","degree":"DEGREE","pg":""}]}';
var x = xmlhttp.responseText;
var json1 = JSON.parse(x);
document.getElementById("mydiv").innerHTML = "ID : "+json1[0].id;
document.getElementById("myd").innerHTML = "NAME : "+json1[0].name;
}
}
xmlhttp.open("GET", "plain.php?edit=" + str, true);
xmlhttp.send();
}
</script>
当我们给出响应时,请确保该 php 文件页面中不包含 html ......甚至注释行......他们会给出响应,我们无法解析响应文本所以要小心......当我们传递来自php dnt 添加或评论 php 中包含的任何 html(服务器访问页面)
我想你们都可以理解我的回答..如果不是很好,请原谅我..我的英语交流非常低,那是..
要检查 html 是否包含包含在响应中,您只需在 javascript 中提醒响应,然后它显示任何 html 包含响应,然后 php 有一些 html 包含我们无法解析它.....
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.