2014-07-06 60 views
0

新编程.i现在我的json数据正确我想要显示它在html中。 这里是我的PHP代码显示收到json数据到html

<?php 
header('Access-Control-Allow-Origin: *'); 
$con=mysqli_connect("host","user","pass","db"); 
// Check connection 
if (mysqli_connect_errno()) { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    exit(); 
} 
$callback=$_GET['callback']; 
$result = mysqli_query($con,"SELECT * FROM demo"); 

$var= array(); 

while($row = mysqli_fetch_assoc($result)) 
     { 
    $var[]=$row; 
    } 
echo $callback."(".json_encode($var).")"; 
mysqli_close($con); 
?> 

ND这里是Java脚本& HTML

<!DOCTYPE html> 
<html> 
<head> 
    <script src="http://code.jquery.com/jquery-1.11.1.min.js"></script> 
<script> 
fname = screen.width; 
$.ajax({ 
    type: 'GET', 
    url: "devangpatel.host56.com/sample.php?jsonp=processResults"+"fname="+fname, 
    dataType: 'jsonp', 
    jsonp: 'callback' ,//jquery will add callback 
    jsonpCallback:'processResults'//name of the callback function which server must return 
    }); 
    window.processResults = function (response){ 
    //it must be called with response data inside server answer 
    console.log(response); 
    } 
</script> 
</head> 
<body> 

<h2>AJAX</h2> 

<div id="myDiv"></div> 

</body> 
</html> 

也提出了一些网站,了解有关从服务器获取数据(AJEX。)

+0

这又回到一个字符串,所以你只是试图让它无论如何显示在页面上?下面是你的页面上的样子,从服务器'{name:'first',age:10,lived:['Dallas','Boston']}' - 显然有不同的数据 – Deryck

回答

0

你可以使用jQuery html http://api.jquery.com/html/显示内容

假设回应为

{ "name":"user1" }

你可以设置DIV HTML内容

$("#myDiv").html("hello"+response.name);

你也可以看看替代模板引擎http://www.sitepoint.com/10-javascript-jquery-templates-engines/

+0

([{ “FirstName”:“devang”,“LastName”:“patel”},{“FirstName”:“chetan”,“LastName”:“sh ah”}])这是我的json响应。我把$(“#myDiv”)。html(“hello”+ response.FirstName);在window.processResult()仍然不起作用 – user3790198