Php 我使用Jquery没有得到任何数据
我使用下面的代码从数据库中获取数据。这里没有问题Php 我使用Jquery没有得到任何数据,php,jquery,mysql,jqxgrid,Php,Jquery,Mysql,Jqxgrid,我使用下面的代码从数据库中获取数据。这里没有问题 $query = "SELECT * FROM takenlijst"; $from = 0; $to = 999; $query .= " LIMIT ".$from.",".$to; $result = mysql_query($query) or die("SQL Error 1: " . mysql_error()); while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { $t
$query = "SELECT * FROM takenlijst";
$from = 0;
$to = 999;
$query .= " LIMIT ".$from.",".$to;
$result = mysql_query($query) or die("SQL Error 1: " . mysql_error());
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$taken[] = array(
'taaknummer' => $row['taaknummer'],
'naam' => $row['naam'],
'taak' => $row['taak'],
'prioriteit' => $row['prioriteit'],
'datum' => $row['datum']
);
}
echo json_encode($taken);
然而,当我使用下面的代码时,我没有得到任何数据。我做错了什么
<html lang="en">
<head>
<link rel="stylesheet" href="../../jquery/jqwidgets/styles/jqx.base.css" type="text/css" />
<script type="text/javascript" src="../../jquery/scripts/jquery-1.10.2.min.js"></script>
<script type="text/javascript" src="../../jquery/jqwidgets/jqxcore.js"></script>
<script type="text/javascript" src="../../jquery/jqwidgets/jqxdata.js"></script>
<script type="text/javascript" src="../../jquery/jqwidgets/jqxbuttons.js"></script>
<script type="text/javascript" src="../../jquery/jqwidgets/jqxscrollbar.js"></script>
<script type="text/javascript" src="../../jquery/jqwidgets/jqxmenu.js"></script>
<script type="text/javascript" src="../../jquery/jqwidgets/jqxgrid.js"></script>
<script type="text/javascript" src="../../jquery/jqwidgets/jqxlistbox.js"></script>
<script type="text/javascript" src="../../jquery/jqwidgets/jqxdropdownlist.js"></script>
<script type="text/javascript" src="../../jquery/jqwidgets/jqxgrid.sort.js"></script>
<script type="text/javascript" src="../../jquery/jqwidgets/jqxgrid.filter.js"></script>
<script type="text/javascript" src="../../jquery/jqwidgets/jqxgrid.selection.js"></script>
<script type="text/javascript" src="../../jquery/jqwidgets/jqxgrid.columnsresize.js"></script>
<script type="text/javascript" src="../../jquery/scripts/demos.js"></script>
<script type="text/javascript">
$(document).ready(function () {
// prepare the data
var source =
{
datatype: "json",
datafields: [
{ name: 'taaknummer', type: 'string'},
{ name: 'naam', type: 'string'},
{ name: 'taak', type: 'string'},
{ name: 'prioriteit', type: 'string'},
{ name: 'datum', type: 'string'}
],
url: 'http://www.site.com/management/takenlijstdata.php',
cache: false
};
var dataAdapter = new $.jqx.dataAdapter(source);
$("#jqxgrid").jqxGrid(
{
source: dataAdapter,
columns: [
{ text: 'Taaknummer', datafield: 'taaknummer', width: 250},
{ text: 'Naam', datafield: 'naam', width: 150 },
{ text: 'Taak', datafield: 'taak', width: 180 },
{ text: 'Prioriteit', datafield: 'prioriteit', width: 200 },
{ text: 'Datum', datafield: 'datum', width: 120 }
]
});
});
</script>
</head>
<body class='default'>
<div id="jqxgrid"></div>
</body>
</html>
$(文档).ready(函数(){
//准备数据
变量源=
{
数据类型:“json”,
数据字段:[
{name:'taaknummer',键入:'string'},
{name:'naam',type:'string'},
{name:'taak',键入:'string'},
{name:'prioriteit',键入:'string'},
{name:'datum',type:'string'}
],
网址:'http://www.site.com/management/takenlijstdata.php',
缓存:false
};
var dataAdapter=new$.jqx.dataAdapter(源);
$(“#jqxgrid”).jqxgrid(
{
来源:dataAdapter,
栏目:[
{文本:'taaknumer',数据字段:'taaknumer',宽度:250},
{文本:“Naam”,数据字段:“Naam”,宽度:150},
{文本:'Taak',数据字段:'Taak',宽度:180},
{文本:'Prioriteit',数据字段:'Prioriteit',宽度:200},
{文本:'Datum',数据字段:'Datum',宽度:120}
]
});
});
回波正确显示数据。即使当我将数据输出到txt文件时,它也可以读取,但当我试图从该代码中读取数据时,它就不能读取。请使用mysqli或PDO和准备好的语句。Mysql在浏览器中点击您的数据url并查找警告/错误是否过期?如果您希望获得json数据以外的任何输出,那么接收端将出现json解析错误。您是否检查了浏览器的javascript控制台是否存在错误?检查你的服务器的访问日志,看看你的JS代码是否曾经访问过服务器?以下是我收到的错误消息。。。“语法错误:使用//@表示sourceMappingURL pragmas已被弃用。请改用//#”,“错误:正在分配一个//#sourceMappingURL,但已经有一个”,“未声明HTML文档的字符编码。如果文档包含US-ASCII范围之外的字符,则在某些浏览器配置中,文档将呈现乱码文本。页面的字符编码必须在文档或传输协议中声明。”