Java paramquery赢得';加载servlet URL JSON字符串?
下面是index.xhtml,我无法获取加载和显示数据的url。servlet正常运行,JSON字符串正确返回为Java paramquery赢得';加载servlet URL JSON字符串?,java,json,function,servlets,Java,Json,Function,Servlets,下面是index.xhtml,我无法获取加载和显示数据的url。servlet正常运行,JSON字符串正确返回为 {"data":[{"LASTNAME":"Leonard","PERSON_ID":"0","FIRSTNAME":"Erick","FULLNAME":"Erick Leonard"}]} 我认为挑战在于var dataModel=区域中的URL或getData <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transition
{"data":[{"LASTNAME":"Leonard","PERSON_ID":"0","FIRSTNAME":"Erick","FULLNAME":"Erick Leonard"}]}
我认为挑战在于var dataModel=区域中的URL或getData
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml"
xmlns:ui="http://xmlns.jcp.org/jsf/facelets"
xmlns:h="http://xmlns.jcp.org/jsf/html"
xmlns:f="http://xmlns.jcp.org/jsf/core"
xmlns:c="http://java.sun.com/jsp/jstl/core">
<h:head>
<link rel="stylesheet" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.9.2/themes/base/jquery-ui.css"/>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"/>
<script src="http://ajax.googleapis.com/ajax/libs/jqueryui/1.9.2/jquery-ui.min.js"/>
<h:outputStylesheet name="css/pqgrid.min.css"/>
<h:outputScript name="js/pqgrid.min.js"/>
<h:outputScript name="js/jquery.ui.touch-punch.js"/>
<script>
$(function()
{
var dataModel =
{ location: "remote",
dataType: "JSON",
method: "GET",
url: "queryreturn",
getData: function (dataJSON)
{ return { data: dataJSON.data }; }
}
var obj = {};
obj.dataModel = dataModel;
obj.width = 700;
obj.height = 400;
obj.colModel = [
{ title: "Person ID", width:100, dataType: "int", dataIndx: "person_id"},
{ title: "Full Name", width:200, dataType: "string", dataIndx: "fullname"},
{ title: "First Name", width:150, dataType: "string", dataIndx: "firstname"},
{ title: "Last Name", width:150, dataType: "string", dataIndx: "lastname"}];
$("div#grid_array").pqGrid( obj );
});
</script>
</h:head>
<h:body>
<div id="grid_array"></div>
</h:body>
</html>
$(函数()
{
var数据模型=
{位置:“远程”,
数据类型:“JSON”,
方法:“获取”,
url:“queryreturn”,
getData:函数(dataJSON)
{return{data:dataJSON.data};}
}
var obj={};
obj.dataModel=数据模型;
物体宽度=700;
物体高度=400;
obj.colModel=[
{标题:“个人ID”,宽度:100,数据类型:“int”,数据索引:“个人ID”},
{标题:“全名”,宽度:200,数据类型:“字符串”,数据索引:“全名”},
{标题:“名字”,宽度:150,数据类型:“字符串”,数据索引:“名字”},
{标题:“姓氏”,宽度:150,数据类型:“字符串”,数据索引:“姓氏”}];
$(“div#grid_数组”).pqGrid(obj);
});
首先,javascript区分大小写,JSON字符串以大写形式返回列名
额外的错误是找不到所需的JSTL jar文件,一旦我将程序更改为Maven并添加了依赖项,它就可以在本地工作。我添加了错误通知代码,并部署了开发版本,请查看错误消息,因为我不理解错误消息,并且查看了Tomcat日志。[链接]缺少的是JSTL jar。它现在运行。