如何修改这个代码来显示名字，姓氏和部门只

-3

，所以我得到“http://coenraets.org/blog/2011/10/sample-application-with-jquery-mobile-and-phonegap/” 的PhoneGap应用此样品前几天包含 file.html如何修改这个代码来显示名字，姓氏和部门只

<!DOCTYPE HTML> 
<html> 
<head> 
<title>Employee Directory</title> 
<meta name="viewport" content="width=device-width,initial-scale=1"/> 
<link rel="stylesheet" href="css/jquery.mobile-1.0rc1.min.css" /> 
<link rel="stylesheet" href="css/styles.css" /> 
</head> 

<body> 

<div id="employeeListPage" data-role="page" > 

<div data-role="header" data-position="fixed"> 
    <h1>Employee Directory</h1> 
</div> 

<div data-role="content"> 
    <ul id="employeeList" data-role="listview" data-filter="true"></ul> 
</div>  

</div> 

<script src="js/jquery.js"></script> 
<script src="js/jquery.mobile-1.0rc1.min.js"></script> 
<script src="js/employeelist.js"></script> 
<script src="js/employeedetails.js"></script> 
<script src="js/reportlist.js"></script> 

</body> 

</html>

file.js

enter code here 
var serviceURL = "http://localhost/services/"; 

var employees; 

$('#employeeListPage').bind('pageinit', function(event) { 
getEmployeeList(); 
}); 

function getEmployeeList() { 
$.getJSON(serviceURL + 'getemployees.php', function(data) { 
    $('#employeeList li').remove(); 
    employees = data.items; 
    $.each(employees, function(index, employee) { 
     $('#employeeList').append 
('<li><a href="employeedetails.html?id=' + employee.id + '">' + 
       '<h4>' + employee.lastName + '</h4>' + 
       '<p>' + employee.title + '</p>' + 
       '<span class="ui-li-count">' 
+ employee.reportCount + '</span></a></li>'); 
    }); 
    $('#employeeList').listview('refresh'); 
}); 
}

file.php

<?php 
include 'config.php'; 

$sql = "select e.id, e.lastName, e.title, count(r.id) reportCount " . 
    "from employee e left join employee r on r.managerId = e.id " . 
    "group by e.id order by e.lastName"; 


try { 
$dbh = new PDO("mysql:host=$dbhost;dbname=$dbname", $dbuser, $dbpass); 
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 
$stmt = $dbh->query($sql); 
$employees = $stmt->fetchAll(PDO::FETCH_OBJ); 
$dbh = null; 
echo '{"items":'. json_encode($employees) .'}'; 
} catch(PDOException $e) { 
echo '{"error":{"text":'. $e->getMessage() .'}}'; 
} 


?>

，这是我的localhost database

"INSERT INTO employee (id,firstName,lastName,managerId,title,department,officePhone,cellPhone,email,city,picture) VALUES (12,'Steven','Wells',4,'Software Architect','Engineering','617-000-0012','781-000-0012','[email protected]','Boston, MA','steven_wells.jpg')"

我想问我应该怎么做才能改变我的代码display简单"lastname,department,and title" only，从我的数据库中webbrowser。

来源

2013-05-03 hardsolidman

$sql = "select e.id, e.lastName, e.title, e.department, count(r.id) reportCount " . 
"from employee e left join employee r on r.managerId = e.id " . 
"group by e.id order by e.lastName";

然后

('<li><a href="employeedetails.html?id=' + employee.id + '">' + 
       '<h4>' + employee.lastName + '</h4>' + 
       '<p>' + employee.title + '</p>' + 
       '<p>' + employee.department + '</p>' + 
       '</li>');

来源

2013-05-03 10:59:05 parnas

actualy我不明白的解释包含 “$的SQL =” 选择e.id，e.lastName，e.title，e.department，计数（r.id ）reportCount“。 ”from employee e left join employee r on r.managerId = e.id“。 ”group by e.id order by e.lastName“;” 什么是“e.id”，“功能和”选择e.id，e.lastName，e.title，e.department，count（r.id）reportCount“ ”from employee e left join employee r on “r.managerId = e.id” “e.id订单由e.lastName”参考？ – hardsolidman 2013-05-03 11:06:30

您只需要将file.php中的当前$ sql =行替换为我提供的那一行，它现在也将selet部门然后用下一个位替换file.js中的部分代码。 – parnas 2013-05-03 11:14:28

如何修改这个代码来显示名字，姓氏和部门只

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