만 - VoidCC

-3

그래서 내가 얻을 file.html과만

<!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

를 포함하는 " http://coenraets.org/blog/2011/10/sample-application-with-jquery-mobile-and-phonegap/" 에서 몇 일 전 폰갭 응용 프로그램의 샘플

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() .'}}'; 
} 


?>

은이는 내 localhost database 내가 webbrowser 내 데이터베이스에서, display 단순히 "lastname,department,and title" only에 내 코드를 수정해야 할 일 물어보고 싶은

"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')"

출처

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";

그리고

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

출처

2013-05-03 10:59:05 parnas

는 actualy 난 (e.lastName이 e.title, e.department는, 계산, "$의 SQL은 ="e.id 선택 r.id를 포함의 설명을 이해 해달라고) reportCount ". "종업원 e에서 왼쪽 r.managerId = e.id " "e.lastName에 의해 e.id 순서로 그룹에 가입; "" 직원 e에서 "e.id", "기능 및"은 무엇입니까? e.id, e.lastName, e.title, e.department, count (r.id) reportCount " " r.managerId = e.id " "e.lastName by e.id order by "그룹을 참조 하시겠습니까? – hardsolidman

현재 제공되는 $ sql = line을 file.php로 바꾸면됩니다. 부서도 지정합니다. 그런 다음 file.js의 코드 부분을 다음 비트로 바꿉니다. – parnas

만

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