我使用phonegap构建iphone应用程序。我正在使用sqlite3数据库在本地存储数据。该应用程序在模拟器上完美工作,但在实际的ios设备上出现错误。它是抛出“Could not prepare statement (1 no such table: table_name)
”错误代码为相同的是Code = 5。IOS数据库应用程序(phonegap) - 在模拟器上工作,但不能在设备上工作
我必须在ios设备上安装sqlite插件吗?该应用程序正处于测试阶段。我遵循this中提供的步骤,在iOS设备上安装应用程序。
我在这里错过了什么?
更新: 这是我使用的一段代码。我的数据库驻留在位置
/用户/ iMac电脑/库/ Application Support/iPhone 模拟器/ 7.0.3 /应用/ 4C7CC11A-8938-479F-B810-86121D3311B7 /图书馆/ WebKit的/本地存储/ File_0
和设备上它驻留在
AppData的/图书馆/ WebKit的/本地存储/ File_0
<html>
<head>
<meta charset="utf-8" />
<meta name="format-detection" content="telephone=no" />
<meta name="viewport" content="user-scalable=no, initial-scale=1, maximum-scale=1, minimum-scale=1, width=device-width, height=device-height, target-densitydpi=device-dpi" />
<link rel="stylesheet" type="text/css" href="css/index.css" />
<script type="text/javascript" src="cordova.js"></script>
<script type="text/javascript" src="js/index.js"></script>
<title>Books | Categories</title>
<link href="css/bootstrap.css" rel="stylesheet" type="text/css">
<link href="css/style.css" rel="stylesheet" type="text/css">
<script type="text/javascript" charset="utf-8" src="js/jquery.min.js"></script>
<script type="text/javascript" charset="utf-8">
var db;
var shortName = 'Books';
var version = '1.0';
var displayName = 'BooksDB';
var maxSize = 200000;
function errorHandler(transaction, error) {
alert('Error: ' + error.message + ' code: ' + error.code);
}
function successCallBack() {
alert("DEBUGGING: success");
}
function nullHandler(){
alert('null handler');
};
function onBodyLoad(){
if (!window.openDatabase) {
alert('Databases are not supported in this browser.');
return;
}
db = window.openDatabase(shortName, version, displayName, maxSize);
alert('db open');
ListDBValues();
}
function ListDBValues() {
var ArrayAlphabet=new Array("A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z");
$('.container').empty();
for (var i = 0; i < ArrayAlphabet.length; i++) {
data='<div class="order" id="'+ArrayAlphabet[i]+'"></div>';
load_books(ArrayAlphabet[i]);
$('.container').append(data);
data="";
}
return;
}
function load_books(bookTitleAlphabet)
{
if (!window.openDatabase) {
alert('Databases are not supported in this browser.');
return;
}
db.transaction(function(transaction) {
transaction.executeSql('SELECT * FROM books where book_title like "'+bookTitleAlphabet+'%" order by book_title desc;', [],
function(transaction, result) {if (result != null && result.rows != null) {
$('#'+bookTitleAlphabet).html(bookTitleAlphabet);
for (var i = 0; i < result.rows.length; i++) {
var data;
var row = result.rows.item(i);
data="<a href='details.html?id="+row.book_id+"'> <div class='book'>";
data +="<div class='book_img'><img src="+row.book_thumb_location+"></div>";
data +="<div class='book_detail'>";
data +="<div class='title'>"+row.book_title+"</div>";
data +="<div class='author'>"+row.book_author+"</div>";
data +="</div>";
data +="<div class='clear'>";
data +="</div>";
data +="</div>";
data +="</a>";
var tempId='#'+bookTitleAlphabet;
$(tempId).append(data);
tempId="";
}}
if (result.rows.length==0)
{
var tempId='#'+bookTitleAlphabet;
$(tempId).hide();
}
},errorHandler);},errorHandler,nullHandler);
return;
}
</script>
</head>
<body onload="onBodyLoad()">
<div id="wrapper">
<div class="overflow_hide">
</div>
<div class="menu">
<div class="header">
<div class="back">
<a href="index.html">
<input class="gobutton" type="button" value="Back" ></a>
</div>
<div class="list_book">BOOKS
</div>
<div class="settings">
<a href="index.html">
<input class="gobutton" type="button" value="Home" ></a>
</div>
</div>
<div class="container">
<div class="clear">
</div>
</div>
</div>
<div class="opac">
<a href="about.html">
<div class="opac1">About Us
</div></a>
<a href="search.html">
<div class="opac1">Search
</div></a>
</div>
<div class="clear">
</div>
</div>
</body>
</html>
这是因为应用程序找不到数据库?
是的我在我的应用程序中使用预先填充的sqlite3 Db。 – Arti
Hi @Matt Ray,我已经更新了我的questoin。请看一下。 – Arti
您好,@ user1650891。我基于此更新了我的答案并批准了其他信息! –