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隨着PhoneGap的幫助,我正在製作一個Android應用程序,其中我已經使用phonegap sqlite數據庫,我將數據存儲在數據庫中,並從數據庫中取回以顯示,但我的圖像未顯示,只有路徑到圖像存儲的位置。請幫我找到我的錯誤。無法從數據庫加載圖像;只有URL顯示
jQuery中: -
function List() {
$.ajax({
type: "GET",
url: "one.html",
contentType: "text/xml",
dataType: "xml",
data: "",
crossDomain: true,
success: function(xml) {
$(xml).find('xyz').each(function() {
var title = $(this).find('title').text();
var Image = $(this).find('image').text();
db.transaction(function(transaction) {
transaction.executeSql('INSERT INTO A
VALUES ("' + title + '","' + Image + '")',
nullHandler, errorHandler);
});
});
Dynamic_List();
return false;
}
}
});
}
/*This Method Create Dynamic Menu Item List*/
function Dynamic_List() {
$('.mylistview').empty();
db.transaction(function(transaction) {
transaction.executeSql('SELECT * FROM A;', [],
function(transaction, results) {
if (results != null && results.rows != null) {
for (var i = 0; i < results.rows.length; i++) {
var image = results.rows.item(i).A_Image;
var Title = results.rows.item(i).A_Title;
$('.mylistview').append(
'<li class = "cat_list">' +
'<div class = "divli">' +
'<div class = "menuImg" ' +
'style = "height:48px; width:48px;">' +
menu_image +
'</div>' +
'<div class = "divbody">' +
'<h3>' + menu_Item_Title + '</h3>' +
'</div>' +
'</div>' +
'</li>');
}
}
}
}, errorHandler);
}, errorHandler, nullHandler);
return;
}
在HTML5: -
<div class = "foodList">
<ul class = "mylistview"
style = "display: block;"
id = "my_dynamic_list_view">
</ul>
</div>