我有一个php文件,我用ajax连接到它,回调值是JSON,当我从php获取数据时dosx显示,当警报数据我看到对象时
我的问题在哪里?
PHP:
if(isset($_SERVER["HTTP_X_REQUESTED_WITH"])){
$query = mysql_query("select * from tab");
for ($i=0;$i<mysql_num_rows($query);$i++){
while($row = mysql_fetch_assoc($query)){
$title['t'][i] = $row['title'];
$title['d'][i] = $row['description'];
}
}
echo(json_encode($title));
exit();
?>
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JS:
$('#button').click(function(){
$.ajax({
url : "test2.php",
data : $("#tab"),
type : "GET",
success : function(b){
b = eval('('+ b +')');
console.log((b['t']));
alert(b);
}
});
});
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如何从这个JSON中获取所有数据并向我展示corect呢?
我想将一个onPreExecute AsyncTask添加到SherlockFragment但它会导致错误.我的代码如下.在preExecute AsyncTask上添加对话框的代码有什么问题?
public class customlist extends SherlockFragment {
static final String URL = "esample";
static final String KEY_SONG = "song"; // parent node
private static final String KEY_TAB_NUM = "key.tab.num";
private ProgressDialog pDialog;
ListView list;
LazyAdapterbeth adapter;
XMLParser parser = new XMLParser();
public void onActivityCreated(Bundle savedInstanceState) {
super.onActivityCreated(savedInstanceState);
}
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
new getFeed().execute();
}
public View onCreateView(LayoutInflater inflater, ViewGroup container,
Bundle savedInstanceState)
{
View thisfragment = inflater.inflate(R.layout.dovomi, container, false);
return thisfragment;
}
private class getFeed …Run Code Online (Sandbox Code Playgroud) 我有一个搜索的地方,当我进入它开始搜索,这是好的,但我想,当我按下输入功能启动2秒后.
我正在使用,setTimeout但它无法正常工作.
我能做什么 ?
表格:
<form id="search" method="get">
<input id="searchplace" type="text" size="40" value="" placeholder="xxxx" />
</form>
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JavaScript
$("#search").keypress(function(event){
if(event.keyCode == 13){
$('.menu-bg').addClass('loading');
var x = $("#searchplace").val();
setTimeout(m(),2000);
function m(){
$('.tabs').remove();
window.location="search.php?page=noajax&q="+x;
return false;
};
}
});
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