我有一個鏈接到一個PHP文件,下面的代碼,目的是發佈一些字符串和PHP文件檢索,所有的東西工作正常,但事情是錯我的代碼,請參閱波紋管:ASIFormDataRequest POST錯誤 - 的iOS 6
的iOS應用程序代碼
NSString *urlString = @"http://www.myurl.com/path/ios.st.addquestion.php?";
NSURL *uploadURL = [NSURL URLWithString:urlString];
ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:uploadURL];
[request setPostValue:askerUsername forKey:@"username"];
[request setPostValue:@"1" forKey:@"questiontype"];
[request setPostValue:title forKey:@"title"];
[request setPostValue:hasAditional forKey:@"hasaditional"];
[request setPostValue:aditional forKey:@"aditional"];
[request setPostValue:hasLocation forKey:@"haslocation"];
[request setPostValue:location forKey:@"location"];
[request setPostValue:latitude forKey:@"latitude"];
[request setPostValue:longitude forKey:@"longitude"];
[request setPostValue:category forKey:@"category"];
[request setPostValue:hasImage forKey:@"hasimage"];
[request startAsynchronous];
PHP文件
$questiontype = mysql_real_escape_string($_GET['questiontype']);
$username_PRO = mysql_real_escape_string($_GET['username']);
$title_PRO = mysql_real_escape_string($_GET['title']);
$hasAditional = mysql_real_escape_string($_GET['hasaditional']);
$aditional_PRO = mysql_real_escape_string($_GET['aditional']);
$hasLocation = mysql_real_escape_string($_GET['haslocation']);
$location_PRO = mysql_real_escape_string($_GET['location']);
$latitude_PRO = mysql_real_escape_string($_GET['latitude']);
$longitude_PRO = mysql_real_escape_string($_GET['longitude']);
$category = mysql_real_escape_string($_GET['category']);
echo "$username_PRO";
當我運行的應用程序和POST字符串到PHP文件,什麼都不會發生,就像PHP收到一個空值,我無法弄清楚什麼是錯在我的代碼,任何人都知道的過程是這樣的?
應用程序=> ASIFormDataRequest(Post values)=>PHP文件接收請求並在MYSQL數據庫上插入字符串(問題在這裏)。
我已經在網上搜索上和ASI的網站,但似乎沒有任何工作。
您是否正在實施'request'的委託?如果是這樣,也許'requestFailed:'方法可以提供更多的線索。事實上,關於問題出在哪裏的信息太少。 – FluffulousChimp