Ios segue不在脚本(xcode)中工作?
嗨,我正在尝试使用php登录程序 这是我的密码:Ios segue不在脚本(xcode)中工作?,ios,xcode,segue,Ios,Xcode,Segue,嗨,我正在尝试使用php登录程序 这是我的密码: - (IBAction)login:(id)sender { if ([username.text isEqualToString:@""] || [password.text isEqualToString:@""]) { UIAlertView *alertsuccess = [[UIAlertView alloc] initWithTitle:@"Error"
- (IBAction)login:(id)sender
{
if ([username.text isEqualToString:@""] || [password.text isEqualToString:@""]) {
UIAlertView *alertsuccess = [[UIAlertView alloc] initWithTitle:@"Error"
message:@"Input Your Value"
delegate:self
cancelButtonTitle:@"OK"
otherButtonTitles:nil, nil];
[alertsuccess show];
}
else {
NSError *err;
NSString *strURL = [NSString stringWithFormat:@"http://192.168.1.5:81/cer_app/cer_login.php?username=%@", username.text];
NSData *dataURL = [NSData dataWithContentsOfURL:[NSURL URLWithString:strURL]];
manage.commonuser = username.text;
NSLog(@"%@", manage.commonuser);
NSString *bb = password.text;
NSDictionary *jsonArray = [NSJSONSerialization JSONObjectWithData:dataURL
options:kNilOptions
error:&err];
NSMutableArray *array1 = [[NSMutableArray alloc] init];
array1 = [jsonArray objectForKey:@"key"];
NSLog(@"%@", array1);
NSString *aa1 = [[array1 objectAtIndex:0]objectForKey:@"password"];
NSString *bb1 = [[array1 objectAtIndex:0]objectForKey:@"temp_password"];
if ([bb1 isEqualToString:@""]) {
if ([bb isEqualToString:aa1]) {
[self performSegueWithIdentifier:@"login" sender:self];
}
else {
UIAlertView *alertsuccess = [[UIAlertView alloc] initWithTitle:@"Error"
message:@"password wrong"
delegate:self
cancelButtonTitle:@"OK"
otherButtonTitles:nil, nil];
[alertsuccess show];
}
}
else {
if ([bb isEqualToString:bb1]) {
[self performSegueWithIdentifier:@"temp_login" sender:self];
}
}
}
}
这里,我从php中获取了arrary1值,并显示在控制台中,但我的PerformsgueWithIdentifier不起作用。是否检查了序列标识符?是的,我检查了。如果else阻塞,也没有错误日志,因为如果[bb IsequalString:bb1]返回false,你什么也不做。我已经做了。但是我的标识符性能不工作。解释不工作。它是被召唤而没有做你期望的事情,还是没有被召唤?您可以使用调试器逐步检查代码并查看哪个是正确的。