小编Res*_*lov的帖子

我想玩UIPopupController,我明白我无法检测到我的popover解雇.我的步骤:
1.从XCode创建示例(文件 - >新建项目 - >实用应用程序)
2.添加到MainViewController.hUIPopoverControllerDelegate

#import "FlipsideViewController.h"

@interface MainViewController : UIViewController <FlipsideViewControllerDelegate,UIPopoverControllerDelegate>

@property (strong, nonatomic) UIPopoverController *flipsidePopoverController;

- (IBAction)showInfo:(id)sender;

@end

1. 在MainViewController中:

    - (IBAction)showInfo:(id)sender
    {
        if ([[UIDevice currentDevice] userInterfaceIdiom] == UIUserInterfaceIdiomPhone) {
            FlipsideViewController *controller = [[FlipsideViewController alloc] initWithNibName:@"FlipsideViewController" bundle:nil];
            controller.delegate = self;
            controller.modalTransitionStyle = UIModalTransitionStyleFlipHorizontal;
            [self presentModalViewController:controller animated:YES];
        } else {
            if (!self.flipsidePopoverController) {
                FlipsideViewController *controller = [[FlipsideViewController alloc] initWithNibName:@"FlipsideViewController" bundle:nil];
                controller.delegate = self;

                self.flipsidePopoverController.delegate = self

                self.flipsidePopoverController = [[UIPopoverController alloc] initWithContentViewController:controller];
            }
            if ([self.flipsidePopoverController isPopoverVisible]) {
                [self.flipsidePopoverController …

我需要抓住特殊区域的滑动动作.但是调试器说unrecognized selector sent to instance

- (void)viewDidLoad
{
    UISwipeGestureRecognizer *swipeLeft = [[UISwipeGestureRecognizer alloc]
          initWithTarget:self.viewName action:@selector(didSwipe:)];
    swipeLeft.direction = UISwipeGestureRecognizerDirectionLeft;
    [self.view addGestureRecognizer:swipeLeft];
}

-(void)didSwipe:(UISwipeGestureRecognizer*)swipe{
   NSLog(@"swiped left");
}

怎么了？

是的,我知道,这是常见的问题,但我在最近3个小时内无法解决.

我这样查询:

$query = 'UPDATE `'.$masterapp_users_table.'` SET `login` = "'.htmlspecialchars(strip_tags($_POST['login'])).'", `pass` = "'.md5(htmlspecialchars(strip_tags($_POST['pass']))).'", `email` = "'.htmlspecialchars(strip_tags($_POST['email'])).'", `firstname` = "'.htmlspecialchars(strip_tags($_POST['fn'])).'", `secondname` = "'.htmlspecialchars(strip_tags($_POST['sn'])).'" WHERE `login` = "'.htmlspecialchars(strip_tags($_POST['previouslogin'])).'"';
            echo $query.'<br />';
            mysql_query($query) or die(mysql_error());

我明白了

UPDATE `masterapp_users` SET `login` = "asdasdasd", `pass` = "a3dcb4d229de6fde0db5686dee47145d", `email` = "asdasdasd", `firstname`
= "asdasdasd", `secondname` = "asdasdasd" WHERE `login` = "88888881"<br />Unknown column 'login' in 'where clause'

但它改变了记录!也许有人能看到我看不到的东西？

哦! 忘了说:如果我将该字符串从浏览器粘贴到PMA,它可以正常工作.

更新:

CREATE TABLE IF NOT EXISTS `masterapp_users` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `login` varchar(64) COLLATE …

在我的脚本中，我使用 cURL 发送数据，并启用 CURLOPT_RETURNTRANSFER。响应是 json 编码的数据。当我尝试 json_decode 时，它​​返回 null。然后我发现响应在字符串的开头包含 utf-8 BOM 符号 ()。

有一些实验：

$data = $data = curl_exec($ch);
echo $data;

结果是 {"field_1":"text_1","field_2":"text_2","field_3":"text_3"}

$data = $data = curl_exec($ch);
echo mb_detect_encoding($data);

结果 - UTF-8

$data = $data = curl_exec($ch);
echo mb_convert_encoding($data, 'UTF-8', mb_detect_encoding($data));
// identical to echo mb_convert_encoding($data, 'UTF-8', 'UTF-8');

结果 - {"field_1":"text_1","field_2":"text_2","field_3":"text_3"}

有帮助的一件事是删除前 3 个符号：

if (substr($data, 0, 3) == pack('CCC', 239, 187, 191)) {
    $data = substr($data, 3);
}

但是，如果还有另一个 BOM 呢？所以问题是：如何检测 cURL 响应的正确编码？或者如何检测 BOM 已到？或者也许如何使用 BOM 转换响应？