小编adi*_*dit的帖子

我有以下代码:

        $rsm = new ResultSetMapping();
        $rsm->addEntityResult('App\MainBundle\Entity\InstagramShopPicture', 'p');
        $rsm->addFieldResult('p', 'id', 'id');
        $rsm->addFieldResult('p','lowresimageurl','lowresimageurl');
        $rsm->addFieldResult('p','medresimageurl','medresimageurl');
        $rsm->addFieldResult('p','highresimageurl','highresimageurl');
        $rsm->addFieldResult('p','caption','caption');
        $rsm->addFieldResult('p','numberoflikes','numberoflikes');
        $rsm->addFieldResult('p','numberofdislikes','numberofdislikes');
        $rsm->addJoinedEntityResult('App\MainBundle\Entity\InstagramShop', 's', 'p', 'shop');
        $rsm->addFieldResult('s', 'id', 'id');
        $rsm->addFieldResult('s', 'username', 'username');

        $query = $em->createNativeQuery('SELECT picture.id, picture.lowresimageurl, picture.medresimageurl, picture.highresimageurl, picture.caption, picture.numberoflikes, picture.numberofdislikes, shop.id AS shop_id , shop.username
                                        FROM App_instagram_picture_category category
                                        INNER JOIN App_instagram_shop_picture picture ON category.picture_id = picture.id
                                        INNER JOIN App_instagram_shop shop ON shop.id = picture.shop_id
                                        WHERE category.first_level_category_id = ?
                                        AND picture.deletedAt IS NULL
                                        AND shop.deletedAt IS NULL
                                        AND shop.isLocked = 0
                                        AND shop.expirydate IS NOT …

我有一个名为post的表,其中有一个名为title的列.我想选择所有只有1个单词作为标题的帖子.所以例如'cat','dog'.有些帖子有多个,比如"猫和狗都不好",这个我不想选择.仅发布一个单词标题.我怎么能用mysql这样做？

我在一个.php文件中使用以下内容存储了一个会话:

session_start();
    $_SESSION['uid'] = $_POST['uid']; 

当我导航到其他.php文件并尝试通过执行以下操作来访问该值:

$_SESSION['uid']

我收到一个错误:

Notice: Undefined variable: _SESSION in C:\wamp\www\saved.php on line 6

第6行是:

$result->execute(array($_SESSION['uid']));

为什么是这样？

一个非常基本的问题,当我有类似的东西:

 TTStyledText * text = [TTStyledText textFromXHTML:message.message lineBreaks:YES URLs:NO];
 text.width = self.frame.size.width - 60;
 text.font = [UIFont fontWithName:@"ArialMT" size:17.0];
 _main_title.text = text;

当我分配text时_main_title.text,是否意味着_main_title.text保留text？

我有时会听到人们不再使用表来格式化他们的形式,这是否已经成为一种被抛弃的趋势？为什么不呢？

所以我用它NSUserDefaults来存储我的FBAccessTokenKey和FBExpirationDateKey.我正在创建一个单例用户对象:

- (id)init
{
    self = [super init];
    if (self != nil) {
         facebook = [[Facebook alloc] initWithAppId:kAppId andDelegate:self];

         NSUserDefaults *defaults = [NSUserDefaults standardUserDefaults];
         if ([defaults objectForKey:@"FBAccessTokenKey"] 
            && [defaults objectForKey:@"FBExpirationDateKey"] ) {
            NSLog(@"ACCESS KEY IS NOT EMPTY");
            facebook.accessToken = [defaults objectForKey:@"FBAccessTokenKey"];
            facebook.expirationDate = [defaults objectForKey:@"FBExpirationDateKey"];
         } else {
              NSLog(@"ACCESS KEY IS EMPTY");
         }


    }
    return self;
}

我还实现了一个在我注销时调用的didLogout方法:

- (void)fbDidLogout {
    [[NSUserDefaults standardUserDefaults] removeObjectForKey:@"FBAccessTokenKey"];
    [[NSUserDefaults standardUserDefaults] removeObjectForKey:@"FBExpirationDateKey"];
}

这基本上清除/刷新了令牌密钥.现在真正的问题是,当我点击退出,退出应用程序,然后再次运行应用程序时,它会检测到该密钥FBAccessTokenKey并且FBExpirationDateKey仍然存在.为什么是这样？

检查委托方法中哪个请求的最佳方法是什么:

- (void)connection:(NSURLConnection *)connection didReceiveResponse:(NSURLResponse *)response
{

}

现在我有一个NSURLConnection,我在发出请求之前设置为NSURLConnection,我在里面didReceiveResponse做:

if (self.tempConnection == connection)

但是有可能这对竞争条件不起作用.有一个更好的方法吗？

所以我有以下代码:

-

 (void) drawLinearGradient:(CGRect) rect
{
    CGContextRef context = UIGraphicsGetCurrentContext();
    CGColorSpaceRef colorSpace = CGColorSpaceCreateDeviceRGB();
    CGFloat locations[] = { 0.0, 0.3, 1.0 };

    CGColorRef grayColor = [UIColor colorWithRed:37/255.f green:37/255.f 
                                            blue:37/255.f alpha:1.0].CGColor; 
    CGColorRef blueColor = [UIColor colorWithRed:23.0/255.0 green:171.0/255.0 
                                            blue:219.0/255.0 alpha:1.0].CGColor;

    NSArray *colors = [NSArray arrayWithObjects: (id) grayColor, (id) grayColor, (id) blueColor, nil];

    CGGradientRef gradient = CGGradientCreateWithColors(colorSpace, 
                                                        (CFArrayRef) colors, locations);

    CGPoint startPoint = CGPointMake(CGRectGetMidX(rect), CGRectGetMinY(rect));
    CGPoint endPoint = CGPointMake(CGRectGetMidX(rect), CGRectGetMaxY(rect));

    CGContextSaveGState(context);
    CGContextAddRect(context, rect);
    CGContextClip(context);
    CGContextDrawLinearGradient(context, gradient, startPoint, endPoint, 0);
    CGContextRestoreGState(context);

    CGGradientRelease(gradient);
    CGColorSpaceRelease(colorSpace);
} …

我在我的xcode项目中做了一些更改,这在某种程度上导致我无法在模拟器和设备中运行应用程序.我现在唯一能看到的选项是My Mac 64位.我该如何解决？目标是正确的.

我有以下NSPredicate:

 fetchRequest.predicate = [NSPredicate predicateWithFormat:@"url = %@ AND type = %@ OR type = %@", newsStory.url, [NSNumber numberWithInt:StoryHighlighted], [NSNumber numberWithInt:StorySavedAndHighlighted]];

我想用这个特定的URL和类型(突出显示或突出显示并保存)来获取故事.但是上面的查询似乎不起作用.我究竟做错了什么？