小编Min*_*er3的帖子

是否可以将图像添加到表格视图中？在左边？如果是这样,那应该是多大？

我试图在屏幕上移动子视图,但我也希望为对象添加惯性或动量.
我已经拥有的UIPanGestureRecognizer代码如下.

提前致谢.

UIPanGestureRecognizer *panGesture = [[UIPanGestureRecognizer alloc] initWithTarget:self       action:@selector(handlePan:)];
[self addGestureRecognizer:panGesture];

(void)handlePan:(UIPanGestureRecognizer *)recognizer
{

    CGPoint translation = [recognizer translationInView:self.superview];
    recognizer.view.center = CGPointMake(recognizer.view.center.x + translation.x,
                                     recognizer.view.center.y + translation.y);
    [recognizer setTranslation:CGPointMake(0, 0) inView:self.superview];

    if (recognizer.state == UIGestureRecognizerStateEnded) {
        [self.delegate card:self.tag movedTo:self.frame.origin];
    }
}

再次谢谢.

我想得到json与PHP编码功能如下

<?php 
  require "../classes/database.php";

  $database = new database();
  header("content-type: application/json");
  $result = $database->get_by_name($_POST['q']);   //$_POST['searchValue']

  echo '{"results":[';
  if($result)
  {
     $i = 1;
     while($row = mysql_fetch_array($result))
     {
        if(count($row) > 1) 
        {
           echo json_encode(array('id'=>$i, 'name' => $row['name']));
           echo ",";
        }
        else 
        {
           echo json_encode(array('id'=>$i, 'name' => $row['name']));
        }
        $i++; 
     }
  }
  else
  {
     $value = "FALSE";
     echo json_encode(array('id'=>1, 'name' => ""));  // output the json code
  }

  echo "]}";

我希望输出json是这样的

{"results":[{"id":1,"name":"name1"},{"id":2,"name":"name2"}]}

但输出json如下所示

{"results":[{"id":1,"name":"name1"},{"id":2,"name":"name2"},]}

当你意识到最后有逗号时,我想删除它所以它可以是正确的json语法,如果我删除了echo ",";当有多个结果时json会生成这样的{"results":[{"id":1,"name":"name1"}{"id":2,"name":"name2"}]}并且语法也是错误的

希望每个人都明白我的意思,任何想法都会受到赞赏

我正在尝试开发一个包含大量单词数组的应用程序.我想从中创建一个新的过滤数组.过滤是基于我使用正则表达式创建的模式完成的.例如,我应该能够用我的数组中的模式"ap_l_"过滤出单词"apple".谁能帮我吗？