我有一个UITableViewController，在我的UITableViewCell中我有一个imageView当用户点击它时，我使用我自己的自定义UIViewControllerContextTransitioning以模拟方式对其进行缩放效果，它就像一个魅力的问题是当动画完成，我打电话给
[transitionContext completeTransition:YES];
现在我的Modal UIViewController背后的UITableView被从视图层次结构中移除，所以我后面有一个黑屏
[containerView addSubview:toViewController.view];
[UIView animateWithDuration:duration delay:0.0 usingSpringWithDamping:damping initialSpringVelocity:1.0 / damping options:0 animations:^{
// blablabla animation stuff
} completion:^(BOOL finished) {
[transitionContext completeTransition:YES];// this line makes my UITableView removed form the view hierarchy
}];
这是我到目前为止所尝试的内容：
在完成块中添加UITableView
[[UIApplication sharedApplication ].keyWindow insertSubview:fromViewController.view belowSubview:containerView];
但这仅适用于iOS8 ......
任何想法如何解决这个问题？谢谢!!
答案 0 :(得分：0)
尝试将展示视图控制器的UIModalPresentationCustom设置为<?php
$a = $_POST['findSchool'];
//1. Enter database credentials
$servername = "****";
$username = "****";
$password = "****";
$dbname = "****";
$flag = 0;
// 2. Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
echo '<form id="myNewForm" name="myNewForm" class="form" method="post" action="">';
echo '<table class="table table-bordered">';
echo '<tr scope="col" >
<th align="center">SchoolName</th>
<th align="center">Details</th>
</tr>';
$sql = "SELECT schName FROM schoolinformation WHERE schName LIKE '%$a%'";
echo $sql;
$i = 0;
$result = mysqli_query($conn, $sql);
while($rowval = mysqli_fetch_array($result))
{
$schName= $rowval['schName'];
echo '<tr scope="col"><td align="center">'.$rowval['schName'].'</td>
<td align="center"><input type="submit" id="shwData" name="shwData" class="btn" value="Show"/></td>';
}
echo '</table>';
echo '</form>';
if($schName == '')
{
echo "Entry not found";
$flag = 1;
}
else
{
//echo $schName;
$flag = 0;
}
?>
<?php
if (isset($_POST['shwData']))
{
echo "this is it";
}
?>