我有一个自定义单元格的表格视图和每个单元格中的一些按钮.单击单元格内的任何按钮将显示该单元格下方的另一个自定义视图.下一步单击同一按钮将折叠视图,并且所有单元格都需要相同我尝试使用insertrow方法单击按钮但是徒劳无功.如何仅使用表视图委托来完成此操作.
这是我试过的:
- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{
static NSString *simpleTableIdentifier = @"CustomCell_For_Dashboard";
CustomCellFor_Dashboard *customCell = (CustomCellFor_Dashboard *)[tableView dequeueReusableCellWithIdentifier:simpleTableIdentifier];
if (customCell == nil)
{
NSArray *nib = [[NSBundle mainBundle] loadNibNamed:@"CustomCellFor_Dashboard" owner:self options:nil];
customCell = [nib objectAtIndex:0];
}
[customCell.howyoulfeelBtn addTarget:self action:@selector(buttonclicked:) forControlEvents:UIControlEventTouchUpInside];
customCell.nameLabel.text = @"test";
customCell.imgView.image = [UIImage imageNamed:@"Default.png"];
// customCell.prepTimeLabel.text = [prepTime objectAtIndex:indexPath.row];
return customCell;
}
-(void)buttonclicked:(id)sender{
NSIndexPath *indexPath = [myTable indexPathForCell:sender];
[myTable beginUpdates];
NSIndexPath *insertPath = [NSIndexPath indexPathForRow:indexPath.row inSection:indexPath.section];
[myTable insertRowsAtIndexPaths:[NSArray arrayWithObject:insertPath] withRowAnimation:UITableViewRowAnimationTop];
}
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谁能帮我?
我需要以数组格式发送NSArray到服务器JSON.我怎样才能将其转换为JSON.这是NSArray我必须通过的一个样本.
array([0] => array('latitude'=>'10.010490',
'longitude'=>'76.360779',
'altitude'=>'30.833334',
'timestamp'=>'11:17:23',
'speed'=>'0.00',
'distance'=>'0.00');
[1] => array('latitude'=>'10.010688',
'longitude'=>'76.361378',
'altitude'=>'28.546305',
'timestamp'=>'11:19:26',
'speed'=>'1.614',
'distance'=>'198.525711')
)`
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并且所需的格式是这样的
[
{ "latitude":"10.010490",
"longitude":"76.360779",
"altitude":"30.833334",
"timestamp":"11:17:23",
"speed":"0.00",
"distance":"0.00"
},
{
"latitude":"10.010688",
"longitude":"76.361378",
"altitude":"28.546305",
"timestamp":"11:19:26",
"speed":"1.614",
"distance":"198.525711"
}
]
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有人有解决方案吗?提前致谢.
我需要从youtube等视频网站下载并保存视频.我想将它们保存到iPhone的视频库中.如何检测视频并将其保存到库中?我已经查看了一些可用的源代码.这是我在下载操作中所做的,但是无法正常工作.
- (IBAction)download {
[downloadButton setEnabled:NO];
[webView setUserInteractionEnabled:NO];
UIUserInterfaceIdiom userInterfaceIdiom = [UIDevice currentDevice].userInterfaceIdiom;
NSString *getURL = @"";
if (userInterfaceIdiom == UIUserInterfaceIdiomPhone) {
getURL = [webView stringByEvaluatingJavaScriptFromString:@"function getURL() {var player = document.getElementById('player'); var video = player.getElementsByTagName('video')[0]; return video.getAttribute('src');} getURL();"];
} else {
getURL = [webView stringByEvaluatingJavaScriptFromString:@"function getURL() {var bh = document.getElementsByClassName('bh'); if (bh.length) {return bh[0].getAttribute('src');} else {var zq = document.getElementsByClassName('zq')[0]; return zq.getAttribute('src');}} getURL();"];
}
NSString *getTitle = [webView stringByEvaluatingJavaScriptFromString:@"function getTitle() {var jm = document.getElementsByClassName('jm'); if (jm.length) {return jm[0].innerHTML;} else {var lp …Run Code Online (Sandbox Code Playgroud) 我在opengl中为我的一个应用程序创建了一个3D对象.物体就像人体一样,可以在触摸时旋转.如何检测此3D对象上的触摸位置.意味着如果用户触摸头部,我必须检测到它是头部.如果手上有触摸,则必须识别.即使将对象旋转到其他方向,它也应该工作.我认为需要3D对象上的触摸坐标.
这是我在视图上获得触摸位置的方法.
- (void) touchesBegan: (NSSet*) touches withEvent: (UIEvent*) event
{
UITouch* touch = [touches anyObject];
CGPoint location = [touch locationInView: self];
m_applicationEngine->OnFingerDown(ivec2(location.x, location.y));
}
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有人可以帮忙吗?提前致谢!
在我的应用程序中,我必须使用NFD Unicode系统规范化NSString.我怎样才能做到这一点?我需要转换用户输入的字符串作为规范化的字符串,我必须计算一个数字.谁能帮我?
我建立一个应用程序,使一个IPA文件,但不能ceated.我有版本的Xcode 3.2.6遵循必要使IPA file.But没有希望的步骤.我存档了应用程序并在组织器中打开然后分发它,但它没有显示任何ipa文件正在创建的警报.
iphone ×5
objective-c ×4
3d ×1
cocoa ×1
cocoa-touch ×1
download ×1
ios ×1
json ×1
nsarray ×1
nsstring ×1
opengl-es ×1
uitableview ×1
unicode ×1
xcode ×1
youtube ×1