我来自JSON
[{ "照片":空}]
我使用这段代码
NSMutableArray *jPhoto = [NSMutableArray arrayWithArray:(NSArray *)[jsonDict valueForKey:@"photo"]];
如果我想使用if(),我怎么检查它？
编辑
这是JSON数据
[{"photo":
[{"image":"http:\/\/www.yohyeh.com\/upload\/shisetsu\/13157\/photo\/1304928459.jpg","title":"test picture","content":"this is description for test picture.\r\n\u8aac\u660e\u6587\u306a\u306e\u306b\u30fb\u30fb\u30fb\u30fb\u30fb\u30fb\u30fb\u30fb\u30fb\u30fb\u30fb\u30fb"}
,{"image":"http:\/\/www.yohyeh.com\/upload\/shisetsu\/13157\/photo\/1304928115.jpg","title":"nothing","content":"iMirai"}
,{"image":"http:\/\/www.yohyeh.com\/upload\/shisetsu\/13157\/photo\/1303276769.jpg","title":"iMirai","content":"Staff"}]}
]
这是我的JSON解析器
NSError *theError = nil;
NSString *URL = [NSString stringWithFormat:@"http://www.yohyeh.com/apps/get_sub_detail.php?id=%@&menu=photo",g_id];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:URL]];
NSURLResponse *theResponse =[[[NSURLResponse alloc]init] autorelease];
NSData *data = [NSURLConnection sendSynchronousRequest:request returningResponse:&theResponse error:&theError];
NSMutableString *string = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
NSDictionary *jsonDict = [string JSONValue];
谢谢你的帮助
我尝试创建闹钟应用程序,但我不知道如何将iTunes中的歌曲设置为本地通知的声音.
现在我使用此代码来调用iTunes
- (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath
{
if (indexPath.section == 0) {
MPMediaPickerController *picker = [[MPMediaPickerController alloc] initWithMediaTypes: MPMediaTypeMusic];
picker.delegate = self;
picker.allowsPickingMultipleItems = NO;
picker.prompt = NSLocalizedString (@"Select any song from the list", @"Prompt to user to choose some songs to play");
//[self presentModalViewController: picker animated: YES];
[self.navigationController pushViewController:picker animated:YES];
NSLog(@"gsudifghukdsf");
[picker release];
}
}
- (void) mediaPicker: (MPMediaPickerController *) mediaPicker didPickMediaItems: (MPMediaItemCollection *) mediaItemCollection
{
[self.navigationController popToRootViewControllerAnimated:YES];
//[self dismissModalViewControllerAnimated: YES];
NSLog(@"%@",mediaItemCollection);
UILocalNotification *local = [[UILocalNotification alloc] init]; …我试图在iPad上隐藏键盘,但我不知道为什么resignFirstResponder不起作用.但popToRoot运作良好.
- (BOOL)textFieldShouldReturn:(UITextField *)textField
{
NSString *desc = [NSString stringWithFormat:@"%@",[descTF text]];
[textField resignFirstResponder];
[self.navigationController popToRootViewControllerAnimated:YES];
return YES;
}
那么你能指导我该怎么办？
我有2个问题
我经常看到这个
$sql->bindParam(':xxx', $yyy, PDO::PARAM_zzz);
$sql->execute();
为什么不使用(1)
$sql->execute(array(':xxx'=>$yyy));
如果有太多的价值我应该使用(2)
$sql->bindParam(':aaa', $bbb, PDO::PARAM_zzz);
$sql->bindParam(':ccc', $ddd, PDO::PARAM_zzz);
$sql->bindParam(':eee', $fff, PDO::PARAM_zzz);
$sql->execute();
要么
$sql->execute(array(':aaa'=>$bbb,':ccc'=>$ddd,':eee'=>$fff));
JSON示例:
[{
"title1":"Hey",
"title2":"Hoh",
"title3":"Hah",
"title4":"Hoo"
}]
如何在(hey,hoh,hah,hoo)不使用的情况下获得价值valueForKey？
有人可以指导我吗？
- - - - - - - 编辑 - - - - - - -
我使用JSON-Framework.现在我使用NSDictionary*jsonDict = [string JSONValue];
我有javascript函数从另一个PHP文件获取JSON但我无法将值传递给html文件.我该怎么做 ？？
请稍等一下
<body>
<script type="text/javascript" src="jquery.min.js"></script>
<script type="text/javascript">
$.getJSON("getquestion.php",function(result)
{
document.getElementById("question").innerHTML = result.test_question;
testform.textquestion.value = result.test_question;
});
</script>
<form name="testform">
<div id="question"></div>
<input id="aaa" type="text" name="textquestion"></input>
</body>
div:问题可以显示来自result.test_question的数据但输入:aaa不能.
能告诉我怎样才能将价值传递给或？我可以指定函数名称.getJSON以及如何？
ios ×4
objective-c ×4
ipad ×2
json ×2
dictionary ×1
html ×1
iphone ×1
itunes ×1
javascript ×1
jquery ×1
mysql ×1
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