2012-04-28 63 views
1

我遇到了以下问题: 我想要将我的iPhone应用程序连接到服务器上的数据库。因此我使用一些(简单).php文件来管理对数据库的访问。插入新的数据已经工作,但我有一些麻烦,提取的数据转换成的NSMutableArray:NSJSONSerialization - 如何正确地将JSON转换为NSArray?

NSURL *contentURL = [NSURL URLWithString:[kHOSTURL stringByAppendingString:kGETBarsURL]]; 
NSLog(@"URL : %@", contentURL); 

NSData *contentData = [NSData dataWithContentsOfURL:contentURL]; 
NSLog(@"Data : %@", contentData); 

NSError *e = nil; 
NSMutableArray *jsonArray = [NSJSONSerialization JSONObjectWithData:contentData 
                  options:kNilOptions 
                   error:&e]; 
NSLog(@"JSON : %@", jsonArray); 
NSLog(@"Error : %@", e); 

输出是这样的(我“XX”缩短“数据:”):

2012-04-28 13:49:37.229 XX[14434:f803] URL : http://xx/getBars.php 
2012-04-28 13:49:37.389 XX[14434:f803] Data : <5b7b2275 6e697175 65223a22 34222c22 4e616d65 223a2254 65737422 2c224465 7461696c 73223a22 54686973 49734154 65737422 7d2c7b22 ...> 
2012-04-28 13:49:37.390 XX[14434:f803] JSON : (null) 
2012-04-28 13:49:37.392 XX[14434:f803] Error : Error Domain=NSCocoaErrorDomain Code=3840 "The operation couldn’t be completed. (Cocoa error 3840.)" (Garbage at end.) UserInfo=0x6daa610 {NSDebugDescription=Garbage at end.} 

如果我在浏览器中打开网页,它看起来像这样:

[{"unique":"4","Name":"Test","Details":"ThisIsATest"}, 
{"unique":"5","Name":"Test","Details":"ThisIsATest"}, 
{"unique":"6","Name":"Test","Details":"ThisIsATest"}, 
{"unique":"7","Name":"Test","Details":"ThisIsATest"}, 
{"unique":"8","Name":"Test","Details":"ThisIsATest"}, 
{"unique":"9","Name":"Test","Details":"ThisIsATest"}, 
{"unique":"10","Name":"Test","Details":"ThisIsATest"}] 

我也尝试过其他选项NSJSONSerialization但没有工作:(任何人可以帮助我在这里

2012-04-28 14:18:30.192 XX[14541:f803] Encoding : [{"unique":"4","Name":"Test","Details":"ThisIsATest"},{"unique":"5","Name":"Test","Details":"ThisIsATest"},{"unique":"6","Name":"Test","Details":"ThisIsATest"},{"unique":"7","Name":"Test","Details":"ThisIsATest"},{"unique":"8","Name":"Test","Details":"ThisIsATest"},{"unique":"9","Name":"Test","Details":"ThisIsATest"},{"unique":"10","Name":"Test","Details":"ThisIsATest"}] 
<script type="text/javascript"> 

    var _gaq = _gaq || []; 
    _gaq.push(['_setAccount', 'UA-16106315-6']); 
    _gaq.push(['_setDomainName', '.xx.de']); 
    _gaq.push(['_trackPageview']); 

    (function() { 
    var ga = document.createElement('script'); ga.type = 'text/javascript'; 
ga.async = true; 
    ga.src = ('https:' == document.location.protocol ? 'https://ssl' : 
'http://www') + '.google-analytics.com/ga.js'; 
    var s = document.getElementsByTagName('script')[0]; 
s.parentNode.insertBefore(ga, s); 
    })(); 

</script> 
+1

错误说最后有垃圾数据。你检查过了吗? – 2012-04-28 12:06:33

+0

我该如何检查? :)正如我所提到的,我打开了一个网页,上面显示的内容,但似乎好吧不是吗? – pasql 2012-04-28 12:13:09

+1

尝试将您的数据转换为NSString,打印并查看它是否为有效的JSON。 'NSLog(@“%@”,[[NSString alloc] initWithData:contentData encoding:NSUTF8StringEncoding]);' – 2012-04-28 12:15:06

回答

2

很明显,你确实在最后有'垃圾'。你有一个JavaScript块,虽然在浏览器中不可见,但它仍然从你的php脚本返回。删除,你应该很好去。

+0

非常感谢:)现在工作! – pasql 2012-04-28 18:05:48

0

我最近遇到了同样的问题,经过将近一个小时后,我发现这是我发送请求的URL的问题。检查URL以查看它是否实际上响应了JSON数据。祝你好运!;