问题描述
我正在尝试使用数据库的应用程序（实际上是在我的localhost中），我尝试使用ASIHTTPRequest但是在iOS 5中遇到了很多问题（我在那里学习了如何使用ASIHTTPRequest表单：
I'm trying do to an app which uses a database (actually in my localhost), I tried with ASIHTTPRequest but having so much troubles with iOS 5 (I learnt how to use ASIHTTPRequest form there : http://www.raywenderlich.com/2965/how-to-write-an-ios-app-that-uses-a-web-service
现在我正在尝试使用API由Apple提供：NSURLRequest / NSURLConnection等，...
Now i'm trying with the API provided by Apple : NSURLRequest / NSURLConnection etc,...
我阅读了Apple在线指南并制作了第一个代码：
I read the Apple online guide and make this first code :
- (void)viewDidLoad
{
[super viewDidLoad];
// Do any additional setup after loading the view, typically from a nib.
NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL
URLWithString:@"http://localhost:8888/testNSURL/index.php"]
cachePolicy:NSURLRequestUseProtocolCachePolicy
timeoutInterval:60.0];
[request setValue:@"Hello world !" forKey:@"myVariable"];
NSURLConnection *theConnection=[[NSURLConnection alloc] initWithRequest:request delegate:self];
if (theConnection) {
receiveData = [NSMutableData data];
}
}
我添加了API所需的代表
I added the delegates needed by the API
- (void)connection:(NSURLConnection *)connection didReceiveResponse:(NSURLResponse *)response
- (void)connection:(NSURLConnection *)connection didReceiveData:(NSData *)data
- (void)connection:(NSURLConnection *)connection
- (void)connectionDidFinishLoading:(NSURLConnection *)connection
这是我的php代码：
<?php
if(isset($_REQUEST["myVariable"])) {
echo $_REQUEST["myVariable"];
}
else echo '$_REQUEST["myVariable"] not found';
?>
那有什么问题？当我启动应用程序时，它会立即崩溃并显示此输出：
So what is wrong? When I start the app, it will immediately crash with this output :
**
**> 2012-04-09 22:52:16.630 NSURLconnextion[819:f803] *** Terminating app
> due to uncaught exception 'NSUnknownKeyException', reason:
> '[<NSURLRequest 0x6b32bd0> setValue:forUndefinedKey:]: this class is
> not key value coding-compliant for the key myVariable.'
> *** First throw call stack: (0x13c8022 0x1559cd6 0x13c7ee1 0x9c0022 0x931f6b 0x931edb 0x2d20 0xd9a1e 0x38401 0x38670 0x38836 0x3f72a
> 0x10596 0x11274 0x20183 0x20c38 0x14634 0x12b2ef5 0x139c195 0x1300ff2
> 0x12ff8da 0x12fed84 0x12fec9b 0x10c65 0x12626 0x29dd 0x2945) terminate
> called throwing an exception**
**
我想，这意味着这一行有些不对劲：
I guess, it means that somethings is wrong with this line :
[request setValue:@"Hello world !" forKey:@"myVariable"];
如果我评论这一行，它实际上有效。
It actually works if I comment this line.
我的问题是：如何使用NSURLRequest和NSURLConnexion将数据发送到PHP API？
My question is : How can I send datas to a PHP API, using NSURLRequest and NSURLConnexion?
感谢您的帮助。
PS顺便说一句，我对服务器，PHP等知识很差......
P.S. By the way, I have poor knowledges about server, PHP ect,...
推荐答案
试试这个：
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL
URLWithString:@"http://localhost:8888/testNSURL/index.php"]
cachePolicy:NSURLRequestUseProtocolCachePolicy
timeoutInterval:60.0];
[request setHTTPMethod:@"POST"];
NSString *postString = @"myVariable=Hello world !";
[request setHTTPBody:[postString dataUsingEncoding:NSUTF8StringEncoding]];
NSURLConnection *theConnection=[[NSURLConnection alloc] initWithRequest:request delegate:self];
if (theConnection) {
receiveData = [NSMutableData data];
}
此处显示
这篇关于请求NSURLRequest的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！