10

我正在尝试使用数据库（实际上在我的本地主机中）的应用程序，我尝试使用 ASIHTTPRequest 但在 iOS 5 上遇到了很多麻烦（我在那里学习了如何使用 ASIHTTPRequest 表单：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 Apple 提供的 API：NSURLRequest / NSURLConnection 等，...

我阅读了 Apple 在线指南并编写了第一个代码：

- (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 所需的委托

- (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';
?>

那么有什么问题呢？当我启动应用程序时，它会立即崩溃并显示以下输出：

**

**> 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，这意味着这条线有问题：

[request setValue:@"Hello world !" forKey:@"myVariable"];

如果我评论这一行，它实际上是有效的。

我的问题是：如何使用 NSURLRequest 和 NSURLConnexion 将数据发送到 PHP API？

感谢您的帮助。

PS 顺便说一句，我对服务器、PHP 等知之甚少……

4

2 回答 2

15

尝试这个：

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];   

}

在这里看到https://stackoverflow.com/a/6149088/1317080

于 2012-04-09T21:42:05.863 回答
5

试试下面的代码，它是使用 Web 服务（json）的简单方法之一

NSURL *url = [NSURL URLWithString:@"yourURL"];

NSMutableURLRequest *urlReq=[NSMutableURLRequest requestWithURL:url];

NSURLResponse *response;

NSError *error = nil;

 NSData *receivedData = [NSURLConnection sendSynchronousRequest:urlReq
                                              returningResponse:&response
                                                         error:&error];
if(error!=nil)
{
   NSLog(@"web service error:%@",error);
}
else
{
 if(receivedData !=nil)
 {
    NSError *Jerror = nil;

    NSDictionary* json =[NSJSONSerialization
                         JSONObjectWithData:receivedData
                         options:kNilOptions
                         error:&Jerror];

   if(Jerror!=nil)
   {
    NSLog(@"json error:%@",Jerror);
   }
 }
}

希望这可以帮助。

于 2013-07-29T09:51:51.377 回答