用于接受从iPhone提交的表单的PHP脚本
嗨,我正在使用此示例将照片上载到我的服务器 但是,我无法使用$\u POST或$\u请求获取参数1和参数2用于接受从iPhone提交的表单的PHP脚本,php,iphone,objective-c,uiimagepickercontroller,Php,Iphone,Objective C,Uiimagepickercontroller,嗨,我正在使用此示例将照片上载到我的服务器 但是,我无法使用$\u POST或$\u请求获取参数1和参数2 NSString*param1=@“参数文本”; [body appendData:[[NSString stringWithFormat:@”--%@\r\n,边界]数据使用编码:NSUTF8StringEncoding]]; [body appendData:[[NSString stringWithFormat:@“内容处置:表单数据;名称=\“参数1\”\r\n\r\n“]data
NSString*param1=@“参数文本”;
[body appendData:[[NSString stringWithFormat:@”--%@\r\n,边界]数据使用编码:NSUTF8StringEncoding]];
[body appendData:[[NSString stringWithFormat:@“内容处置:表单数据;名称=\“参数1\”\r\n\r\n“]dataUsingEncoding:NSUTF8StringEncoding];
[body appendData:[[NSString stringWithString:param1]dataUsingEncoding:NSUTF8StringEncoding];
[body appendData:[[NSString stringWithString:@“\r\n”]dataUsingEncoding:NSUTF8StringEncoding]代码>
除了$\u POST和$\u REQUEST之外,我还需要使用其他东西来获取参数吗?尝试以下方法:
-(void) httpRequestSender{
NSString *par1 = @"test1";
NSString *par2 = @"test2";
NSString *requestedURL=@"http://something/post.php";
NSString *urlString =[NSString stringWithFormat:@"%@?par1=%@&par2=%@",requestedURL,par1,par2];
NSURL *url = [NSURL URLWithString:urlString];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:url];
[request setHTTPMethod:@"GET"];
NSURLResponse *response = nil;
NSError *error = nil;
NSData *result = [NSURLConnection sendSynchronousRequest:request
returningResponse:&response error:&error];
NSString *resultString = [[NSString alloc] initWithData:result encoding:NSUTF8StringEncoding];
//resultString is the returned buffer from calling the page
[request release];
}