我对 Objective-C 相当陌生,并且希望使用 POST 将许多键值对传递给 PHP 脚本。我正在使用以下代码,但数据似乎没有被发布。我也尝试使用 NSData 发送内容,但似乎都不起作用。
NSDictionary* data = [NSDictionary dictionaryWithObjectsAndKeys:
@"bob", @"sender",
@"aaron", @"rcpt",
@"hi there", @"message",
nil];
NSURL *url = [NSURL URLWithString:@"http://myserver.com/script.php"];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];
[request setHTTPMethod:@"POST"];
[request setHTTPBody:[NSData dataWithBytes:data length:[data count]]];
NSURLResponse *response;
NSError *err;
NSData *responseData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&err];
NSLog(@"responseData: %@", content);
这将被发送到这个简单的脚本来执行数据库插入:
<?php $sender = $_POST['sender'];
$rcpt = $_POST['rcpt'];
$message = $_POST['message'];
//script variables
include ("vars.php");
$con = mysql_connect($host, $user, $pass);
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("mydb", $con);
mysql_query("INSERT INTO php_test (SENDER, RCPT, MESSAGE)
VALUES ($sender, $rcpt, $message)");
echo "complete"
?>
有任何想法吗?
谢谢大家的建议。最后我设法通过使用给定的东西解决了这个问题here https://stackoverflow.com/questions/330060/problem-using-nsurlrequest-to-post-data-to-server.
Code:
NSString *myRequestString = @"sender=my%20sender&rcpt=my%20rcpt&message=hello";
NSData *myRequestData = [ NSData dataWithBytes: [ myRequestString UTF8String ] length: [ myRequestString length ] ];
NSMutableURLRequest *request = [ [ NSMutableURLRequest alloc ] initWithURL: [ NSURL URLWithString: @"http://people.bath.ac.uk/trs22/insert.php" ] ];
[ request setHTTPMethod: @"POST" ];
[ request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"content-type"];
[ request setHTTPBody: myRequestData ];
NSURLResponse *response;
NSError *err;
NSData *returnData = [ NSURLConnection sendSynchronousRequest: request returningResponse:&response error:&err];
NSString *content = [NSString stringWithUTF8String:[returnData bytes]];
NSLog(@"responseData: %@", content);
本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系:hwhale#tublm.com(使用前将#替换为@)