我正在尝试快速发布请求。我在Objective-C中成功完成,但是使用Swift却无法发送请求。 PHP中的$emailOK
无法获取信息。问题出在哪里?谢谢你的帮手。这是快速代码
func httpPost1(url:String, postData: String, completion: String -> Void) {
let request = NSMutableURLRequest(URL: NSURL(string: url)!)
request.HTTPMethod = "POST"
let postString = postData
request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding)
let task = NSURLSession.sharedSession().dataTaskWithRequest(request) { data, response, error in
guard error == nil && data != nil else { // check for fundamental networking error
print("error=\(error)")
return
}
if let httpStatus = response as? NSHTTPURLResponse where httpStatus.statusCode != 200 { // check for http errors
print("statusCode should be 200, but is \(httpStatus.statusCode)")
print("response = \(response)")
}
let responseString = NSString(data: data!, encoding: NSUTF8StringEncoding)
print("responseString = \(responseString)")
}
task.resume()
}
这是执行代码
httpPost1("xxxxx", postData: "emailOK=Hello") { result in
print(result)//result is your string-response from server
}
这是我的PHP代码
<?php
$link = mysqli_connect($dbhost, $username, $dbpass, $database);
if (!$link) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}
echo "Success: A proper connection to MySQL was made! The my_db database is great." . PHP_EOL;
echo "Host information: " . mysqli_get_host_info($link) . PHP_EOL;
$emailOK = isset($_GET["emailOK"]) ? $_GET["emailOK"] : '';
echo $emailOK;
$query = "INSERT INTO UserInfo VALUES ('', '$emailOK')";
mysqli_query($link, $query) or die (mysqli_error("error"));
mysqli_close($link);
?>
最佳答案
我发现了问题,感谢@OOPer的评论。问题出在PHP $emailOK = isset($_GET["emailOK"]) ? $_GET["emailOK"] : '';
中,应该是$emailOK = $_POST['emailOK'];
关于php - 无法快速发布请求,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/38460624/