[英]How to post data from iOS app to MySQL database?
我看到了與我的問題類似的帖子,但是由於某種奇怪的原因,他的解決方案對我不起作用,這使我的年齡比奧巴馬快。
基本上,我想將數據從iOS應用發布到MySQL數據庫。
iOS代碼
NSString *strURL = [NSString stringWithFormat:@"http://www.example.com/phpfile.php?dishname=%@&description=%@",textfieldTwo.text, textfieldTwo.text];
NSData *data = [NSData dataWithContentsOfURL:[NSURL URLWithString:strURL]];
NSString *strResults = [[NSString alloc]initWithData:data encoding:NSUTF8StringEncoding];
NSLog(@"%@", strResults);
PHP代碼
<?php
$servername="localhost";
$username="admin";
$password="admin";
$dbname="dbname";
$conn=mysqli_connect($servername, $username, $password, $dbname);
if (!$conn)
{
die("Connection failed: " . mysqli_connect_error());
}
$dishname=$_POST['dishname'];
$description=$_POST['description'];
echo "Name : " . $dishname;
//echo "Mail : ". $mail;
$sql="insert into RecipeFeed (DishName, Description) values ('" . $dishname . "', '" . $description . "')";
$result=mysqli_query($conn, $sql);
?>
我不知道為什么我遇到這個問題。 任何幫助將不勝感激,謝謝!
您正在通過GET傳遞參數。 所以你必須改變
$dishname = $_POST['dishname'];
$description = $_POST['description'];
至
$dishname = $_GET['dishname'];
$description = $_GET['description'];
在您的PHP腳本中。
因為mapek已經發布了PHP代碼,所以讓我僅發布iOS部分的答案。 您可以像下面這樣在POST中傳遞參數。
方法:1
NSData* submitData = [[NSString stringWithFormat:@"dishname=%@&description=%@",textfieldOne.text, textfieldTwo.text] dataUsingEncoding:NSUTF8StringEncoding];
NSMutableURLRequest *submitrequest = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:@"http://www.example.com/phpfile.php"]];
NSString *request = [[NSString alloc]initWithData:submitData encoding:NSUTF8StringEncoding];
NSLog(@"request is %@",request);
[submitrequest setHTTPMethod:@"POST"];
[submitrequest setHTTPBody:submitData];
[NSURLConnection sendAsynchronousRequest:submitrequest
queue:[NSOperationQueue mainQueue]
completionHandler:^(NSURLResponse *response, NSData *data, NSError *error)
{
NSString *jsonString = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
NSLog(@"jsonString values=%@",jsonString);
id values = [NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingAllowFragments error:nil];
NSLog(@"json values=%@",values);
}];
方法二
NSMutableDictionary *dictionnary = [NSMutableDictionary dictionary];
[dictionnary setObject:textfieldOne.text forKey:@"dishname"];
[dictionnary setObject:textfieldTwo.text forKey:@"description"];
NSError *error = nil;
NSData *submitData = [NSJSONSerialization dataWithJSONObject:dictionnary
options:kNilOptions
error:&error];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:@"http://www.example.com/phpfile.php"]];
[request setHTTPMethod:@"POST"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[request setValue:@"json" forHTTPHeaderField:@"Data-Type"];
[request setValue:[NSString stringWithFormat:@"%d", [jsonData length]] forHTTPHeaderField:@"Content-Length"];
[request setHTTPBody:jsonData];
[NSURLConnection sendAsynchronousRequest:submitrequest
queue:[NSOperationQueue mainQueue]
completionHandler:^(NSURLResponse *response, NSData *data, NSError *error)
{
NSString *jsonString = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
NSLog(@"jsonString values=%@",jsonString);
id values = [NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingAllowFragments error:nil];
NSLog(@"json values=%@",values);
}];
我知道了,這是下面的代碼。
PHP
<?php
// Create connection
$servername = "localhost";
$username = "admin";
$password = "admin";
$dbname = "db";
$con=mysqli_connect("localhost","admin","admin","db");
if (!$con) {
die("Connection failed: " . mysqli_connect_error());
echo "Nothing happened";
}else{
}
if (isset ($_GET["firstname"]))
$firstname = $_GET["firstname"];
else
$firstname = "Null";
if (isset ($_GET["lastname"]))
$lastname = $_GET["lastname"];
else
$lastname = "Null";
if (isset ($_GET["email"]))
$email = $_GET["email"];
else
$email = "Null";
if (isset ($_GET["password"]))
$password = $_GET["password"];
else
$password = "Null";
if (isset ($_GET["timestamp"]))
$timestamp = $_GET["timestamp"];
else
$timestamp = "Null";
$id = "null";
echo "FirstName : ". $firstname;
echo "LastName : ". $lastname;
$sql = "insert into Users (FirstName, ID, LastName, Email, Password, TimeStamp) values ('".$firstname."', '".$id."', '".$lastname."', '".$email."', '".$password."', '".$timestamp."')";
$result = mysqli_query($con, $sql);
?>
iOS版
NSString *strURL = [NSString stringWithFormat:@"http://www.example.com/register.php?firstname=%@&lastname=%@&password=%@&email=%@&",firstName.text, lastName.text, password.text, email.text];
NSData *dataURL = [NSData dataWithContentsOfURL:[NSURL URLWithString:strURL]];
NSString *strResult = [[NSString alloc] initWithData:dataURL encoding:NSUTF8StringEncoding];
NSLog(@"%@", strResult);
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