Question

我在问题中看到了一个类似的帖子，但他的解决方案因为一些奇怪的原因对我不起作用，这让我的年龄比奥巴马快。

基本上我想将数据从iOS应用程序发布到MySQL数据库。

iOS代码

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代码

<?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);


?>

数据库图片

我不知道为什么我遇到这个问题。任何帮助将不胜感激，谢谢！

Answer 1

您正在通过GET传递参数。所以你必须改变

$dishname = $_POST['dishname'];
$description = $_POST['description'];

到

$dishname = $_GET['dishname'];
$description = $_GET['description'];

在PHP脚本中。

Answer 2

由于mapek已经发布了PHP代码，让我发布仅限iOS部分的答案。您可以在POST中传递参数，如下所示。

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2

 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);
}];

Answer 3

我想出来了，这是下面的代码。

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);
?>

的iOS

  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);

如何将数据从iOS应用程序发布到MySQL数据库？

3 个答案: