问

用于从iOS代码中恢复数据的PHP文件无效

玫瑰编辑1轶事发布于 2023-02-09 19:41

ios

我在从iOS应用程序获取帖子时遇到问题.我认为这是PHP文件.我知道它连接到数据库确定,但它没有将值插入SQL数据库.

这是我的PHP文件:

我的iOS应用程序很好,因为我知道并发送信息正常.但这是我第一次使用PHP.

我在网上显示错误

       mysqli_query INSERT INTO "test" ("", "name", "message",)

但不确定,如何解决.

我的IOS代码如下

-(void) postMessage: (NSString*) message withName: (NSString *) name{

if (name != nil && message != nil){

    NSMutableString *postString = [NSMutableString stringWithString:kPostURL];

    [postString appendString:[NSString stringWithFormat:@"?%@=%@", kName, name]];
    [postString appendString:[NSString stringWithFormat:@"&%@=%@", kMessage, message]];
    [postString appendString:[postString    stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];

    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL  URLWithString:postString]];


    [request setHTTPMethod:@"POST"];

    postConnection = [[NSURLConnection alloc] initWithRequest:request delegate:self];


}


}


-(IBAction)post:(id)sender{

[self postMessage:messageText.text withName:nameText.text];
[messageText resignFirstResponder];
messageText.text = nil;

nameText.text = nil;



}

新的PHP代码

$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
  die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';

$name =$_GET["name"];
$message = $_GET["message"];

mysql_query("INSERT INTO test ('', $name, $message)") or die(mysql_error()); 


mysql_close();

?>

额外信息

该应用程序只是一个视图控制器,带有TextField,TextView和.h文件中的按钮

#define kPostURL @"http://www.dmk-media.com/taxi/tuttest.php"
#define kName @"name"
#define kMessage @"message"




@interface DMKFirstViewController : UIViewController{


IBOutlet UITextField *nameText;
IBOutlet UITextView  *messageText;
NSURLConnection *postConnection;


}

-(IBAction)post:(id)sender;

然后发布消息

-(void) postMessage: (NSString*) message withName: (NSString *) name{

if (name != nil && message != nil){

    NSMutableString *postString = [NSMutableString stringWithString:kPostURL];

    [postString appendString:[NSString stringWithFormat:@"?%@=%@", kName, name]];
    [postString appendString:[NSString stringWithFormat:@"&%@=%@", kMessage, message]];
    [postString appendString:[postString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];

    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL  URLWithString:postString]];


    [request setHTTPMethod:@"POST"];

    postConnection = [[NSURLConnection alloc] initWithRequest:request delegate:self];


}


}

然后是按钮动作

-(IBAction)post:(id)sender{

[self postMessage:messageText.text withName:nameText.text];
[messageText resignFirstResponder];
messageText.text = nil;

nameText.text = nil;



}

多数民众赞成在代码中它只是一个基本的应用程序,尝试测试发布到我的SQL服务器

1 个回答

去掉:

$name =$_GET["name"];
$message = $_GET["message"];

...

mysqli_query INSERT INTO "test" ("", "name", "message",)
mysql_query($query) or die(mysql_error());

它应该是:

$name = $_POST["name"];
$message = $_POST["message"];
mysql_select_db('DATABASE_NAME', $conn) or die(mysql_error());

...

mysql_query("INSERT INTO test(name, mesage) VALUES ('$name', '$message');", $conn) or die(mysql_error());

2023-02-09 19:42 回答

旭峰fd_817

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