如何將圖像插入到mysql數據庫中

Question

我想用mysql數據庫中的其他細節插入圖像。我能夠保存數據值，但圖像沒有被插入。請檢查我的代碼如下。

(IBAction) addData:(id)sender; { 

    //Image upload 
    NSData *imageData = UIImageJPEGRepresentation(imageView.image, 90); 
    NSString *urlString = @"http://localhost/myservice.php?"; 

    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init]; 
    [request setURL:[NSURL URLWithString:urlString]]; 
    [request setHTTPMethod:@"POST"]; 

    NSString *boundary = @"---------------------------14737809831466499882746641449"; 
    NSString *contentType = [NSString stringWithFormat:@"multipart/form-data; boundary=%@",boundary]; 
    [request addValue:contentType forHTTPHeaderField: @"Content-Type"]; 

    NSMutableData *body = [NSMutableData data]; 
    [body appendData:[[NSString stringWithFormat:@"\r\n--%@\r\n",boundary] dataUsingEncoding:NSUTF8StringEncoding]]; 
    [body appendData:[@"Content-Disposition: form-data; name=\"userfile\"; filename=\"ipodfile.jpg\"\r\n" dataUsingEncoding:NSUTF8StringEncoding]]; 
    [body appendData:[@"Content-Type: application/octet-stream\r\n\r\n" dataUsingEncoding:NSUTF8StringEncoding]]; 
    [body appendData:[NSData dataWithData:imageData]]; 
    [body appendData:[[NSString stringWithFormat:@"\r\n--%@--\r\n",boundary] dataUsingEncoding:NSUTF8StringEncoding]]; 
    [request setHTTPBody:body]; 

    NSData *returnData = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil]; 
    NSString *returnString = [[NSString alloc] initWithData:returnData encoding:NSUTF8StringEncoding]; 
    NSLog(@"Image upload string%@",returnString); 


    //Upload data 
    NSString *url = [NSString stringWithFormat:[urlString stringByAppendingString:@"userName=%@&age=%@&phone=%@&image=%@"],txtName.text, txtAge.text, txtPhone.text,returnString]; 
    NSData *dataUrl = [NSData dataWithContentsOfURL:[NSURL URLWithString:url]]; 

    NSString *strResult = [[NSString alloc] initWithData:dataUrl encoding:NSUTF8StringEncoding]; 
    NSLog(@"%@",strResult); 


    if (![strResult isEqual: @""]) { 
     UIAlertView *alert = [[UIAlertView alloc] 
           initWithTitle: @"Record Saved" 
           message: @"Your Record is saved successfully" 
           delegate: nil 
           cancelButtonTitle:@"OK" 
           otherButtonTitles:nil]; 
     [alert show]; 


     //Clear fields 
     txtName.text=@""; 
     txtAge.text=@""; 
     txtPhone.text=@""; 
     imageView.image= nil; 

    }else { 

     UIAlertView *alert = [[UIAlertView alloc] 
           initWithTitle: @"Record not Saved" 
           message: @"Your Record does not saved successfully" 
           delegate: nil 
           cancelButtonTitle:@"OK" 
           otherButtonTitles:nil]; 
     [alert show]; 

    }   
} 

Answer 1

幾個問題：

1. 你的PHP代碼是不正確的。 userfile您應該使用$_FILES。見Handling File Uploads。

2. 你不能採取二進制數據，只是從它建立一個SQL語句。您可能希望在您的SQL中使用?佔位符，然後手動將與上傳圖像關聯的斑點與mysqli_stmt::bind_param或其他等效物綁定。

   坦率地說，無論如何這樣做是謹慎的，以防止SQL注入攻擊。

3. PHP代碼引用的$_GET是針對您的請求未設置的一堆變量。首先，應該是$_POST，而不是$_GET，其次，如果你的服務器需要這些變量，你應該在你的請求中設置它們。