Question

我想在mysql数据库中插入带有其他详细信息的图像。我能够保存数据值但是没有插入图像。请检查下面的代码。

(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

有几个问题：

您的PHP代码不正确。您应该$_FILES使用userfile。请参阅Handling File Uploads。
您无法获取二进制数据，只需从中构建SQL语句即可。您可能希望在SQL中使用?占位符，然后手动将与上载图像关联的blob与mysqli_stmt::bind_param或类似的东西绑定。

坦率地说，无论如何，这样做是谨慎的，以保护自己免受SQL注入攻击。
对于您的请求未设置的一堆变量，PHP代码引用$_GET。首先，它应该是$_POST，而不是$_GET，其次，如果您的服务器需要这些变量，您应该在请求中设置它们。

如何将图像插入mysql数据库

1 个答案: