Question

Warning: Unsupported style property aria-label. Did you mean ariaLabel? in input (created by ForwardRef(InputBase)) in div (created by ForwardRef(InputBase)) in ForwardRef(InputBase) (created by WithStyles(ForwardRef(InputBase))) in WithStyles(ForwardRef(InputBase)) (created by ForwardRef(Input)) in ForwardRef(Input) (created by WithStyles(ForwardRef(Input)))如注释部分所建议，我已经更新了代码并立即使用PDO。我仍然有同样的错误。我正在尝试将多个图像存储到数据库（phpmyadmin）。当我尝试上传时，出现错误UPDATED:

我的数据库：

我有一个名为image_upload的数据库，并且在数据库中有一个名为car_detailss的表，该表具有一个id（int11），car_name（varchar 255）和imageOfcar（longblob）。

这是我的图像详细信息：

Error: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'file.png' in 'field list'

根据评论部分的建议，我现在正在使用PDO 那是我的代码：（更新）

Array
(
    [0] => Array
        (
            [name] => file.png
            [type] => image/png
            [tmp_name] => /opt/lampp/temp/phpJYyrQn
            [error] => 0
            [size] => 77776
        )

    [1] => Array
        (
            [name] => files.png
            [type] => image/png
            [tmp_name] => /opt/lampp/temp/phpXOLvzL
            [error] => 0
            [size] => 84710
        )

)

如何解决我提到的错误（在顶部）？

Answer 1

如评论部分所建议。我首先将代码更改为PDO，而不是插入没有任何反引号或“”的数据库，而是遵循注释部分的建议，并在将图像插入数据库时添加了反引号和“”。

$conn->exec( "INSERT INTO car_details ( car_name , imageOfcar ) VALUES ('$carname', '$imagename')");

现在，我可以将图像插入数据库了。我也在这里更新了代码。

我的代码：

<!doctype html>
<html>
    <head>
        <title>Test</title>
    </head>
    <body>
    <form   method="post" enctype="multipart/form-data">
           <input type="file" name="userfile[]" multiple="" /> 
           <input type="submit" name="submit" value="upload" />
    </form>

    <?php
      $servername="localhost";
      $username="root";
      $password = "";
      $dbname="image_upload";
      try {
      $conn = new PDO("mysql:host=$servername;dbname=$dbname",$username, $password);
      $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

      // begin the transaction
      $conn->beginTransaction();


      
      

$phpFileUploadErrors = array(
    0 => "There is no error, the file uploaded with success",
    1 => "The uploaded file exceeds the upload_max_filesize directive in php.ini",

    2 => "The upload file exceeds the MAX_FILE_SIZE directive that was specified in the HTML form",

    3 => "The upload file was only partially uploaded",


    4 => "No file was uploaded",

    6 => "Missing a temporary folder",
    7 => "Failed to write file to disk.",
    8 => "A php extension stopped the file upload"
);



    if(isset($_FILES['userfile'])) {
      
       $file_array = reArrayFiles($_FILES['userfile']);
       pre_r($file_array);
        for($i=0;$i<count($file_array);$i++) {

    $carname = $file_array[$i]["size"];
    $tablename = "car_detailss";

            $imagename = $file_array[$i]["name"];
           
            $conn->exec( "INSERT INTO `car_detailss` (`car_name`, `imageOfcar`) VALUES ('$carname', '$imagename')");

          
          
          
            }
           
         
         

        }
        $conn->commit();
        echo "new records created succesfully";
    } catch(PDOException $e) {
        // roll back the transaction if something failed
        $conn->rollback();
        echo "Error: " . $e->getMessage();
      }
      
      $conn = null;
       
    

      
 
    

 function reArrayFiles($file_post) {
     $file_ary = array();
     $file_count = count($file_post['name']);
     $file_keys = array_keys($file_post);

     for ($i = 0; $i < $file_count; $i++){
     foreach($file_keys as $key) {
         $file_ary[$i][$key] = $file_post[$key][$i];
     
     }
 }
 return $file_ary;
}

    function pre_r($array) {
        echo '<pre>';
        print_r($array);
        echo '</pre>';
    }

    
    ?>
    
    </body>
</html>

尝试将图像存储到数据库时发生sql错误（已更新）

1 个答案: