Question

我有一个PHP表单，有五个输入供用户上传五个图像。用户必须至少选择一个图像，并且必须在第一个输入中，以便提交表单。表单将正确地将所有信息提交到MySQL表中，但图像本身除外。例如，我提交了一个图像，图像列显示[BLOB-14 B]，当我认为它应该至少为300 KB。我的理论是，当图像被提交到MySQL表中时，图像被压缩成其他格式。

这是我的完整PHP页面：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

  <html>
  <head><title>File Upload To Database</title></head>
  <body>
  <h2>Please Choose a File and click Submit</h2>
  <form enctype="multipart/form-data" action="<?php echo htmlentities($_SERVER['PHP_SELF']);?>" method="post">
  <input type="hidden" name="MAX_FILE_SIZE" value="99999999" />
  <div><input name="userfile[]" type="file" /></div>
    <div><input name="userfile[]" type="file" /></div>
      <div><input name="userfile[]" type="file" /></div>
        <div><input name="userfile[]" type="file" /></div>
          <div><input name="userfile[]" type="file" /></div>
  <div><input type="submit" value="Submit" /></div>
  </form>

</body></html>

<?php
/*** check if a file was submitted ***/
if(!isset($_FILES['userfile']))
    {
    echo '<p>Please upload a display picture.</p>';
    }
else
    {
    try    {
        upload();
        /*** give praise and thanks to the php gods ***/
        echo '<p>Thank you for submitting</p>';
        }
    catch(Exception $e)
        {
        echo '<h4>'.$e->getMessage().'</h4>';
        }
    }

/*
 * Check the file is of an allowed type
 * Check if the uploaded file is no bigger thant the maximum allowed size
 * connect to the database
 * Insert the data
 */

/**
 *
 * the upload function
 * 
 * @access public
 *
 * @return void
 *
 */
function upload(){

$maxsize = 99999999;
$columnNames = '';
$columnValues = '';
$paramsToBeBound = array();

echo '<pre>' . print_r($_FILES, TRUE) . '</pre>';

/*** check if a file was uploaded ***/
for($i = 0; ($i < count($_FILES['userfile']['tmp_name']) && $i < 5); $i++) {
    if($_FILES['userfile']['tmp_name'][$i] != '') { // check if file has been set to upload
        if($_FILES['userfile']['error'][$i] == 0 && is_uploaded_file($_FILES['userfile']['tmp_name'][$i]) && getimagesize($_FILES['userfile']['tmp_name'][$i]) != false) {
            /***  get the image info. ***/
            $size = getimagesize($_FILES['userfile']['tmp_name'][$i]);
            /*** assign our variables ***/
            $type = $size['mime'];
            $imgfp = fopen($_FILES['userfile']['tmp_name'][$i], 'rb');
            $size = $size[3];
            $name = $_FILES['userfile']['name'][$i];


             /***  check the file is less than the maximum file size ***/
            if($_FILES['userfile']['size'][$i] < $maxsize)
                {
                    if($i > 0) {
                        $columnNames .= ', image_type' . $i . ', image' . $i . ', image_size' . $i . ', image_name' .$i;
                        $columnValues .= ', ?, ?, ?, ?';
                    } else {
                        $columnNames .= 'image_type, image, image_size, image_name';
                        $columnValues .= '?, ?, ?, ?';
                    }

                    $paramsToBeBound[] = $type;
                    $paramsToBeBound[] = $imgfp;
                    $paramsToBeBound[] = $size;
                    $paramsToBeBound[] = $name;
                } else
                    throw new Exception("File Size Error"); //throw an exception is image is not of type
            }
        else
            {
            // if the file is not less than the maximum allowed, print an error
            throw new Exception("Unsupported Image Format of image!");
            }
        }
    }
    if(count($paramsToBeBound) > 0) {
        $dbh = new PDO("mysql:host=dsa.com;dbname=s_gbm", 'kss', 'Kisr'); 

        $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

        $stmt = $dbh->prepare('INSERT INTO testblob (' . $columnNames . ') VALUES (' . $columnValues . ')');

        $i = 0;
        foreach($paramsToBeBound as &$param) {
            $i++;
            if($i == 2 || $i - floor($i / 4) == 2) {
                $stmt->bindParam($i, $param, PDO::PARAM_LOB);
            } else {
                $stmt->bindParam($i, $param);
            }
        }

        $stmt->execute();
    }
}


?>

以下是我在PHP MyAdmin SQL中用来创建MySQL表的代码：

CREATE TABLE testblob ( image_id tinyint(3) NOT NULL AUTO_INCREMENT, image_type varchar(25) NOT NULL, image longblob NOT NULL, image_size varchar(25) NOT NULL, image_name varchar(50) NOT NULL, image_type1 varchar(25) NOT NULL, image1 longblob NOT NULL, image_size1 varchar(25) NOT NULL, image_name1 varchar(50) NOT NULL, image_type2 varchar(25) NOT NULL, image2 longblob NOT NULL, image_size2 varchar(25) NOT NULL, image_name2 varchar(50) NOT NULL, image_type3 varchar(25) NOT NULL, image3 longblob NOT NULL, image_size3 varchar(25) NOT NULL, image_name3 varchar(50) NOT NULL, image_type4 varchar(25) NOT NULL, image4 longblob NOT NULL, image_size4 varchar(25) NOT NULL, image_name4 varchar(50) NOT NULL, image_ctgy varchar(25) NOT NULL, KEY image_id (image_id) ) ENGINE=MyISAM DEFAULT CHARSET=latin1;

感谢您的帮助。我感谢所有给予的帮助。

Answer 1

首先，我建议不要在MySQL中存储图像，除非你绝对必须这样做。

你绝对要吗？是？那真不幸。这应该有所帮助......

$imgfp = fopen($_FILES['userfile']['tmp_name'][$i], 'rb');
$imageData = fread($imgfp, filesize($_FILES['userfile']['tmp_name'][$i]));
$imageData = addslashes($imageData);
fclose($imgfp);
…
…
$paramsToBeBound[] = $imageData;

Answer 2

你永远不应该在MySQL数据库中存储任何媒体类型，你应该做的是将媒体文件（无论是图片还是视频）存储在本地硬盘上，然后让MySQL存储一个文件，在需要时调用该媒体类型。我希望这有帮助（=

PHP表单未将图像提交到MySQL表

2 个答案: