在单行中将多个图像存储到SQL数据库

Question

我使用以下代码将多个图像存储到数据库中，每件事都可以很好地存储和显示，当我一次上传3张照片时，我需要上传的内容是创建三行以供一次上传时如何将多张图像的路径存储在一行中

class Upload_Rename {
const ALLOWED_TYPES = "jpg,gif,png";
public static function generate_new_name($extension, $uppercase = true, $prefix = '', $sufix = '') {
$new_name = $prefix . uniqid() . '_' . time() . $sufix;
return ($uppercase ? strtoupper($new_name) : $new_name) . '.' . $extension;
}

public static function check_and_get_extension($file) {
$file_part = pathinfo($file);
$allowed_types = explode(",", Upload_Rename::ALLOWED_TYPES);
if (!in_array($file_part['extension'], $allowed_types)) {
    throw new Exception('Not ok.. bad bad file type.');
}
return $file_part['extension'];
}

public function upload($file, $target_destination) {
if (!isset($file['tmp_name'])) {
    throw new Exception('Whaaaat?');
}
$_name = $file['name'];
$_tmp = $file['tmp_name'];
$_type = $file['type'];
$_size = $file['size'];
$file_extension = '';
try {
    $file_extension = Upload_Rename::check_and_get_extension($_name);
} catch (Exception $e) {
    throw new Exception('Ops.. file extension? what? ' . $e->getMessage());
}
$new_name = Upload_Rename::generate_new_name($file_extension, true, 'whaat_', '_okey');
$destination = $target_destination . DIRECTORY_SEPARATOR . $new_name;
move_uploaded_file($_tmp, $destination);
return $destination;
}

public function multiple_files($files, $destination) {
$number_of_files = isset($files['tmp_name']) ? sizeof($files['tmp_name']) : 0;
$errors = array();
$paths=array();
for ($i = 0; $i < $number_of_files; $i++) {
    if (isset($files['tmp_name'][$i]) && !empty($files['tmp_name'][$i])) {
        try {
            $path=$this->upload(array(
                'name' => $files['name'][$i],
                'tmp_name' => $files['tmp_name'][$i],
                'size' => $files['size'][$i],
                'type' => $files['type'][$i]
                    ), $destination);
             $paths[]=$path;
        } catch (Exception $e) {
            array_push($errors, array('file' => $files['name'][$i], 'error' => $e->getMessage()));
        }
    }
}
return $paths;
}

}

if ($_FILES) {
$upload = new Upload_Rename();
$destination = 'upload';
$paths=$upload->multiple_files($_FILES['myfile'], $destination);

//Fill this with correct information
$mysql_hostname = "";
$mysql_user = "";
$mysql_password = "";
$mysql_database = "";
$tbl_name="test";
$pathfield_name='path';
//
$mysql= new mysqli($mysql_hostname,$mysql_user,$mysql_password,$mysql_database);
foreach($paths as $path){
$query='INSERT INTO `'.$tbl_name.'` (`'.$pathfield_name.'`) VALUES ("'.$mysql->escape_string($path).'");';
$mysql->query($query);}
$mysql->close();
}
?>

<form  method="post" enctype="multipart/form-data">
<?php for ($i = 0; $i < 10; $i++): ?>
file: <input type="file" name="myfile[]"><br>
<?php endfor; ?>
<input type="submit">
</form>

我使用此代码在sql中创建表

CREATE TABLE IF NOT EXISTS `test` (
`path` text(50) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

Answer 1

首先获取路径字段并存储在一个变量中，然后插入

foreach($paths as $path){  $pathfield_name.=$pathfield_name."/";   }


$query='INSERT INTO `'.$tbl_name.'` (`'.$pathfield_name.'`) VALUES ("'.$mysql->escape_string($path).'");';
$mysql->query($query);