我从单个输入上传多个图像并重命名所有上传的文件。我的问题是;我需要使用单独的变量显示所有上传的图像
例如:
If user upload 3 images, i need to echo as
$upload1 = 1st file name
$upload2 = 2nd file name
$upload3 = 3rd file anme
No of images uploaded (upload number of file)
这是我目前的代码:
<?php
if (isset($_FILES['files'])) {
$uploadedFiles = array();
foreach ($_FILES['files']['tmp_name'] as $key => $tmp_name) {
$errors = array();
$file_name = md5(uniqid("") . time());
$file_size = $_FILES['files']['size'][$key];
$file_tmp = $_FILES['files']['tmp_name'][$key];
$file_type = $_FILES['files']['type'][$key];
if($file_type == "image/gif"){
$sExt = ".gif";
} elseif($file_type == "image/jpeg" || $file_type == "image/pjpeg"){
$sExt = ".jpg";
} elseif($file_type == "image/png" || $file_type == "image/x-png"){
$sExt = ".png";
}
if (!in_array($sExt, array('.gif','.jpg','.png'))) {
$errors[] = "Image types alowed are (.gif, .jpg, .png) only!";
}
if ($file_size > 2097152000) {
$errors[] = 'File size must be less than 2 MB';
}
$desired_dir = "user_data/";
if (empty($errors)) {
if (is_dir($desired_dir) == false) {
mkdir("$desired_dir", 0700); // Create directory if it does not exist
}
if (move_uploaded_file($file_tmp, "$desired_dir/" . $file_name . $sExt)) {
$uploadedFiles[$key] = array($_FILES['files']['name'][$key], 1);
} else {
echo "Couldn't upload file " . $_FILES['files']['name'][$key];
$uploadedFiles[$key] = array($_FILES['files']['name'][$key], 0);
}
} else {
print_r($errors);
}
}
foreach ($uploadedFiles as $key => $row) {
if (!empty($row[1])) {
$codestr = '$file' . ($key+1) . " = $row[0];";
eval ($codestr);
} else {
$codestr = '$file' . ($key+1) . " = NULL;";
eval ($codestr);
}
}
}
echo $file1;
echo $file2;
echo $file3;
echo $file4;
echo $file5;
?>
我试过了,我收到了错误
Parse error: syntax error, unexpected 'png' (T_STRING) in C:\Users\logon\Documents\NetBeansProjects\upload file rename single and multiple\multiple file rename\method 2\process.php(43) : eval()'d code on line 1
我做错了什么?可以有人帮助我
答案 0 :(得分:0)
我认为第42行的" = $row[0];"
可能无法按照您的意愿行事。由于双引号,变量$row[0]
的值将打印在代码中。
通过eval运行的代码如下所示:
$file0 = FILENAME.png;
而不是:
$file0 = $row[0];
要解决此问题,您只需将第42行转换为以下内容:
$codestr = '$file' . ($key+1) . ' = $row[0];';
有关单引号和双引号之间差异的详细信息:http://php.net/manual/en/language.types.string.php
编辑:对于文件的正确名称,您可以将$_FILES['files']['name'][$key]
更改为$file_name . $sExt
。希望它有所帮助!