【发布时间】：2020-08-18 17:40:54
【问题描述】：
我正在尝试通过变量将 URL 婴儿车“链接”传递给我的上传脚本。我在调试问题时遇到了很多麻烦，并且没有任何运气将其传递给upload.php。任何帮助或建议将不胜感激。第一段代码是我的 index.php，它应该始终带有一个 URL 参数，例如 '?link=9288'。
示例https://exampledomain.com/videos/?link=9288
<!DOCTYPE html>
<html>
<body>
<?php
$_SESSION['varname'] = $_GET['link'];
?>
<form action="upload.php" method="post" enctype="multipart/form-data">
Select image to upload:
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form>
</body>
</html><?php
$target_dir = "../../uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
echo "File is a Video- " . $check["mime"] . ".";
$uploadOk = 1;
} else {
$uploadOk = 1;
}
}
// Check if file already exists
if (file_exists($target_file)) {
echo "Sorry, file already exists.";
$uploadOk = 0;
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 400000000000000000) {
echo "Sorry, your file is too large.";
$uploadOk = 0;
}
// Allow certain file formats
if($imageFileType != "webm" && $imageFileType != "mpg" && $imageFileType != "mpeg" && $imageFileType != "mp4" && $imageFileType != "m4p" && $imageFileType != "m4v" && $imageFileType != "avi" && $imageFileType != "wmv" && $imageFileType != "mov" ) {
echo "Sorry, only WEBM, MPG, MPEG, MP4, M4P, M4V, AVI, WMV, MOV files are allowed.";
$uploadOk = 0;
}
// Check if $uploadOk is set to 0 by an error
if ($uploadOk == 0) {
echo "Sorry, your file was not uploaded.";
// if everything is ok, try to upload file
} else {
if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) {
echo "The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded.";
} else {
echo "Sorry, there was an error uploading your file.";
}
}
$var_value = $_SESSION['varname'];
echo $var_value;
?>【问题讨论】：
-
您在使用会话之前是否已开始会话？如果您希望
$_SESSION中的数据是持久的，则需要在它之前有start_session();，无论是在您读取还是写入会话之前。否则，数据将不会在调用之间保留。 -
在这种情况下，为什么不直接在带有链接的表单中添加一个隐藏的输入呢？比如：
<input type="hidden" name="link" value="<?= $_GET['link'] ?? '' ?>">。然后你应该能够在你的 PHP 中使用$_POST['link']获取值。
标签： php html file-upload upload