PHP FileToUpload 未定义索引错误

Question

这个问题已经有答案了:

"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP (29 个回答)

已关闭 5 年前。

我想上传文件到uploads ，但我每次都会收到此错误消息:

Notice: Undefined index: fileToUpload in /path/to/dir/upload.php on line 3

HTML 文件:

<form action="php/upload.php" method="post" enctype="multipart/form-data">
    <input type="file" name="fileToUpload" id="fileToUpload">
    <input type="submit" value="Upload Image" name="submit">
</form>

PHP 文件:

<?php
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($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 an image - " . $check["mime"] . ".";
        $uploadOk = 1;
    } else {
        echo "File is not an image.";
        $uploadOk = 0;
    }
}
?>

Answer 1

首先，欢迎来到 Stack Overflow。

您必须将第 3 行移动到 if 语句中，因为如果没有 post，则不会设置全局 $_FILES 变量。

例如:

<?php
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
    $target_dir = "uploads/";
    $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
    $uploadOk = 1;
    $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);

    $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
    if($check !== false) {
        echo "File is an image - " . $check["mime"] . ".";
        $uploadOk = 1;
    } else {
        echo "File is not an image.";
        $uploadOk = 0;
    }
}
?>

即使如此，也可能没有发布文件

所以:

if(isset($_POST["submit"], $_FILES["fileToUpload"])) {