Question

在我的项目中，我试图通过$ _FILES上传文件。但是，当我检查$ _FILES的值时，它不包含任何内容。我还附上我的PHP和HTML代码 HTML：

<form id="insert_form" method="post" enctype="multipart/form-data">
   <input type="file"  name="fileToUpload" id="fileToUpload" />
    <input class="btn btn-info" name="submit" type="submit" value="Insert" />

</form>

PHP：

$filename = $_FILES["fileToUpload"]['name'];

这是我基于表单提交点击的ajax调用

     $('#insert_form').on('submit', function (event) {
                event.preventDefault();
                 var form_data = $(this).serialize();

                    $.ajax({
                        url: 'next.php',
                        method: "POST",
                        data: form_data,
                        success: function (data) {

                        }
                      });
                    });

Answer 1

您可以尝试以下代码

index.php

<!DOCTYPE html>
<html>
<body>

<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>

然后在upload.php

<?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 an image - " . $check["mime"] . ".";
        $uploadOk = 1;
    } else {
        echo "File is not an image.";
        $uploadOk = 0;
    }
}
// Check if file already exists
if (file_exists($target_file)) {
    echo "Sorry, file already exists.";
    $uploadOk = 0;
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 500000) {
    echo "Sorry, your file is too large.";
    $uploadOk = 0;
}
// Allow certain file formats
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg"
&& $imageFileType != "gif" ) {
    echo "Sorry, only JPG, JPEG, PNG & GIF 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.";
    }
}
?>

上面是带有各种验证的完整代码，您可以删除不需要的代码。

无法通过php中的$ _FILES获取文件名

1 个答案: