php - 如何在成功处理 PHP 脚本时打开模态框？

Question

我有 2 个文件，分别是 upload.php 和 process.php。

在我的 upload.php 文件中，我有一个这样的表单:

<form action="upload.php" method="post" enctype="multipart/form-data">
    <h5>Upload File</h5>
    <p>.jpg and .png only.</p>
    <input type="file" name="myfile" required> <br><br>

    <h5>Choose action</h5>
        <select class="form-control" name="action"required>
            <option>No action (Finished)</option>
            <option>Please Check</option>
        </select><br>
    <h5>Who are you</h5>
        <select class="form-control" name="uploader" required>
            <option>Person A</option>
            <option>Person B</option>
            <option>Person C</option>
        </select><br>
    <h5>Remark</h5>
        <textarea name="remark" style="resize:none; width:100%; height:100px;">-</textarea><br><br>

    <button type="submit" name="save" class="btn btn-dark btn-block">Upload</button><br>
</form>

在我的 process.php 中:

if (isset($_POST['save'])) { // if save button on the form is clicked
    // name of the uploaded file
    $filename = $_FILES['myfile']['name'];
    $uploader = $_POST['uploader'];
    $action = $_POST['action'];
    $remark = $_POST['remark'];

    // destination of the file on the server
    $destination = 'uploads/' . $filename;

    // get the file extension
    $extension = pathinfo($filename, PATHINFO_EXTENSION);

    // the physical file on a temporary uploads directory on the server
    $file = $_FILES['myfile']['tmp_name'];
    $size = $_FILES['myfile']['size'];

    if (!in_array($extension, ['jpg','png'])) {
        echo '<script language="javascript">';
        echo 'alert("Upload failed! Your file extension must be .jpg or .png")';
        echo '</script>';
    } elseif ($_FILES['myfile']['size'] > 1000000) { // file shouldn't be larger than 1Megabyte
        echo "File too large!";
    } else {
        // move the uploaded (temporary) file to the specified destination
        if (move_uploaded_file($file, $destination)) {
            $sql = "INSERT INTO files (name, size, uploader, action, remark, downloads) VALUES ('$filename', $size, '$uploader', '$action', '$remark', 0)";
            if (mysqli_query($conn, $sql)) {
                echo '<script language="javascript">';
                echo 'alert("File uploaded successfully.")';
                echo '</script>';
            }
        } else {
            echo '<script language="javascript">';
            echo 'alert("Failed to upload file.")';
            echo '</script>';
        }
    }
}

如何实现:如果文件上传成功，一个模态框即looks like this会显示上传成功吗？现在，就像您在代码中看到的那样，我只让它显示一个警报。

Answer 1

我想我只是解决了我自己的问题。

我在 upload.php 中添加了模式，然后我添加了这个 javascript，这样我就可以通过直接 url 触发模式框。

<script type="text/javascript">

$(document).ready(function() {

  if(window.location.href.indexOf('#success') != -1) {
    $('#success').modal('show');
  }

});
</script>

然后在process.php中，如果上传成功，重定向到mywebsite.com/upload.php#success

header(Location: 'upload.php#success')

如果谁有更好的解决方案，欢迎分享。 :)