PHP 文件上传-重命名文件

Question

Using www.w3schools.com/php/php_file_upload.asp as a base, I want to upload a file but change the file name to include a field in my form, plus specific test.

How can I do this? (I very new to this)

EG. file name being uploaded = TestFile.jpeg

I want the file renamed to be [first_name field in form]_invoice.jpeg

This is my form:

 <form action="upload.php" method="post" enctype="multipart/form-data">
First name
<input type="text" name="first_name" id="first_name" />

  Select image to upload:
  <input type="file" name="fileToUpload" id="fileToUpload">
  <input type="submit" value="Upload Image" name="submit">
</form>

This is my PHP:

    <?php
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));

// 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 != "pdf"  && $imageFileType != "txt"  && $imageFileType != "png" && $imageFileType != "jpeg" && $imageFileType != "gif" ) {
  echo "Sorry, only PDF, TXT, 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 ". htmlspecialchars( basename( $_FILES["fileToUpload"]["name"])). " has been uploaded.";
  } else {
    echo "Sorry, there was an error uploading your file.";
  }
}
?>

Answer 1

Change the first lines in the the code like this:

<?php
$target_dir = "uploads/";
$imageFileType = strtolower(pathinfo($_FILES["fileToUpload"]["name"],PATHINFO_EXTENSION));
$target_file = $target_dir . trim($_POST['your_name']) . '_invoice.' . $imageFileType;
$uploadOk = 1;

But you should also have to have <input type="text" name="your_name"> in your form because according your code and the article , there's no text-typed input.