如何从php中的表单获取文件名？ $_FILES[“filename”][“name”] 似乎不起作用

Question

I have a form below which I'm trying to use to extract some details into a mySQL database. However, upon trying to retrieve the file in the form, everything always comes up as null. I'm not so sure why this is happening, so any help would be greatly appreciated!

 <form method="post" action="myfile.php" enctype="multipart/form-data"> <table class="form-table"> <tr> <td><b>Document (Optional)</b></td> <td> <select name="d" class="form-control input input-sm"> <option value="companyDoc">Company Registration Form</option> <option value="businessCard">Business Card</option> </select> </td> </tr> <tr> <td><b>Upload Document (Optional)</b></td> <td><input type="file" class="form-control input input-sm" id="ud" name="ud"></td> </tr> <tr> <td></td> <td><input type="submit" value="Submit" name="submit" class="btn btn-primary pull-right"></td> </tr> </table> </form>

Now my php snippet is below, where I try and call the results of the form, in the same file as the html.

 if(isset($_POST['submit'])) { $target_dir = "../../public_html/clientarea/docs/"; $fileName = basename($_FILES["ud"]["name"]); $target_file = $target_dir . $fileName; if($fileName!="") { $uploadOk = 1; $imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION)); $errorMsg = ""; if (file_exists($target_file)) { $errorMsg = "Some file already exists. "; $uploadOk = 0; } else { if (move_uploaded_file($_FILES["ud"]["tmp_name"], $target_file)) { $errorMsg = "The file ". basename( $_FILES["ud"]["name"]). " has been uploaded. "; } else { $errorMsg = "There was an error uploading your some of the file."; } } } }

Answer 1

Tested your HTML and PHP code. Worked just fine with $_FILES['ud']['name'] to get the filename.

You can try to print_r($_FILES) or maybe print_r($_FILES['ud']) to find out the values of your form that you sent through the POST method.