Question

I have a table in which every row has an option to upload a CSV file and a button to generate the report.
So every row has a file input with same name “file” and a button to submit the form. This file should be uploaded corresponding to the other values in the row. But when I was trying to upload the file on server side PHP unable to detects a uploaded file.
The current workaround that is working is creating a different form for every row.But isn’t there a way to get it done by one form? Is it good to have too many forms on a single HTML page?

Answer 1

The way to do this is to turn the input name attributes into arrays with []:

<input type='text' name='input[]' />

Then in PHP, access them as:

// $_POST['input'] is an array...
foreach ($_POST['input'] as $input_value) {
}

For file inputs, they’ll appear as an array in $_FILES:

<input type='file' name='infile[]' />

// $_FILES['infile'] is an array
foreach ($_FILES['infile'] as $f) {
  echo $f['tmpname'];
  echo $f['size']';
  // etc....
}