php read file and create drop down menu

Question

I have following php code which create Drop-Down menu reading text file which contain name

files.txt

JAMES
MARK
TONY

drop-down.php

<?php
$path = "files.txt";
$file = fopen($path, 'r');
$data = fread($file, filesize($path));
fclose($file);

$lines =  explode(PHP_EOL,$data);
echo '<select name="file">';
foreach($lines as $line) {
  echo '<option value="'. urlencode($line).'">'.$line.'</option>';
}
echo '</select>';
?>

Here is the POST section of code, it is a loop

foreach ($_POST["bname"] AS $id => $value)
{
...
...   
USERNAME: "'.$_POST["file"][$id].'"

Everything working but when i submit data to other form i am getting only first letter of users like if i select JAMES and submit data i am only getting J letter in .$_POST. I want Full name JAMES, what is wrong in my code?

What is probably happening is you are saying I only want the first character of the string `$_POST["file"][0]` or something similar somewhere, but we need to see the part where you echo it out. — Get Off My Lawn, May 31 '13 at 15:28
Yeah this worked fine for me. It must be what ever you're posting. — TheSnooker, May 31 '13 at 15:31

score 1 · Accepted Answer · answered May 31 '13 at 15:32

1

The name attribute of your select is wrong, must be:

echo '<select name="file[]">';

or

echo '<select name="file[someNumber]">';

answered May 31 '13 at 15:32

Marcelo Pascual

810
8
20

score 1 · Answer 2 · answered May 31 '13 at 15:34

You don't need the Id, because that is saying "I want this value, and the second part says which character" (at least the way that you are using it). Try this:

"USERNAME " . $_POST["file"]

Strings are treated like arrays in PHP.

So if you had the string kitty

$string = "Kitty";
echo $string[0]; // prints "K"
echo $string[4]; // prints "y"

score 1 · Answer 3 · answered May 31 '13 at 15:35

Try read the txt file with file(), it adds the rows directy into an array, then use for() to traverse the array.

<?php
$path = "files.txt";
$lines = file($path, FILE_IGNORE_NEW_LINES | FILE_SKIP_EMPTY_LINES);
$nrl = count($lines);
echo '<select name="file[]">';
for($i=0; $i<$nrl; $i++) {
  echo '<option value="'. urlencode($lines[$i]).'">'.$lines[$i].'</option>';
}
echo '</select>';

php read file and create drop down menu

3 Answers3

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