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Editorial Team
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Asked: 2026-05-22T16:06:15+00:00

Please could you check this code and see why its returning an Undefined offset:

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Please could you check this code and see why its returning an Undefined offset: and how it can be fixed.

options.php

<?php
$options = array();
$options["PC 1"] = array("year"=>"2000","colour"=>"blue");
$options["PC 2"] = array("year"=>"2003","colour"=>"pink");
$options["PC 3"] = array("year"=>"2006","colour"=>"orange"); 
?>

index.php

<html>
<body>
<form name="input" action="test.php" method="post">
Device Name: <?php
include("options.php");
echo '<select name="option">';
foreach($options as $name => $values)
{    echo '<option value="' .$name .'">' .$name .'</option>';
}
echo '</select>';
?>
<input type="submit" value="Submit" />
</form> 
</body>
</html>

test.php

<?php
include("options.php");
$chosenValue = $_POST['option'];
list($year,$colour) = $options[$chosenValue];  ---- here is the error line

echo $year;
echo $colour;

?>

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1 Answer

  1. Editorial Team

    I think it’s because the key’s in your array $options have spaces in them. Which is completely valid/legal in PHP, but in HTML, when the form is submitted it doesn’t like it.

    Try changing them to $options["PC1"] and so on and see if it fixes it.

    Edit: From the PHP manual – list() only works on numerical arrays and assumes the numerical indices start at 0.

    Try changing test.php to:

    <?php
include("options.php");
$chosenValue = $_POST['option'];
$year = $options[$chosenValue]['year'];
$colour = $options[$chosenValue]['colour'];

echo $year;
echo $colour;
?>
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