PHP removing empty value from dropdown of an array

Question

I am creating an array and then displaying the elements of it with a dropdown using php and html. The array is varying size, so I have a foreach statement to display it, the problem is I can't figure out why the first value in the dropdown is always empty (I want the first value to actually be from my array). Here's the code, any help is much appreciated!

<html>
  <form id="form1" name="form1" method="post" onsubmit="" onreset="" action="thing.php">
<?php

$data = array();
$data[4] = 20.4541;
$data[5] = 32.5631;
$data[6] = 41.9715;
$data[7] = "text";
$data[10] = 2231;
$bestValue = 5;

echo " " . $key;
echo "<select name='dropdown'>";
echo "<option value=''></option>";

if($_POST['dropdown'] !=0){
    foreach($data AS $key=>$value) {
        $sel = '';
        if($key==$_POST['dropdown'] ) {
            $sel = "selected";
        }

        echo "<option value='".$key."' ".$sel.">".$key."</option>"; 
    }}      
else{

    foreach($data AS $key=>$value) {
        $sel = '';
        if($key==$bestValue ) {
                $sel = "selected";
        }
        echo $value;

        echo "<option value='".$key."' ".$sel.">".$key."</option>";             
    }}      

echo "</select>";


foreach($data AS $key2=>$value){
    if($_POST['dropdown'] == $key2){
    echo "  " . $data[$key2] . " degrees";}}
?>

<script>
function submitForm(action)
{
    document.getElementById('form1').action = action;
    document.getElementById('form1').submit();
}
</script>
<center>
<input type="button" onclick="submitForm('thing.php')" value="Find Angle" />
</center>

</html>

As you can see the first value in the dropdown is empty. How do I remove this? Thanks.

Cyril Tourist · Accepted Answer · 2012-05-29 20:57:52Z

4

Just remove the third echo line.

echo "<option value=''></option>";

edited May 29, 2012 at 20:57

Cyril Tourist

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answered May 29, 2012 at 20:18

rekire

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Comments

None · Accepted Answer · 2012-05-29 20:20:34Z

2

The first option you are printing is an empty option.

 echo "<select name='dropdown'>";
 echo "<option value=''></option>";

Remove the line

echo "<option value=''></option>";

answered May 29, 2012 at 20:20

None

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1 Comment

Joegramming Over a year ago

Thank you, as you guys can see I'm new to this. Why is this getting minus on the question? Is this website really so elitist that nooby questions are discouraged? I'm sure you guys were just starting out learning php for the first time at some point too...

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