Why can't I create a dropdown list with variables? - php

Question

I'm trying to create a dropdown list with all the values that i have inside my table, but I can't do it, the select box stay blank.
My PHP code, I'm calling a function
Job category
<select>
<?php
function category($category_id, $category_name){
echo "<option value='".$category_name."'>".$category_name."</option>";
}
?>
</select>
And my function
function selectcategory(){
connect();
$category = ("SELECT * FROM job_categories");
$val = DB_array($category, 'a+');
$ii = count($val);
$ii = $ii - 1;
for ($i = 0; $i <= $ii; $i++) {
$category_id = $val[$i]['category_id'];
$category_name = $val[$i]['category_name'];
category($category_id, $category_name);
}
}

Answer 1

Assuming the flow of your code , that 2nd part executes first , i.e. connection is made , data is fetched and you are calling category() , it will execute
function category($category_id, $category_name){
echo "<option value='".$category_name."'>".$category_name."</option>";
}
$ii times and you are returning to the calling function .
So that's why < select > does not populate the data because it is interpreted after running the above code.

Answer 2

<select>
<?php
$result = category($category_id, $category_name);
foreach($result as $key => $value){
echo "<option value='".$key."'>".$value."</option>";
}
?>
</select>

Answer 3

$sql = mysql_query("SELECT * FROM job_categories");
$result = mysql_db_query($dbname, $sql) or die("Failed Query of " . $sql);
while ($row = mysql_fetch_row($result)){
echo "<option value='".$row["category_id"]."'>".$row["category_name"]."</option>"; // the value should be id
}