How to Create a Dynamic Drop Down List in PHP populated from MySQL Database

Question

I am trying to create a dynamic drop down list using PHP and mysql database. I have written the following code and its giving me the output, but the problem is that its showing different drop-down menu for each item, I want all items in a single drop down list. Kindly check it and guide me.

        $select_query=          "Select name from category";
        $select_query_run =     mysql_query($select_query);
        while   ($select_query_array=   mysql_fetch_array($select_query_run) )
        {
            foreach ($select_query_array as $select_query_display)
            {
                echo "  
                    <select>
                        <option value='' >$select_query_display</option>                        
                    </select>
                ";


                }

            }

Thanks

Orangepill · Accepted Answer · 2013-07-17 05:45:45Z

4

Get rid of the inner foreach loop... it is doing nothing for you and move the start and end select tags outside of the while loop.

$select_query=          "Select name from category";
$select_query_run =     mysql_query($select_query);
echo "<select name='category'>";
while ($select_query_array=   mysql_fetch_array($select_query_run) )
{
   echo "<option value='' >".htmlspecialchars($select_query_array["name"])."</option>";
}
echo "</select>";

answered Jul 17, 2013 at 5:45

Orangepill

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1 Comment

Taha Kirmani Over a year ago

Thank you so much for the prompt response! It is working !! :)

Amir Habibzadeh · Accepted Answer · 2013-07-17 05:46:21Z

1

take look at this code.

$select_query=          "Select name from category";
$select_query_run =     mysql_query($select_query);

echo "<select>";
while   ($select_query_array=   mysql_fetch_array($select_query_run) )
{
        echo "<option value='' >".$select_query_array['name']."</option>";                        
}
echo "</select>";

answered Jul 17, 2013 at 5:46

Amir Habibzadeh

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Comments

Rob · Accepted Answer · 2017-08-01 05:41:55Z

1

<?php

$res = mysqli_query($conn, "SELECT DISTINCT coloumn_name FROM table_name;" );
while($row = mysqli_fetch_array($res))    
{
    echo "<option value='" . $row['selected_coloumn']. "'>" . $row['selected_coloumn'] . "</option>";
}
?>

In this example please select your 'coloumn_name' and 'table_name'.

edited Aug 1, 2017 at 5:41

Rob♦

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answered Aug 1, 2017 at 5:41

Aryan Arora

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Comments

Yogesh · Accepted Answer · 2013-07-17 05:43:50Z

0

$select_query= "Select name from category";
$select_query_run =     mysql_query($select_query);
$selectTag = "<select>";
while   ($select_query_array=   mysql_fetch_array($select_query_run) ){
    foreach ($select_query_array as $select_query_display){
        $selectTag .="<option value='' >$select_query_display</option>";
    }
}
$selectTag .= "</select>";

   echo $selectTag;

answered Jul 17, 2013 at 5:43

Yogesh

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1 Comment

Taha Kirmani Over a year ago

It is working, but its showing duplicate records. I didn't give it negative points. Thanks :)

vjy tiwari · Accepted Answer · 2017-08-01 05:51:01Z

0

You may try it

    $select_query=          "Select name from category";
    $select_query_run =     mysql_query($select_query);
    $select_query_array=   mysql_fetch_array($select_query_run)
    $select = "<select>";
    foreach ($select_query_array as $val)
    {
        $select .= "<option value='".$val['name']."' >".$val['name']."</option>"; 


    }

    $select = "</select>";

    echo $select;

Hope It will work

answered Aug 1, 2017 at 5:51

vjy tiwari

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