CODEWITHSUNDEEP
phpdrop-down-menu
RickTicky
RickTicky

Reputation: 114

populate a drop down in php

I have a mysql database of games that you can hire. I created a query in php so that the user is able to type in a title or game description and the result will be shown in a table as a simple list. However, I want to display the results of my array in a drop down. But how? I've tried to apply code snippets I've found online(including Stack Overflow) but failed miserably to make them work. Any help would be most appreciated.

<table>
    <tr>
        <td>Game ID</td>
        <td>Game Name</td>
        <td>Game Description</td>
    </tr>
<?php
    if(isset($_REQUEST['submit'])){
        $search=$_POST['search'];
        $sql="SELECT* FROM gamestbl WHERE game_name LIKE '%{$search}%' OR game_description LIKE '%{$search}%'";
        $q=mysqli_query($conn,$sql);
        while($res=mysqli_fetch_array($q)){
?>
<tr>
    <td><?php echo $res['game_id'];?></td>
    <td><?php echo $res['game_name'];?></td>
    <td><?php echo $res['game_description'];?></td>
</tr>
    <?php }
}?>
</table>

Upvotes: 0

Views: 64

Answers (2)

user169600
user169600

Reputation:

Replace your table with select:

<select name="game">
<?php
    if(isset($_REQUEST['submit'])){
        $search=$_POST['search'];
        $sql="SELECT* FROM gamestbl WHERE game_name LIKE '%{$search}%' OR game_description LIKE '%{$search}%'";
        $q=mysqli_query($conn,$sql);
        while($res=mysqli_fetch_array($q)){
?>
<option value="<?php echo $res['game_id'];?>"><?php echo $res['game_name'];?></option>
    <?php }
}?>
</select>

Upvotes: 2

Tim Lewis
Tim Lewis

Reputation: 29278

This is a pretty common problem, and you've got the right idea. Right now, you're populating these into a <table> element's <tr> and <td> elements. So, why not change that to a <select> and <option> layout?

Here's the basic idea. During your while loop, echo an <option> for each game. Set the value attribute to the game's id and it's display to the game's name and description.:

<option value="<?php echo $res["game_id"]; ?>"><?php $res["game_name"]."-".$res["game_description"]; ?></option>

Like you have with your <table>, make sure to define the <select> outside of the php while loop, or you'll end up with multiple dropdowns.

Hope that helps!

Upvotes: 2

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