Dynamically displaying option values using PHP

Question

I'm new to PHP and have created a very basic HTML form. As you can see in my form, the option values are all done by hand (there are more, I just simplified this example). What I want is for these to be generated dynamically using just PHP, so that I would physically have to add every single year etc.

I've done some searching but I can't seem to find exactly what I'm after so thought I'd ask here. From what I gather I need to create a query and echo out the option value somehow, although I'm not sure how to do this.

SELECT gameYear from games

I'd guess the above would be the correct query as all the form would need is the bookYear from the table?

<form id = "gameYear" method="get" action="result.php">
<label> 
    Game Year
    <select name="gameYear">
       <option value="2000">2000</option>
       <option value="2001">2001</option>
       <option value="2002">2002</option>
    </select>
</label>
   <input type = "submit" name="search" value = "Search">
</form>

Thanks, any help/guidance is appreciated.

Ace · Accepted Answer · 2014-11-26 02:49:34Z

2

<form id = "gameYear" method="get" action="result.php">
<label> 
Game Year
<select name="gameYear">
<option value=''>--Select Year--</option>
<?php
$link = mysqli_connect("localhost", "my_user", "my_password", "world");
$SqlResult = mysqli_query($link, "SELECT gameYear from games");
while($Row = mysqli_fetch_array($SqlResult))
{
  ?>
  <option value="<?php echo $Row['gameYear'] ?>"><?php echo $Row['gameYear'] ?></option>
}
?>
</select>
</label>
<input type = "submit" name="search" value = "Search">
</form>

answered Nov 26, 2014 at 2:49

Ace

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Comments

ryvasquez · Accepted Answer · 2014-11-26 02:59:20Z

1

<?php $sql = "SELECT gameYear from games order by gameYear ASC";
    $result = mysql_query($sql, $connection) or die ("Couldn't perform query $sql <br />".mysql_error()); ?>


<form id="gameYear" method="get" action="result.php">
<label>Game Year
<select name="gameYear">
<?php while($row = mysql_fetch_array($result)){ ?>                        
<option value="<?php echo $row['gameYear'] ?>"><?php echo $row['gameYear']?></option>
<?php } ?>
</select>
</label>
<input type = "submit" name="search" value = "Search">
</form>

edited Nov 26, 2014 at 2:59

answered Nov 26, 2014 at 2:36

ryvasquez

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1 Comment

dpDesignz Over a year ago

You may also want to add ORDER BY gameYear ASC. Also your <option value="<?php echo $row['gameYear'] ?>"><?php echo $row['fullname']?></option> should be <option value="<?php echo $row['gameYear'] ?>"><?php echo $row['gameYear']?></option>

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