drop down box

slimjim

I am using a drop down box to add to a table that will chose a category.

Like this...

<td><? 

$sql="SELECT category FROM category"; 
$result=mysql_query($sql); 

$options=""; 

while ($row=mysql_fetch_array($result)) { 

//$id=$row["id"]; 
$category=$row["category"]; 
$options.="<OPTION VALUE=\"$category\">".$category; 
} 
?> 
<SELECT NAME=categoryg> 
<OPTION VALUE=0>Choose 
<?=$options?> 
</SELECT> 
</td>

But when it comes time for me to edit it, i have to select the category again. Which i do not want to do. How can i make it so that the category that is listed in the table is already showing instead of showing "choose"

I know that this is easy but for the life of me I cant figure it out.

PallaviDalvi

echo "<select name=category><OPTION VALUE=0>Choose</option>";
while ($row=mysql_fetch_array($result)) 
{  

    echo "<OPTION VALUE='" . $row["category"] . "' ";
    if (it is the value selected) // ur logic to find that out
    	echo " selected ";
    echo " >" . $category . "</option>";  

}

hopefully this is what u wanted

slimjim

I think it is, but I cant get it to work.

TheDefender

You have to determine how you are making the comparison to find out which one is the selected option...

If you are posting the form back to itself, for example, then you'd do something like this:

echo "<select name=category><OPTION VALUE=0>Choose</option>"; 
while ($row=mysql_fetch_array($result))  

{   

    echo "<OPTION VALUE='" . $row["category"] . "' "; 
    if (isset($_POST['category']) && $row['category'] == $_POST['category']) // ur logic to find that out 
        echo " selected "; 
    echo " >" . $category . "</option>";   

}

Without seeing more code, all of the help we offer is just speculation based on our perceptions of what we think your code should be. Post what you have, along with an explanation of what kind of comparison you want to make, and we can help from there.