I am using the following query to pull categories from the DB then creating a select list.
Can some tell me how to make the current value selected in the list when they open the category?
<?php
include("mysql.php");
$queryc = "select * from tbl_category WHERE active = 'yes' ORDER by name ASC";
$resultc = mysql_query($queryc);
while ($rowc = mysql_fetch_array($resultc)) {
$catIDc = $rowc;
$namec = $rowc;
$activec = $rowc;
echo "<option value=$catIDc>$namec</option>";
}
?>
I am not quite sure what you are wanting here.
Remember you can only select one item from a drop down by default and since you outputing all that are equal to yes I get confused as to the purpose of the while loop (unless there are many rows...which wouldnt work).
To set the default selected item on a drop down do it like so...
<option value="..." selected="selected"></option>Could you offer more of an explantion?
-------------------------------------------------------------------
<?php
//connection
//select db
$selected_id=$_POST['select'];
$query="Select * from table";
$result=mysql_query($query);
$options="";
while($row=mysql_fetch_array($result)){
$name=$row['name'];
$id=$row['id'];
if($selected_id==$id){
$selected="selected":
}
else {
$selected="";
}
$options.="<option value='$id' $selected>$name</option>";
}
?>
<html>
<body>
<form method="post" action="test.php">
Select something: <select name="select">
<?php echo $options; ?>
</select><br />
<input type="submit" name="submit" value="submit">
</form>
</body>
</html>Cheers,
Naveen
Worked great, thanks.
You are welcome :)
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