[Resolved] Selected Value NOT in db...

[ndorfnz]

I am filling a select box from an array...

If the value matches another value from elsewhere I am selecting this value... not a problem.

However, if the value DOES NOT match, I want to echo out "Select a Value" and still have all the results from the db appear in the select box...

Below is my code, I am just missing how to echo out the "Select A Value" text while keeping all the values to select from...

==============================================
//STEP THROUGH THE FOR LOOP POPULATING THE SELECT BOX BASED ON THE VALUES OF THE ARRAY
for ($i=0; $i<$WOCount; $i++)
{
//LOAD VALUES FROM ARRAY INTO VARIABLE TO ECHO OUT IN THE OPTION VALUES
while ($workout_type = mysql_fetch_array($sql_result))
{
if ($uworkout_type==$workout_type[$i])
{
echo $sel = ' selected';
}
else
{
echo $sel = ' Select A Workout...';

	}		
echo '<option value="'.$workout_type[$i].'"'.$sel.'>'.$workout_type[$i].'</option>';
}

}

echo "</SELECT>";

Thanks much!!!

[tommynanda]

Change your code to

==============================================
//STEP THROUGH THE FOR LOOP POPULATING THE SELECT BOX BASED ON THE VALUES OF THE ARRAY
?>
<select name=wkt>
<option value="nomatch">Select A Workout
<?
for ($i=0; $i<$WOCount; $i++)
{
//LOAD VALUES FROM ARRAY INTO VARIABLE TO ECHO OUT IN THE OPTION VALUES
while ($workout_type = mysql_fetch_array($sql_result))
{
if ($uworkout_type==$workout_type[$i])
{
echo $sel = ' selected';
}
else
{
echo $sel = ' ';

}
echo '<option value="'.$workout_type[$i].'"'.$sel.'>'.$workout_type[$i].'</option>';
}
}

echo "</SELECT>";

By default, browser selects first value

Good Luck
TommYNandA

[ndorfnz]

Thanks for your input...

However, doesn't work...

Something to do with the $sel in the if statement.

Here's is an example of what occurs...

http://www.ndorfnz.com/selectupdate.php

with the following code:

<?
/////////////////////////////////////////////////////////////////////
//ECHO OUT THE SELECT NAME DROP DOWN
echo "<SELECT NAME=workout>";
echo "<option value=novalue>Select a Workout...";

//GET THE VALUES FROM THE TABLE FOR THE SELECT BOX
$sql = "SELECT workout_type FROM workout";

//GET THE TOTAL COUNT OF THE RECORDS FROM THE TABLE TO USE FOR THE FOR LOOP
$workout_count = "SELECT COUNT(*) FROM workout";

//QUERY THE SQL
$workout_count_result = mysql_query($workout_count, $connection)
or die ("Couldn't execute query.");

//LOAD THE TOTAL COUNT INTO A VARIABLE TO USE IN THE FOR LOOP
list ($WOCount) = mysql_fetch_row($workout_count_result);
//echo $WOCount;

// execute SQL query and get result
$sql_result = mysql_query($sql,$connection)
or die("Couldn't execute query.");
//$uworkout_type = "Race";
//STEP THROUGH THE FOR LOOP POPULATING THE SELECT BOX BASED ON THE VALUES OF THE ARRAY
for ($i=0; $i<$WOCount; $i++)
{
//LOAD VALUES FROM ARRAY INTO VARIABLE TO ECHO OUT IN THE OPTION VALUES
while ($workout_type = mysql_fetch_array($sql_result))
{
if ($uworkout_type==$workout_type[$i])
{
echo $sel = ' selected';
}
else
{
echo $sel = '';
}
echo '<option value="'.$workout_type[$i].'"'.$sel.'>'.$workout_type[$i].'</option>';
}
}

echo "</SELECT>";

Thanks again!

[tommynanda]

When you look at your HTML source you can find this

<SELECT NAME=workout>
<option value=novalue>Select a Workout... selected
<option value="" selected></option>
<option value="Easy run">Easy run</option>
....
....

Look at the second one its selected and its empty
verify ur DB if your first row is empty... Try to get rid of that empty option with selected in there...

[ndorfnz]

That's It!!!

Thanks much for your help!