Dynamic Drop - display selected

[bradw]

I am modifying a shopping cart with the following. After they select an option the cart automatically updates and refreshes the page but does not show the selected value. I have had no luck echoing back the selected form the database. Is it possible to echo this from same page. Sessions are used by the cart.

Here is the code:

<select name="radio" id="RATE" onclick="updateshiprate(this,0)" >
<option value=""><strong>View Delivery Zip Codes</strong></option>
<?php do { ?><option value="RATE" <?php if (!(strcmp("RATE", $row_rsZipCode['zip_code']))) {echo "selected=\"selected\"";} ?>><?php echo $row_rsZipCode['zip_code']; ?></option><?php } while ($row_rsZipCode = mysql_fetch_assoc($rsZipCode)); ?>
</select>

<?php echo $row_rsZipCode['zip_code']; ?>

[etully]

I have had no luck echoing back the selected form the database.

I'm not trying to be picky or a grammar Nazi... but I have no idea what you were trying to say here. I simply don't know what you are trying to accomplish.

[bradw]

That's what happens when I try and work around screaming kids .

Simply Put.

When a user selects an option from the dropdown, the form is submitted and immediately returned to the page but the selected option is not listed. In other words the dropdown reverts to its original state. This will obviously cause confusion with the using wondering if the correct value was selected.

Hope this clears it up.

[etully]

do { 
     ?><option value="RATE" <?php 
     if (!(strcmp("RATE", $row_rsZipCode['zip_code']))) {
         echo "selected=\"selected\"";
     } ?>><?php 
     echo $row_rsZipCode['zip_code']; ?></option><?php 
} 
while ($row_rsZipCode = mysql_fetch_assoc($rsZipCode)); ?>

Ok. So this part of the code loops through all the rows returned by MySQL and provides all the options in the select drop down menu like this:

<option value="RATE">90210</option>
<option value="RATE">90211</option>
<option value="RATE">90212</option>
<option value="RATE">90213</option>
<option value="RATE">90214</option>
<option value="RATE">90215</option>

And that doesn't make any sense... is it possible that you modified the PHP before pasting it here? (It doesn't make sense because no matter which Zip Code the user chooses, the value is always the same... the word "RATE"... and that's pretty useless).

In general, here's how to do what you're trying to do:

$_REQUEST['drop_down_name'] is going to contain the VALUE of the OPTION that the user selected.

For example, if the user chooses zip code 90213 from the select statement below, then $_REQUEST['zip_xyz_field'] will be set to "4".

<select name="zip_xyz_field">
<option value="1">90210</option>
<option value="2">90211</option>
<option value="3">90212</option>
<option value="4">90213</option>
<option value="5">90214</option>
<option value="6">90215</option>
</select>

So what you want to do is this: As you are looping through all the database values and printing out each OPTION tag, you want to check to see if $_REQUEST['drop_down_name'] is equal to the VALUE that you are currently displaying and IF SO, then you want to print the word, "selected" because: The value that the user chose is the same as the value of the option that we are currently displaying. Like this:

do {
    echo "<option value='" . $row_rsZipCode['cost'] . "'";
    if ($_REQUEST['zip_xyz_field']==$row_rsZipCode['cost']) {
        print " selected";
    }
    echo ">".$row_rsZipCode['zip_code']."</option>";
} 
while ($row_rsZipCode = mysql_fetch_assoc($rsZipCode));

You can use this as a model where the IF statement simply compares the value passed in by the user to the value of the current option.