Viewing Thread:
"PHP: Form submission problem"

You are here: Freeola's Archived Forums→Freeola Customer Forum→PHP:

Form submission problem

Hello,

I'm having a problem with the following code.
The current code works great, when an option is chosen from the 'referrer' drop down list it is returned selected if any other fields are blank. (I have if statements handling whether any fields are blank i.e. 'name' when the form is submitted)


<?php
$sel_referrer .= "
<option value=\"\"></option>";

$get_referrers = mysql_query("SELECT * FROM advertisers WHERE hide = '1'");
$i = 0;
$total_referrers = mysql_num_rows($get_referrers);

while($i < $total_referrers) {
$selected = ($_POST['referrer'] == mysql_result($get_referrers,$i,"id"))?" selected":"";
$sel_referrer .= "
<option value=\" ".mysql_result($get_referrers,$i,"id")."\" $selected> ".stripslashes(mysql_result($get_referrers,$i," advertiser"))." </option>";
$i++;
}
?>



<form action="index.php" method="post">
<input type="text" name="name" value="<? $_POST[name] ?>">
<select name="referrer"><? echo $sel_referrer ?></select>
<input type="submit" name="Submit" value="Submit Enquiry">
</form>


However, when changing the to a 'multiple' listing (shown below), the form is not returning mutliple selected options from the list, just one option that is selected.
I've tried a few different changes to the php code above but nothing seems to be working.


<form action="index.php" method="post">
<input type="text" name="name" value="<? $_POST[name] ?>">
<select name="referrer" [B]multiple[/B]><? echo $sel_referrer ?></select>
<input type="submit" name="Submit" value="Submit Enquiry">
</form>


Any ideas would be greatly appreciated.

Regards

Drazen

 

Report

Thu 08/11/07 at 21:51

Drazen

Regular

"LFXX98763X"

Posts: 7

Hello,

I'm having a problem with the following code.
The current code works great, when an option is chosen from the 'referrer' drop down list it is returned selected if any other fields are blank. (I have if statements handling whether any fields are blank i.e. 'name' when the form is submitted)


<?php
$sel_referrer .= "
<option value=\"\"></option>";

$get_referrers = mysql_query("SELECT * FROM advertisers WHERE hide = '1'");
$i = 0;
$total_referrers = mysql_num_rows($get_referrers);

while($i < $total_referrers) {
$selected = ($_POST['referrer'] == mysql_result($get_referrers,$i,"id"))?" selected":"";
$sel_referrer .= "
<option value=\" ".mysql_result($get_referrers,$i,"id")."\" $selected> ".stripslashes(mysql_result($get_referrers,$i," advertiser"))." </option>";
$i++;
}
?>



<form action="index.php" method="post">
<input type="text" name="name" value="<? $_POST[name] ?>">
<select name="referrer"><? echo $sel_referrer ?></select>
<input type="submit" name="Submit" value="Submit Enquiry">
</form>


However, when changing the

to a 'multiple' listing (shown below), the form is not returning mutliple selected options from the list, just one option that is selected.
I've tried a few different changes to the php code above but nothing seems to be working.


<form action="index.php" method="post">
<input type="text" name="name" value="<? $_POST[name] ?>">
<select name="referrer" [B]multiple[/B]><? echo $sel_referrer ?></select>
<input type="submit" name="Submit" value="Submit Enquiry">
</form>


Any ideas would be greatly appreciated.

Regards

Drazen

Change your select tag to
<select name="referrer[]" multiple>

You'll then find $_POST['referrer'] is an array of your selected items.

Thanks Garin, I've done that and have also added the following to the php code, not sure if its correct, but seems to work to a certain extent:


$selected = ($_POST['referrer'][B][$i][/B] == mysql_result($get_referrers,$i,"id"))?" selected":"";


This returns more than one selected option form the list but only if you have selected all options in order.
E.G. If you select option 1, 2, 4 from the referrer list (missing 3), it returns the form with 1, 2 selected but not 4.

If you arent sure whats going on, put print_r($_POST); at the top of your script and see what the array contains.

It looks like you're mixing your variables incorrectly there anyway. $i is the current mysql row you're on. That value has no relevance in the $_POST['referrer'] array you've got, its just a list of the values that were selected. The index in the array doesnt correspond to the index on your mysql result set.

Not tested or anything but I suspect what you really want is something like

$selected = (in_array( mysql_result($get_referrers, $i, "id"),$_POST['referrer']))?" selected":"";

Basically you need to check if the record id appears anywhere in that array.