Validate Selected Option in PHP Only

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kouki0196

I'm looking to validate whether or not a user selects an option from a drop down menu:

<select name="state[]">
<option value="NULL">-- Please select a state --</option>
<option value="AK">AK - Alaska</option>
<option value="AL">AL - Alabama</option>
...
<option value="WY">WY - Wyoming</option>
</select>

How do I check if they leave the default state selected? I really am having a hard time with how PHP interprets selected options (Please no Javascript.)

Thanks!

Paul_Jr

Something like this?

<?php
	if($_POST) {
		if(isset($_POST['state'])) {
			if($_POST['state'] == 'NULL') {
				echo '<p>Please select an option from the select box.</p>';
			}
			else {
				echo '<p>You have selected: <strong>', $_POST['state'], '</strong>.</p>';
			}
		}
	}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
	<fieldset>
		<legend>Please select a state</legend>
		<select name="state">
			<option value="NULL">-- Please select a state --</option>
			<option value="AK">AK - Alaska</option>
			<option value="AL">AL - Alabama</option>
			<option value="WY">WY - Wyoming</option>
		</select>
		<input type="submit" name="submit">
	</fieldset>
</form>

kouki0196

Something like that. Thank you very much.

Out of curiosity, why didn't you name the "name" field in select tag to "state[]"? How would you validate $_POST['state'] if it were stored in an array?

NogDog

Originally posted by kouki0196
Something like that. Thank you very much.

Out of curiosity, why didn't you name the "name" field in select tag to "state[]"? How would you validate $POST['state'] if it were stored in an array?

The string you assign in the HTML form's SELECT element's NAME parameter is simply an identifier that is associated with the value that the user selects. Depending on which method the form uses (POST or GET), PHP adds it to the $POST or $_GET array, using the NAME as the array element key and the VALUE as the array element value.

If you used the [font=courier]<select name="state[]">[/font] syntax from your example, PHP would create an array element called [font=courier]$_POST['state[]'][/font], which I think you would agree would be a bit confusing?

kouki0196

Originally posted by NogDog
The string you assign in the HTML form's SELECT element's NAME parameter is simply an identifier that is associated with the value that the user selects. Depending on which method the form uses (POST or GET), PHP adds it to the $POST or $GET array, using the NAME as the array element key and the VALUE as the array element value.

If you used the [font=courier]<select name="state[]">[/font] syntax from your example, PHP would create an array element called [font=courier]$_POST['state[]'][/font], which I think you would agree would be a bit confusing?

Thanks, NogDog. I appreciate you taking the time to explain this part to me.

So, for sake of simplicity, it is better to use "state" instead of "state[]", right? However, if I wanted to provide multiple selections to the drop down, then "state[]" would become vital since you need to store all of the selected values in the menu, right?

NogDog

If you want multiple selections, in that case your HTML needs the "multiple" keyword, and you do want to use the array-style name (I just learned something new today ):

<select multiple name="state[]">

See this page and this FAQ for a bit more info.

In your example, I guess validation would become more complex:

if(! isset($state) or count($state) == 0 or in_array("NULL", $state))
{
  echo "<p>ERROR: You must select a state.\n";
}

kouki0196

Double Thank you!!!

aries302

This code is working fine for me. But i could not understand of using three if loops in start. Can i use only one if loop instead of three?

if($_POST)
		{
			if(isset($_POST['category'])=='NULL')
			{
				if($_POST['category']=='NULL')
				{
					echo " <font color=red><b>Please select one category</b></font>";
				}else
				{
					if(!empty($_POST['subcat']))
					{			
						$category = $_POST['category'];			
						$subcategory = $_POST['subcategory'];
						if(!empty($subcategory))
						{
							$query1 = "select * from subcategories where catname = '$category' and subcat = '$subcategory'";
							$run = mysql_query($query1);
							$nums = mysql_num_rows($run);
							if($nums > 0)
							{
								echo "<center><font color=red>Sub Category Already Exists.</font></center>";
							}
							else
							{
								$query2 = "insert into subcategories values('$category','$subcategory')";
								mysql_query($query2);
								echo "<center><font color=green>Sub-Category Added.</font></center>";
							}
						}
						else
						{
							echo "<center><font color=red>Please fill Sub-Category field.</font></center>";
						}
					}
				}
			}
		}

NogDog

You can do whatever makes logical sense and works for you.

Certainly in you example it could be cleaned up and simplified a bit to something like:

if(empty($_POST['category']) or $_POST['category'] === 'NULL') {
    // error condition
}
else {
    // continue processing...
}

PS: Please use [noparse]

...

[/noparse] tags around your code in this forum.

Luca_tall

Something like this?

php code:

<select name="favFruit[]" size="4" multiple="">
<?php
$options = array("apple", "banana", "plum", "pomegranate", "strawberry", "watermelon");
foreach ($options as $option) {
echo '<option value="' . $option . '"';
if (in_array($option, $favFruit)) {
echo " selected";
}
echo ">" . ucfirst($option) . "</option>";
}
?>
</select>

DivineMi

Hi Nogdog

I am having serious problems sorting a table according to the users option. I already have a working php code to display the table, I just have problems with the code to display the table with options the user has selected.

The user can select an option of either asc or desc order from a drop box and the column they want arranged in the specified way.

I have been able to develop the following code and keep getting an error message.Below is the code I generated n the error:
<?php
if (isset($_POST['submitted']))

$walkIn = $_POST['WalkIn'];

$walkOut = $_POST['WalkOut'];

$result = mysql_query("SELECT * FROM Results ORDER BY $walkIn $walkOut ")
or die(mysql_error());

the html

<tr>

<td>
Display table in =
<select name="WalkIn">
<option value="Please Choose">Please Choose</option>
<option value="asc">Ascending order</option>
<option value="desc">Descending order</option>
</select>
</td>

<td>

on =
<select name="WalkOut">
<option value="Please Choose">Please Choose</option>
<option value="Goliath">Goliath</option>
<option value="Samson">Samson</option>
</select>
</td>

</tr>

would appreciate a guide in the right direction