Variable Form Question

thedilleyo

I have this form:

<FORM action="vote.php" method="post">
<P>
<SELECT name="sname">
<OPTION selected>Please Select</OPTION>
<OPTION value="xx">xx</OPTION>
<OPTION value="xx">xx</OPTION>
<OPTION value="xx">xx</OPTION>
<OPTION value="xx">xx</OPTION>
<OPTION value="xx">xx</OPTION>
</SELECT>
<BR>
<TABLE>
<TR>
</TR>
<TR>
<TD>
<INPUT type="submit" value="Submit">
</TD>
<TD>
<INPUT type="reset" value="Reset">
</TD>
</TR>
</TABLE>

</FORM>

I'm using a php script to process multiple forms and I want to make the name of the form a variable to pass in so I don't have to write 5 php scripts. Here's my php code:

$db = mysql_connect ("xxx", "xxx", "") or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db ("xxx");

	// process form

	$db = mysql_connect("localhost", "root");

	mysql_select_db("mydb", $db);

	$fname = I NEED A FORM VARIABLE NAME

	$sql = "INSERT INTO $fname (flname) VALUES (WHAT IS SELECTED ON THE DROP DOWN FORM NO MATTER WHAT FORM IT IS!!)";

	$result = mysql_query($sql);

if($result)
{
echo "Thank you! Vote Submited.\n";
}

All help would be great!!! Thanks in advance.

matt_4013

Hi there,

I'd just use a hidden field in your form, with the name in it.

<FORM action="vote.php" method="post"> 
<P> 
<SELECT name="sname"> 
<OPTION selected>Please Select</OPTION> 
<OPTION value="xx">xx</OPTION> 
<OPTION value="xx">xx</OPTION> 
<OPTION value="xx">xx</OPTION> 
<OPTION value="xx">xx</OPTION> 
<OPTION value="xx">xx</OPTION> 
</SELECT> 
<BR> 
<TABLE> 
<TR> 
</TR> 
<TR> 
<TD> 

<!-- This is what I added -->
<input type="hidden" name="fname" value="name">
<!-- End -->

<INPUT type="submit" value="Submit"> 
</TD> 
<TD> 
<INPUT type="reset" value="Reset"> 
</TD> 
</TR> 
</TABLE> 

</FORM>

Then, set the variable from the posted form:

<? 

$db = mysql_connect ("xxx", "xxx", "") or die ('I cannot connect to the database because: ' . mysql_error()); 
mysql_select_db ("xxx"); 

// process form 

$db = mysql_connect("localhost", "root"); // my suggestion is to not use root, set up a different user with only needed privileges

mysql_select_db("mydb", $db); 

// Make sure they haven't tried to screw up your database by sending false post information
$fname = addslashes(trim($_POST['fname']));

$sql = "INSERT INTO $fname (flname) VALUES (WHAT IS SELECTED ON THE DROP DOWN FORM NO MATTER WHAT FORM IT IS!!)"; 

$result = mysql_query($sql); 

if($result) 
{ 
echo "Thank you! Vote Submited.\n"; 
} 

?>

thedilleyo

How do I INSERT the value of my drop down box with my sql command ($sql = "INSERT INTO $fname (flname) VALUES (??)") sname and fname should be the same names, but when I do that it enters in the value of the hidden input and not what is chosen in the drop down box? How do I get the INSERT command to insert what's selected in the drop down box and not what's in the value of the hidden input?