drop down form + mysql

mamuni

Hello all. Excuse me if the query is so basic. I am writing a script that allows me to take data from a table myql to a lists drop down form to allow to select a product finally and to show its characteristics (not images) and a Link that will be made up of a fixed URL + a code of product (Web Part naumber) to enter to its corresponding webpage. May be, somebody can help me to follow this? Many7 thanks in advance!!!

Im transcribing the code:

<?PHP
$Host = "";
$User = "";
$Passwd = "";
$DBName = "";
$TableName = "";

$Link = mysql_connect($Host, $User, $Passwd) or die("Could not connect: " . mysql_error());

$Query = "SELECT Distinct manufacturer ".
"FROM $TableName";
$Result = mysql_db_query( $DBName, $Query, $Link );
?>
<HTML>
<HEAD>
<SCRIPT language="JavaScript">
<!--
function BodyLoad() {
var select = document.FormName.manufacturer;
select.options[0] = new Option("Choose One");
select.options[0].value = 0;
<?PHP
$ctr = 1;
While( $Row = mysql_fetch_array($Result) ) {
echo "select.options[" . $ctr . "] = new Option(\"" . $Row["manufacturer"] . "\");\n";
echo "select.options[" . $ctr . "].value = \"" . $Row['model'] . "\";\n";
$ctr++;
}
?>
}

function Fill_Sub() {
var main_select = document.FormName.manufacturer;
var sub_select = document.FormName.model;
if( main_select.options[main_select.selectedIndex].value != 0 ) {
sub_select.length = 0;
}
<?PHP
$Query = "SELECT Distinct manufacturer ".
"FROM $TableName";
$Result = mysql_db_query( $DBName, $Query, $Link );

while( $Row = mysql_fetch_array($Result) ) {
?>
if( main_select.options[main_select.selectedIndex].text == "<?PHP echo $Row[manufacturer]; ?>" ) {
<?PHP
$Query2 = "SELECT model ".
"FROM $TableName ".
"WHERE manufacturer = '$Row[manufacturer]'";
$Result2 = mysql_db_query( $DBName, $Query2, $Link );

$ctr = 0;
While( $Row2 = mysql_fetch_array($Result2) ) {
echo "sub_select.options[" . $ctr . "] = new Option(\"" . $Row2["model"] . "\");\n";
echo "sub_select.options[" . $ctr . "].value = \"" . $Row['model'] . "\";\n";
$ctr++;
}
?>
}

<?PHP
}

mysql_close($Link);
?>
}
-->
</SCRIPT>
</HEAD>
<BODY onload="BodyLoad();">
<FORM name="FormName" method="POST" action="">
<TABLE border="1">
<TR>
<TD>Manufacturer</TD>
<TD>Model</TD>
</TR>
<TR>
<TD>
<SELECT name="manufacturer" onchange="Fill_Sub();"></SELECT>
</TD>

</HTML>

Table:

manufacturer model prefix webpartnumber Suffix Lamp Type Wattage Lamp hours Display Trade Canx
Juan X 15e 1 X 15e Lamp UHP 250 watt, 2000 hour, projector 286,50 y
Pedro X 15i 2 X 15i Lamp UHP 250 watt, 2000 hour, projector 286,50 y
Pedro X 30e 8 X 30e Lamp UHP 250 watt, 2000 hour, projector 286,50
Tito X 30i 7 X 30i Lamp UHP 250 watt, 2000 hour, projector 286,50
Roberto SX 15e 7 SX 15e Lamp UHP 250 watt, 2000 hour, projector 286,50 y

Mamu

Aybabtu

Here goes.

<?PHP 
$Host = ""; 
$User = ""; 
$Passwd = ""; 
$DBName = ""; 
$TableName = ""; 

$Link = mysql_connect($Host, $User, $Passwd) or die("Could not connect: " . mysql_error());

These are your connection variables, you would need to know these. I do not think that you need the $Link variable twice in the code at the top.

$Query = "SELECT Distinct manufacturer ". 
"FROM $TableName"; 
$Result = mysql_db_query( $DBName, $Query, $Link ); 
?> 
<HTML> 
<HEAD> 
<SCRIPT language="JavaScript"> 
<!-- 
function BodyLoad() { 
var select = document.FormName.manufacturer; 
select.options[0] = new Option("Choose One"); 
select.options[0].value = 0; 
<?PHP 
$ctr = 1; 
While( $Row = mysql_fetch_array($Result) ) { 
echo "select.options[" . $ctr . "] = new Option(\"" . $Row["manufacturer"] . "\");\n"; 
echo "select.options[" . $ctr . "].value = \"" . $Row['model'] . "\";\n"; 
$ctr++; 
} 
?> 
}

This is a section of hybred code and I do not pretend to fully understand javascript. What this section does is take all of the results it gets from the query and outputs options and values in the while statement. So basically as long as there is data in $Row then output options and values. $Row['model'] and $Row['manufacturer'] are what the current values are for the row of data it is on in the array.

<?PHP 
} 

mysql_close($Link); 
?> 
} 
--> 
</SCRIPT> 
</HEAD> 
<BODY onload="BodyLoad();"> 
<FORM name="FormName" method="POST" action=""> 
<TABLE border="1"> 
<TR> 
<TD>Manufacturer</TD> 
<TD>Model</TD> 
</TR> 
<TR> 
<TD> 
<SELECT name="manufacturer" onchange="Fill_Sub();"></SELECT> 
</TD> 

<TD> 
<SELECT name="model" onchange="submit();"></SELECT> 
</TD> 
</TR> 
</TABLE> 
</FORM> 
</BODY> 
</HTML>

Closes the connection to the database and closes all tags that are still open. It also does some javascript in there but I am not sure exactly what it does. It looks like it populates a feild when you select an option from a drop down.

Hope this helped.

Aybabtu.

mamuni

Aybabtu.

Many thanks!
Now.. how can I show the rest of the data of this model on the same or another page?

Best regards

Mamu

Aybabtu

Well it should pass the manfacutrer and model data VIA post to whatever php page you specify in the "action" of the form so lets say you put action="http://your.website.com/display.php" in the form tag. Then in display.php you would use $POST['manufacturer'] and $POST['model'] to get what was passed. so you could use the following code on display.php

<? echo $_POST['manufacturer']." makes the ".$_POST['model']; ?>

So lets say that they picked Ford as a manufacturer and Mustang as the model. This would result:

Ford makes the Mustang

Note: In the form method in this example they used POST if they use GET then you can use $_GET['model'] and so on.

Let me know if this helps.

Aybabtu.

mamuni

Aybabtu:

Many many thanks...

Yes you are helping me a lot.. but I must be doing something wrong.

Please could you debug my code.. (I know I am not a well coder) sorry!! 🙁

Code:

<?PHP
$Host = "";
$User = "";
$Passwd = "";
$DBName = "";
$TableName = "";

$Link = mysql_connect($Host, $User, $Passwd) or die("Could not connect: " . mysql_error());

	function Fill_Sub() {
		var main_select = document.FormName.manufacturer;
		var sub_select = document.FormName.model;
		if( main_select.options[main_select.selectedIndex].value != 0 ) {
			sub_select.length = 0;
		}
		<?PHP
			$Query = "SELECT Distinct manufacturer ".
					 "FROM $TableName";
			$Result = mysql_db_query( $DBName, $Query, $Link );

			while( $Row = mysql_fetch_array($Result) ) {
		?>
		if( main_select.options[main_select.selectedIndex].text == "<?PHP echo $Row[manufacturer]; ?>" ) {

<?PHP
$Query2 = "SELECT model, prefix, webpartnumber, suffix, lamptype, wattage, lamphours, display, trade ".
"FROM $TableName ".
"WHERE manufacturer = '$Row[manufacturer]' ";
$Result2 = mysql_db_query( $DBName, $Query2, $Link );

<?PHP
}

mysql_close($Link);
?>
}
-->
</SCRIPT>
</HEAD>
<BODY onload="BodyLoad();">

<FORM name="FormName" method="POST" action="http://www.hotav.uk.com/testdemo/testdemo2.php">

<TABLE border="1">
<TR>
<TD>Manufacturer</TD>
<TD>Model</TD>
</TR>
<TR>
<TD>
<SELECT name="manufacturer" onchange="Fill_Sub();"></SELECT>
</TD>

			<TD>
                <SELECT input name="model" onchange="this.form.submit()"></SELECT>
			</TD>
		</TR>
	</TABLE>

cyberlew15

its the "FROM $tablename"; part that is stopping the mysql query it just won't work try

$Query = "SELECT Distinct manufacturer "."FROM ".$TableName;

This will allow this query to work. I know sometimes it's really annoying to have worked so hard on the code only to have a slight error or small typo screw everything up for you 🙂

mamuni

cyberlew

Many any thanks!!!

You are right.

Do I have to change this on all the script ? 😕
I am asking you because it is not passing the variables to the other php.

Sorry but I am going to paste the code again... 😃

$Link = mysql_connect($Host, $User, $Passwd) or die("Could not connect: " . mysql_error());

$Query = "SELECT Distinct manufacturer "."FROM ".$TableName;

$Result = mysql_db_query( $DBName, $Query, $Link );

?>
<HTML>
<HEAD>
<SCRIPT language="JavaScript">
<!--
function BodyLoad() {
var select = document.FormName.manufacturer;
select.options[0] = new Option("Choose One");
select.options[0].value = 0;
<?PHP
$ctr = 1;
While( $Row = mysql_fetch_array($Result) ) {
echo "select.options[" . $ctr . "] = new Option(\"" . $Row["manufacturer"] . "\");\n";
echo "select.options[" . $ctr . "].value = \"" . $Row['model'] . "\";\n";
$ctr++;
}
?>
}

function Fill_Sub() {
var main_select = document.FormName.manufacturer;
var sub_select = document.FormName.model;
if( main_select.options[main_select.selectedIndex].value != 0 ) {
sub_select.length = 0;
}
<?PHP
$Query = "SELECT Distinct manufacturer "."FROM ".$TableName;
$Result = mysql_db_query( $DBName, $Query, $Link );

while( $Row = mysql_fetch_array($Result) ) {
?>
if( main_select.options[main_select.selectedIndex].text == "<?PHP echo $Row[manufacturer]; ?>" ) {
<?PHP
$Query2 = "SELECT model, prefix, webpartnumber, suffix, lamptype, wattage, lamphours, display, trade ".
"FROM $TableName ".
"WHERE manufacturer = '$Row[manufacturer]' ";
$Result2 = mysql_db_query( $DBName, $Query2, $Link );

<?PHP
}

mysql_close($Link);
?>
}
-->
</SCRIPT>

</HEAD>
<BODY onload="BodyLoad();">

<FORM name="FormName" method="POST" action="http://www.hotav.uk.com/testdemo/testdemo2.php">

<TABLE border="1">
<TR>
<TD>Manufacturer</TD>
<TD>Model</TD>
</TR>
<TR>
<TD>
<SELECT name="manufacturer" onchange="Fill_Sub();"></SELECT>
</TD>

			<TD>
                <SELECT input name="model" onchange="this.form.submit()"></SELECT>
			</TD>
		</TR>
	</TABLE>

cyberlew15

is this for projectors I just got myself one with 3000hours in ECO mode strange thing tho pron looks weird. but thats neither here nor there could you attach as files then I will sandbox test and reattach to another message.

mamuni

Yes. Is for a projector's site.
Here you have the files.

Many thanks in advance. And good luck with your new proyector.

Mamuni