selectbox + textbox + submitbutton

rix

Hi!:bemused:
i'm trying for hours to put my forms together on one page without the expected result.
there should be one "textbox" and one "select box" wich values should be inserted to the Product table on "submit". How ever I know thats a simple thing but for me a bloody newbie is this a big challenge. thx for giving attention to my prob's!!!
I wish you a nice day!

First table "ProdGrp"

Id | ProdGrpName

1 | GroupA
2 | GroupB
3 | GroupC

the second one is "Product"

Id | Name | ProdGrpName

1 | Item1 | GroupA
2 | Item2 | GroupB

<?php
if ($HTTP_POST_VARS['submit']) {
mysql_connect("mysql.database.xx", "user", "passwd");
mysql_select_db("db1");

$ProdGrpName=$HTTP_POST_VARS['ProdGrpName'];
$Id=$HTTP_POST_VARS['Id'];
$query ="INSERT INTO ProdGrp (ProdGrpName)";
$query.=" VALUES ('$ProdGrpName','$myselect')";
$result=mysql_query($query);
$res=mysql_query("SELECT ProdGrpName FROM ProdGrp");

if ($result) echo "<b>Eintrag erfolgreich übernommen!</b>";
else echo "<b>Fehler: Datensatz nicht eingefügt.</b>";

echo "<select name=myselect>";
while($row=mysql_fetch_assoc($res)) {
  echo "<option value=$row[Id]>$row[ProdGrpName]</option>";
}
echo "</select>";
}
?>
<form method="POST" action="<? echo $PHP_SELF ?>">
Produktgruppe Hinzuf&uuml;gen
  <input name="Id" type="hidden" value="" size="4">
<br>
<input name="ProdGrpName" type="text" size="25" maxlength="50">
<input type="submit" name="submit" value="Submit">
</form>
[code=php]

jernhenrik

Just to get this straight ..

a) you wish to display a form with a select box (with the different product groups) and a text box into which you'll type a product name, and when this is done hit the submit button and put the product and productgroup into the product table? And then display the form again.

b) display a form with selectbox (product grps) and text box into which you type a productgroup and when submitting update the productgroup table?

rix

hey thanks a lot!

the option a) is exactly what I want to do!

rix!

jernhenrik

Try and have a look at this - please note i've change your product database.

First table "prodgrp"

id | grpname

1 | GroupA
2 | GroupB
3 | GroupC

the second one is "Product"

Id | GrpId | Name

1 | 1 | Item1
2 | 1 | Item2

<?php 
	//connect to database:
	//it's always a good idea to use 'or die' to check if you succeed
	mysql_connect("localhost", "root", "mypass") or die(mysql_error());
	mysql_select_db("product") or die(mysql_error()); 

//check if the button has been pressed
if (isset($HTTP_POST_VARS['submit'])) 
{ 
	//let's just check if there is anything in the textbox submitted
	if(isset($HTTP_POST_VARS['prodname']) && strlen($HTTP_POST_VARS['prodname']) > 0)
	{
		$prodname = $HTTP_POST_VARS['prodname'];
		$prodgrp = $HTTP_POST_VARS['prodgrp'];

		$qry = "INSERT INTO product VALUES ('',$prodgrp,'$prodname')"; 
		$res = mysql_query($qry) or die("<b>Fehler: Datensatz nicht eingefügt.</b>".mysql_error()); 
		echo "<b>Eintrag erfolgreich übernommen!</b>"; 
	}else
	{
		echo "<b>Fill out product name!!</b>"; 
	} 

}

?> 
	<form method="POST" action="<? echo $HTTP_SERVER_VARS['PHP_SELF']; ?>"> 
		Produktgruppe Hinzufügen:<br/> 
		<?
			$qry = "SELECT id, grpname FROM prodgrp";
			$res = mysql_query($qry) or die(mysql_error()); 

		echo "<select name=prodgrp>"; 
		while($row=mysql_fetch_assoc($res)) 
		{ 
			echo "<option value=".$row['id'].">".$row['grpname']."</option>"; 
		} 
		echo "</select><br/>"; 
	?>
	<input name="prodname" type="text" size="25" maxlength="50"> <br/>
	<input type="submit" name="submit" value="Submit"> 
</form>

This is just to get you started, try and compare this to your old code.

rix

thanks a lot! Ihave solved this prob.
have a nice day!