2 drop down menus (gonna pass up)

alternative

does anyone know how to

2 drop down menus, the 1st ddm will query from database, and then according to the 1st ddm, the second ddm will prompt according to 1st ddm.

| FIRST DDM | X|

like this..

leatherback

Hi,

There are two approaches to this:

You get every combination of ddm1 and ddm2 from your database, and return there in a javascript array/xml objects and load the second select list after the first has been loaded using a javascript (Search this forum for populate, and you will find lots of ideas).

This gives problems when
- The user has java switched off
- You have a lot of potential combiations

The other is:

You select from the dop-down menu.
You submit the page to itself, and do a query on the database for the secons ddm2.

$result = mysql_query("Select ddm2 from database where ddm1= '$ddm1'");
  while ($row = mysql_fetch_array($result))
  {
  $option .= "<option value=\"$row[ddm2]\">
  ";
  }

echo "<select name=\"ddm2\">";
echo $option;
echo "</select>";

alternative

i think this is not what i want

i need to select the ddm2 approaching to the ddm1 data

means when i select 'general' from ddm1
then the ddm2 will prompt out all data under the 'general' category.

kevinsequeira

something like this, i havent tested for parse or syntax errors, but this should point u in the righ direction

<form name="test_form" action="<? echo $PHP_SELF; ?>" method="POST">
<?php
//db connections
if(!isSet($_POST['ddm1'])){
		$ddm1 = $_POST['ddm1'];
		$result = mysql_query("select * from table1");
		echo "<select name=ddm1 onChange=\"javascript:document.test_form.submit();\">";
  while ($row = mysql_fetch_array($result)){
    echo ="<option value=".$row['ddm1'];
 		echo ">".$row['ddm1']."</option>";
  }
		echo "</select>";	
}else{
		$result = mysql_query("select * from table1");
		echo "<select name=ddm1 onChange=javascript:\"document.test_form.submit();\">";
  while ($row = mysql_fetch_array($result)){
    echo ="<option value=".$row['ddm1'];
			if($ddm1 == $row['ddm1'])
				echo " SELECTED";
		 echo ">".$row['ddm1']."</option>";
  }
		echo "</select>";
		$result = mysql_query("Select * from table2 where somecol= $ddm1");
		echo "<select name=ddm2>";
  while ($row = mysql_fetch_array($result)){
   $option .= "<option value=".$row['ddm2'].">".$row['ddm2']."</option>";
  }
		echo "</select>";		
?>
</form>

reg
kevin

alternative

this error occured after i paste ur codes inside

Parse error: parse error, expecting ','' or';'' in c:\phpdev\www\dropdownmenu\ddm6.php on line 9

but i quite dun understand ur coding, would u mind if u correct that error for me?

DeltaRho2K

Find line 9 and change this:

echo ="<option value=".$row['ddm1'];

to this:

echo "<option value=".$row['ddm1'];

Notice how there is no equal sign (=) in the second one.

You will need to do the same in line 17 too... That line is the same, so it should be easy to spot.

alternative

i fixed some of the parser already..but error still occured man

Parse error: parse error in c:\phpdev\www\dropdownmenu\ddm6.php on line 31

<form name="test_form" action="<? echo $PHP_SELF; ?>" method="POST"> 
<?php 
//db connections 
if(!isSet($_POST['ddm1'])){ 
        $ddm1 = $_POST['ddm1']; 
        $result = mysql_query("select * from table1"); 
        echo "<select name=ddm1 onChange=\"java script:document.test_form.submit();\">"; 
  while ($row = mysql_fetch_array($result)){ 
    echo "<option value=".$row['ddm1']; 
        echo ">".$row['ddm1']."</option>"; 
  } 
        echo "</select>";     

}else{ 
        $result = mysql_query("select * from table1"); 
        echo "<select name=ddm1 onChange=java script:\"document.test_form.submit();\">"; 
  while ($row = mysql_fetch_array($result)){ 
    echo "<option value=".$row['ddm1']; 
            if($ddm1 == $row['ddm1']) 
                echo " SELECTED"; 
         echo ">".$row['ddm1']."</option>"; 
  } 
        echo "</select>"; 
        $result = mysql_query("Select * from table2 where somecol= $ddm1"); 
        echo "<select name=ddm2>"; 
  while ($row = mysql_fetch_array($result)){ 
   $option .= "<option value=".$row['ddm2'].">".$row['ddm2']."</option>"; 
  } 
        echo "</select>";         

?> 
</form>

DeltaRho2K

Is there more to this file? because the code you posted only has 30 lines, and it all looks pretty good... I am just wondering if there is other code on your page that might be giving this error... it is a pretty vague error.

alternative

oh..this is the code..
the parser error should be hidden in the center actually..if the error page written that error occured in the last line

this is only the codes. it will connect to the database.

DeltaRho2K

Well, I don't see the problem... YOu will probably have to wait until Kev comes back online, since it is his code... From what I can see, it looks good. Sorry I can't be more help at this point.

alternative

and i change to this

according to my database.

<form name="test_form" action="<? echo $PHP_SELF; ?>" method="POST"> 
<?php 
//db connections 
if(!isSet($_POST['ddm1'])){ 
        $ddm1 = $_POST['ddm1']; 
        $result = mysql_query("select * from category"); 
        echo "<select name=ddm1 onChange=\"java script:document.test_form.submit();\">"; 
  while ($row = mysql_fetch_array($result)){ 
    echo "<option value=".$row['categoryID']; 
        echo ">".$row['cate']."</option>"; 
  } 
        echo "</select>";     

}else{ 
        $result = mysql_query("select * from category"); 
        echo "<select name=ddm1 onChange=java script:\"document.test_form.submit();\">"; 
  while ($row = mysql_fetch_array($result)){ 
    echo "<option value=".$row['categoryID']; 
            if($ddm1 == $row['cate']) 
                echo " SELECTED"; 
         echo ">".$row['cate']."</option>"; 
  } 
        echo "</select>"; 
        $result = mysql_query("Select * from category2 where categoryID= $ddm1"); 
        echo "<select name=ddm2>"; 
  while ($row = mysql_fetch_array($result)){ 
   $option .= "<option value=".$row['category2ID'].">".$row['cate2']."</option>"; 
  } 
        echo "</select>";         

?> 
</form>

alternative

i fix the parser..but the database havent connected yet

        echo "</select>";          

?>  

</form>

------------ to-------------

        echo "</select>";          

}  //add more this
?>  

</form>

alternative

by the way, how to check wat username or password that used by my phpmyadmin?