Drop Down Menu

cavolks

My drop down menu works, my question is, how do i get a 2nd menu filled with different information from the equipment table and have the selections from both of the drop down menus submit with only one button?

<?php
include("misc.php");

$cxn = mysqli_connect($host,$user,$password,$database)

or die ("couldn't connect to server");
$head = "SELECT * FROM equipment where wear = 'head'";
$result = mysqli_query($cxn,$head) or die ("Couldn't execute query.");

/ Create form containing selection list /
echo "<form action='display.php' method='POST'>\n";
echo "<select name='eq'>\n";
$counter=1;

while($row = mysqli_fetch_assoc($result))

{
extract($row);
echo "<option>$name</option>";

$counter++;

}
?>
<input name='submit' type='submit' value='Submit'>
</select></form>

Scriptmaker

If you only want to call your db once, you can fill a couple of arrays with the data from the name column and the equip column and then build your select menu afterwards.

(For future reference, please note how I used the PHP tags to show this code. This formats the code so that it is easier to read, and doesn't affect phpbuilder's layout)



<?php
include("misc.php");

$cxn = mysqli_connect($host,$user,$password,$database)
or die ("couldn't connect to server");
$head = "SELECT * FROM equipment where wear = 'head'";
$result = mysqli_query($cxn,$head) or die ("Couldn't execute query.");

/* Create form containing selection list */
$counter=1; //why the counter?

$name_arr = array();
$equip_arr[]  = array();

while($row = mysqli_fetch_assoc($result))
{
extract($row);
$name_arr[] = $name;
$equip_arr[] = $equipment;
}

echo "<form action='display.php' method='POST'>\n";
echo "<select name='name'>\n";
foreach ($name_arr as $name){
echo "<option>".$name."</option>";
}
echo "</select>";
echo "<select name='equip'>\n";
foreach ($equip_arr as $equipment){
echo "<option>".$equipment."</option>";
}
echo "</select>;

?>
<input name='submit' type='submit' value='Submit'>
</select></form>

cavolks

Thanks for your reply. I now have two drop down menus and right now they are populating the same information. The information in my database is divided up by wear location(head or feet). I can't figure out how to split these up after I've made my query and put each into their own drop down box.

<?php 
include("misc.php"); 

$cxn = mysqli_connect($host,$user,$password,$database) 
or die ("couldn't connect to server"); 
$query = "SELECT * FROM equipment";  

$result = mysqli_query($cxn,$query) or die ("Couldn't execute query."); 

/* Create form containing selection list */ 
$name_arr = array(); 
$equip_arr = array(); 

while($row = mysqli_fetch_assoc($result))
{ 
extract($row); 
$name_arr[] = $name; 
$equip_arr[] = $name; 
} 
echo "<form action='display.php' method='POST'>\n"; 
echo "<select name='head'>\n"; 
foreach ($name_arr as $name){ 
echo "<option>".$name."</option>";
} 
echo "</select><BR>"; 
echo "<select name='feet'>\n"; 
foreach ($equip_arr as $name){ 
echo "<option>".$name."</option>"; 
} 
echo "</select><BR>"; 

?> 
<input name='submit' type='submit' value='Submit'> 
</select></form>

Scriptmaker

<?php
include("misc.php");

$cxn = mysqli_connect($host,$user,$password,$database)
or die ("couldn't connect to server");
$query = "SELECT * FROM equipment";  

$result = mysqli_query($cxn,$query) or die ("Couldn't execute query.");

/* Create form containing selection list */
$name_arr = array();
$equip_arr = array();

while($row = mysqli_fetch_assoc($result))
{
extract($row);
if ($wear == "head"){
$head_arr[] = $name;
}
elseif ($wear == "feet"){
$feet_arr[] = $name;
}
echo "<form action='display.php' method='POST'>\n";
echo "<select name='head'>\n";
foreach ($head_arr as $head){
echo "<option>".$head."</option>";
}
echo "</select><BR>";
echo "<select name='feet'>\n";
foreach ($feet_arr as $feet){
echo "<option>".$feet."</option>";
}
echo "</select><BR>";

?>
<input name='submit' type='submit' value='Submit'>
</select></form>

note: you don't really need "elseif" if you only have two options (head or feet), you could just use ELSE and not have the stuff in the parenthesis next to it.

cavolks

Thanks for your help. My drop down menus now work. The final step is getting it to display the results on another page, which I have working... but I know i'm not using the best method and I can't figure out how to get it to work.
my query looks like this...

$query = "SELECT * FROM equipment WHERE name='$head'";
  $query1 = "SELECT * FROM equipment WHERE name='$feet'";
  $query2 = "SELECT * FROM equipment WHERE name='$hands'";
  $result = mysqli_query($cxn,$query) or die ("Couldn't execute query.");
  $result1 = mysqli_query($cxn,$query1) or die ("Couldn't execute query.");
  $result2 = mysqli_query($cxn,$query2) or die ("Couldn't execute query.");

and then i have a couple of whiles to display it $result,$result1 ect

while($row = mysqli_fetch_assoc($result))
  {
     extract($row);
     echo "
	 		<tr>\n
			<td>$wear</td>\n
			<td>$head</td>\n
            <td>$hr</td>\n
            <td>$dr</td>\n
			<td>$ac</td>\n
           </tr>\n";
  }

Thanks in advance for the help

bradgrafelman

$query = "SELECT * FROM equipment WHERE name='$head' OR name='$feet' OR name='$hands'";

Though I would also recommend you never use SELECT * and instead name only the columns you're going to use.

Scriptmaker

Make sure you also have some sort of sql injection protection, for example, http://www.php.net/function.mysql-real-escape-string

cavolks

Thanks for your replies.

My next and hopefully final issue is...
I populate two drop down menus, finger1 and finger2 that populate with the same data, but you can make separate selections.
When I go to display the selections using

  while($row = mysqli_fetch_assoc($result))
  {
     extract($row);
     echo "
	 		<tr>\n
			<td>$wear</td>\n
			<td>$name</td>\n
                        </tr>\n";

it goes through and finds finger in my database, and displays it once.

if this isn't clear let me know

cavolks

On my drop down page I have this...

 while($row = mysqli_fetch_assoc($result))
{
extract($row);
if ($wear == "finger"){
$finger_arr[] = $name;
$finger1_arr[] = $name;
}

and

echo "  <tr>\n";
echo "    <td> Finger </td>\n";
echo "    <td>\n";
echo "<select name='finger'>\n";
foreach ($finger_arr as $finger){
echo "<option>".$finger."</option>";
}
echo "</td>\n";
echo "  </tr>\n";
echo "  <tr>\n";
echo "    <td> Finger </td>\n";
echo "    <td>\n";
echo "</select>\n";
echo "<select name='finger1'>\n";
foreach ($finger1_arr as $finger1){
echo "<option>".$finger1."</option>";
}

and on my display page I have...

$query = "SELECT * FROM equipment WHERE 
 name='$finger' OR
 name='$finger1'";

$result = mysqli_query($cxn,$query) or die ("Couldn't execute query.");
while($row = mysqli_fetch_assoc($result))
  {
     extract($row);
     echo "
	 		<tr>\n
			<td>$wear</td>\n
			<td>$name</td>\n
           </tr>\n";
  }

On the select page you can pick two different things for finger and with this code, on the display page, its only displaying one selection. How do I get it to display both selections?

halojoy

you have to do 2 searches
one for the finger selection
and one for the finger1 selection
then you get 2 results
and you display them separate

or you have to test each row result, if it matches finger/finger1
and display accordingly