Query options using a drop down menu in PHP?

[smonsivaes]

Hello everyone...im sorta a newb still in PHP but here it goes...

This is basically what id like to do...
Is there a way to create drop down menus with options that a user can select which would then change the value in the MySQL query?

....basically i have a database with 6 tables (parts lists) and would like to create a search page where the user can select which table he would like to search in and then also have another drop down menu where the user would define what he is searching for: Item# or a Description....then this would dynamically change the values for the table and filed in the query?

ex.
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<FORM ACTION="parts2.php" METHOD=POST>
<P>Select Manufacturer:
<P>
<select name="select">
<option>AB_DICK</option>
<option>ITEK</option>
<option>MULTIGRAPHICS_1</option>
<option>MULTIGRAPHICS_2</option>
<option>RYOBI</option>
<option>TOWNSEND</option>
</select>
<P>Select: Item # or Description
<P>
<select name="select2">
<option>ITEM</option>
<option>DESCRIPTION</option>
</select>
<P>&nbsp;
<P>Keywords:<br>
<input type=text name="search" size=30>
<p><input type=submit name="submit" value="Search"></p>
</FORM>

<?

$conn = mysql_connect("host", "user", "pass") or die("Could not connect");

mysql_select_db("T51partsDB",$conn);

$sql = "select ITEM, DESCRIPTION, UOM, QTY_PER_PCKG, LIST_PRICE, NOTES from (option here) where (option here) like '%$search%'";

any help appreciated....

[_SuperString]

yes:

<select name="selectname">
<option value="Table Name">Table Name</option>

and

SQL = "SELECT * FROM $selectname LIKE whatever%";

See what is happening her ?, you refer to the name of the select (form element) as this will hold the selected value.

[smonsivaes]

I understand what your doing with the code now and how it works.

i used your method and changed around my values etc.
now when i run it i get an error:
Warning: Supplied argument is not a valid MySQL result resource in /var/www/html/php/parts2withmenu2.php
any ideas?...here is the code:

<FORM ACTION="parts2.php" METHOD=POST>
<P>Select Manufacturer:
<P>
<select name="selectname">
<option value="ab_dick">AB DICK</option>
<option value="itek">ITEK</option>
<option value="multigraphics_1">MULTIGRAPHICS 1</option>
<option value="multigraphics_2">MULTIGRAPHICS 2</option>
<option value="ryobi">RYOBI</option>
<option value="townsend">TOWNSEND</option>
</select>

<P>&nbsp;
<P>Keywords:<br>
<input type=text name="search" size=30>
<p><input type=submit name="submit" value="Search"></p>
</FORM>

<?

$conn = mysql_connect("localhost", "user", "pass") or die("Could not connect");

mysql_select_db("T51partsDB",$conn) or die ("Could not connect to the Database");

$color1 = "#66CCCC";
$color2 = "#999999";
$newArray_count = 0;

$sql = "select ITEM, DESCRIPTION, UOM, QTY_PER_PCKG, LIST_PRICE, NOTES from $selectname like '%$search%'";

?>

<TABLE BORDER="0" WIDTH="60%" CELLPADDING="1">

<?

$result = mysql_query($sql, $conn);

$number_of_rows = mysql_numrows($result);
if ($number_of_rows == 0) {
echo "No Results Found!";
exit;
}

echo "$number_of_rows matches found for your search of <b>$search</b>!";

echo "<TR><TD><b>Item #</b></td>
<TD><b>Description</b></td>
<TD><b>UOM</b></td>
<TD><b>Qty</b></td>
<TD><b>List Price</b></td>
<TD><b>Notes</b></td>
</tr>";

while ($newArray = mysql_fetch_array($result)) {

$id = $newArray['ID'];
$item = $newArray['ITEM'];
$desc = $newArray['DESCRIPTION'];
$uom = $newArray['UOM'];
$qty = $newArray['QTY_PER_PCKG'];
$price = $newArray['LIST_PRICE'];
$notes = $newArray['NOTES'];

$newArray_color = ($newArray_count%2)?$color1:$color2;

echo "<TR><tr bgcolor= $newArray_color ><TD> $item </TD> <TD> $desc </TD> <TD> $uom </TD> <TD> $qty </TD> <TD> $price </TD> <TD> $notes </TD></TR>";

$newArray_count++;

}

?>

[smonsivaes]

NM got it to work!!!! yippe...Thanks for your help $Super String!!!!