Passing dropdown list value to another SELECT statement on same page
Hi and thanks for looking at this with me. I am COMPLETELY new to using PHP to run MySQL select statements. That being said, I have managed to run a SELECT statement to populate a drop down list...and another SELECT statement to populate an HTML table. (this is for a roleplaying game)
But this is where3 I get stuck...
I would like for the dropdown selected value to be the "WHERE racename = " value in the second select statement that populates the table so that only one row is returned instead of all the data.
Here's the page: [gamehermit.com...]
Here's my code so far:
<?php
// Make a MySQL Connection
mysql_connect("localhost", "db_username", "password") or die(mysql_error());
mysql_select_db("db_name") or die(mysql_error());
$query="SELECT * FROM Races";
$result = mysql_query($query);
echo "<select name=racename>";
while($nt=mysql_fetch_array($result))
{
if ($nt[racename]==$_POST["racename"])
$selected="selected";
else
$selected="";
echo "<option ".$selected."value=$nt[racename]>$nt[racename]</option>";
}
echo "</select>";
echo "<br />";
// Get all the data from the "Race" table and create table
$result2 = mysql_query("SELECT * FROM Races")
or die(mysql_error());
echo "<table border='1'>";
echo "<tr> <th>Race Name</th> <th>Might Modifier</th> <th>Valor Modifier</th> <th>Deftness
Modifier</th> <th>Insight Modifier</th> <th>Dweomer Modifier</th> </tr>";
// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result2 )) {
// Print out the contents of each row into a table
echo "<tr><td>";
echo $row['racename'];
echo "</td><td>";
echo $row['modmight'];
echo "</td><td>";
echo $row['modvalor'];
echo "</td><td>";
echo $row['moddeftness'];
echo "</td><td>";
echo $row['modinsight'];
echo "</td><td>";
echo $row['moddweomer'];
echo "</td></tr>";
}
echo "</table>";
?>
I hope this is simple...thanks so much :)
~ Jack
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Passing dropdown list value to another SELECT statement on same page
August 01, 2012 07:20AM
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