I want to pass the title_id value to another page, how do i make it dynamic in the action statement?
<HEAD>
<TITLE>Sample Form</TITLE>
</HEAD>
<BODY>
<?
$connection = mysql_connect("localhost","root","*******")
or die("Couldn't make connection.");
$db = mysql_select_db("db", $connection)
or die("Couldn't select database.");
$sql = "SELECT title_name, title_id FROM title";
$sql_result = mysql_query($sql,$connection)
or die("Couldn't execute query.");
while ($row = mysql_fetch_array($sql_result)) {
$id = $row["title_id"];
$name = $row["title_name"];
$option_block .= "<OPTION value=http://www.phpbuilder.com/board/archive/index.php/\"$id\">$name</OPTION>";
}
?>
<FORM method="post" action="profile.php?title_id=<? echo $id; ?>">
<P>Operating System:<br>
<SELECT name="Job Type">
<? echo "$option_block"; ?>
</SELECT>
<P><INPUT type="submit" value="submit"></p>
</FORM>
</BODY>
ALSO can someone help me with the correct SQL for the results page
$sql = "SELECT * FROM title WHERE title_id= $title_id";
Is this right
TIA
dom
<HEAD>
<TITLE>Sample Form</TITLE>
</HEAD>
<BODY>
<?
$connection = mysql_connect("localhost","root","*******")
or die("Couldn't make connection.");
$db = mysql_select_db("db", $connection)
or die("Couldn't select database.");
$sql = "SELECT title_name, title_id FROM title";
$sql_result = mysql_query($sql,$connection)
or die("Couldn't execute query.");
while ($row = mysql_fetch_array($sql_result)) {
$id = $row["title_id"];
$name = $row["title_name"];
$option_block .= "<OPTION value=http://www.phpbuilder.com/board/archive/index.php/\"$id\">$name</OPTION>";
}
?>
<FORM method="post" action="profile.php?title_id=<? echo $id; ?>">
<P>Operating System:<br>
<SELECT name="Job Type">
<? echo "$option_block"; ?>
</SELECT>
<P><INPUT type="submit" value="submit"></p>
</FORM>
</BODY>
ALSO can someone help me with the correct SQL for the results page
$sql = "SELECT * FROM title WHERE title_id= $title_id";
Is this right
TIA
dom