Like this. Here's the html code. I'm sure you can convert to the appropriate PHP
this is totally not PHP but...
<html> <body> <form> <select size="1" name="site" onChange="window.location=this.value">
shouldn't that be something like:
<select size="1"
name="site"
onChange="if
(this.options[this.selectedIndex].value) {
window.location=this.options[this.selectedIndex].value;
}">which is evil and crude, but does 2 things: uses the selectedIndex to get the value of the selection options and only sets the window location is the value if not empty.
<option value="http://www.google.com";>Google</option> <option value="http://www.yahoo.com";>Yahoo</option> <option value="http://www.microsoft.com";>MS</option> </select> </form> </body> </html>
On Wed, 19 Jan 2005 20:10:25 +0200, William Stokes <kalles@xxxxxxxxxxxxx> wrote:
Hello,
Hope someone can give some directions...
I'm trying to create a list box of links to other pages. I just can't figure
out how to move to the other page whe user selects one of the options in the
list box.
here's the code...
$sql="SELECT team_name FROM teams"; $result=mysql_query($sql); $num = mysql_num_rows($result); $cur = 1; print "<form name=\"form1\" method=\"post\" action=\"select_team.php\">"; print "<select name=\"select_team\" size=\"1\" >"; while ($num >= $cur) { $row = mysql_fetch_array($result); $team_name = $row["team_name "]; print "<option>$team_name </option>"; $cur++; } print "</select>"; print "<input type=\"submit\" value=\"Go!\">"; print "</form>";
This propably can be don without the "Go" button. But i Don't know how...
Thank a lot -Will BTW thanks for the debugging stuff earlier. Got that workin...
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