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Re: [PHP] Re: select-option link list

Posted by Jochem Maas on 01/19/05 23:23

Ricky Morley wrote:
> 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@operamail.com> 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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