Many Buttons on 1 form

Mitch · October 25th, 2003, 09:13 AM

I have made a form that has 2 dropdowns, I want to have a button for each of 4 different results pages that all use the selections from those dropdowns as criteria for queries, but I can't get any of them to work.

How do I write the post statements so that each button posts to a different page?

Below is what I have thus far, which opens the correct page, but that page always says no data was passed.



<select name="searchteam">
<?php
$query = mysql_query('select TeamID, TeamName, Nickname, TeamCity from PH_TEAMS') or die
(mysql_error());
while($row = mysql_fetch_array($query)) {
echo("<option value=\"${row['TeamID']}\">

${row['TeamName']}, ${row['Nickname']}, ${row['TeamCity']}\n");
}
?>
</select>

<select name="searchseason">

<?php
$qSEASON = mysql_query('select SeasonID, SeasonName from PH_SEASON') or die
(mysql_error());
while($row2 = mysql_fetch_array($qSEASON)) {
echo("<option value=\"${row2['SeasonID']}\">

${row2['SeasonName']}\n");
}
?>
</select>
<!-- **These are the buttons I want and the pages to send data to**
<FORM action="goal_rankings_new.php" method="POST">
<INPUT type=submit name="Goals" value=" Goals ">
</FORM>

<FORM action="penalties_rankings.php" method="POST">
<INPUT type=submit name="Penalties" value=" Penalties">
</FORM>

<FORM action="button3.php" method="POST">
<INPUT type=submit name="button3" value=" Button3 ">
</FORM>

joefawcett · October 25th, 2003, 09:25 AM

Basically you need four standard buttons. In the only click of each have:
[code]
doSubmit(this.form, 'target.htm');
[code]
Replace "target.htm" by the page to submit to. This is your doSubmit function:

Code:

function doSubmit(Form, Action)
{
  Form.action = Action;
  Form.submit();
}

--

Joe

Mitch · October 25th, 2003, 10:56 AM

Thanks, but what is a "standard" button? Is it:
<input type="button" value="Button" name="B3">

Jonax · October 25th, 2003, 11:21 AM

Yep, end the 'only click' is onClick...

<input type="button" value="Button" name="B3" onClick="doSubmit(this.form, 'target.htm');">

Mitch · October 25th, 2003, 11:28 AM

I see the function code on my page in the browser, do I need to enclose it in some sort of tag so it is not treated as text? Right now I have it in the header?

Jonax · October 25th, 2003, 11:38 AM

Yup:

Code:

<head>
<script language=javascript>
...
</script>
</head>

Mitch · October 25th, 2003, 11:47 AM

Thought so. Now the code is not showing up on the page.

But, I get a "error on page" when I click on it.

Here is where the page is: http://www.pennhockey.org/stats/test_buttons_2.php

Here is what I am using:

<HEAD>

<script language=javascript>

function doSubmit(Form, Action)
{
  Form.action = Action;
  Form.submit();
}
</SCRIPT>

<TITLE>PENN Hockey Club - Scoreing Leaders</TITLE>

<p align="center">
<?
include ("http://www.pennhockey.org/header.php");
?>

<h1 align="center">Select a Team to View Goal Rankings</h1>

<?
$connection = mysql_connect("xxxx", "xxxx", "xxxx");
$fields = mysql_list_fields("xxxx", "PH_PLAYERS", $connection);
?>


<select name="searchteam">

<?php
$query = mysql_query('select TeamID, TeamName, Nickname, TeamCity from PH_TEAMS') or die
(mysql_error());
while($row = mysql_fetch_array($query)) {
    echo("<option value=\"${row['TeamID']}\">

${row['TeamName']}, ${row['Nickname']}, ${row['TeamCity']}\n");
}
?>
</select>

<select name="searchseason">

<?php
$qSEASON = mysql_query('select SeasonID, SeasonName from PH_SEASON') or die
(mysql_error());
while($row2 = mysql_fetch_array($qSEASON)) {
    echo("<option value=\"${row2['SeasonID']}\">

${row2['SeasonName']}\n");
}
?>
</select>

<input type="button" value="Goals" name="B3" onClick="doSubmit(this.form, 'goal_rankings_new.php');">

<input type="button" value="Penalties" name="B3" onClick="doSubmit(this.form, 'penalties_rankings.php');">

</form>

<p align="center">
<?
include ("http://www.pennhockey.org/footer.php");
?>
************************

Any ideas?

Jonax · October 25th, 2003, 11:55 AM

It seems to me that you have commented out the opening <form>-tag ?

Mitch · October 25th, 2003, 12:03 PM

I went back to an older version and inserted the new stuff and it worked. I must have done just that.

Here is the code that worked:

<HEAD>

<script language=javascript>

function doSubmit(Form, Action)
{
  Form.action = Action;
  Form.submit();
}
</SCRIPT>

<TITLE>PENN Hockey Club - Stats</TITLE>

<p align="center">
<?
include ("http://www.pennhockey.org/header.php");
?>

<h1 align="center">Select a Team and Season to View Stats</h1>

<?
$connection = mysql_connect("xxxxx", "xxxx", "xxxxx");
$fields = mysql_list_fields("xxxxx", "PH_PLAYERS", $connection);
?>

<form method="post" action="goal_rankings_new.php">
<select name="searchteam">

<?php
$query = mysql_query('select TeamID, TeamName, Nickname, TeamCity from PH_TEAMS') or die
(mysql_error());
while($row = mysql_fetch_array($query)) {
    echo("<option value=\"${row['TeamID']}\">${row['TeamName']}, ${row['Nickname']}, ${row['TeamCity']}\n");
}
?>
</select>

<input type=submit value="Select">

<select name="searchseason">

<?php
$qSEASON = mysql_query('select SeasonID, SeasonName from PH_SEASON') or die
(mysql_error());
while($row2 = mysql_fetch_array($qSEASON)) {
    echo("<option value=\"${row2['SeasonID']}\">${row2['SeasonName']}\n");
}
?>
</select>

<input type=submit value="Select">

<input type="button" value="Goals" name="B3" onClick="doSubmit(this.form, 'goal_rankings_new.php');">

<input type="button" value="Penalties" name="B4" onClick="doSubmit(this.form, 'penalties_rankings.php');">

</form>

<p align="center">
<?
include ("http://www.pennhockey.org/footer.php");
?>

Mitch · October 25th, 2003, 12:08 PM

Thanks everyone for your help! Now my players can see their stats. (I'm a high school hockey coach, who just happens to like to program, albeit web programming is new to me.)