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Passing Parm in URL with PHP

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USADarts

USADarts

Programmer
Dec 20, 2004
21
US
First, is this correct to pass the parms name and nameid?
<form action="tournaments.php?$name&$nameid" method="POST">

Second, on the tournament.php page, how to I go about retrieving the name and nameid value to use?
I have tried:
var pname = getParameter("name");
var pnameid = getParameter("nameid");

Not working

Thanks
David

 
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<form action="tournaments.php?mname=$name&nameid=$nameid" method="POST">

Known is handfull, Unknown is worldfull
 
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Receiving the following error:

Parse error: parse error, unexpected T_VAR in /home/hascupd/public_html/members/tournaments.php on line 9


Line 9 code is this:
var pname = getParameter("name");
 
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Still Getting same error:
Parse error: parse error, unexpected T_VAR in /home/hascupd/public_html/members/tournaments.php on line 9


tournaments.htm
<form action="tournaments.php?name=$name&nameid=$nameid" method="POST">


tournaments.php
<?php
var $pname = getParameter("name");
var $pnameid = getParameter("nameid");
....
 
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Are you invoking "var" outside of a class definition? That's a no-no.

If you want to instantiate a bare variable and assign it a value, just do:

$pname = getParameter("name");




Want the best answers? Ask the best questions!

TANSTAAFL!!
 
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Thank you very much for the help. I am fairly new at some of the PHP procedures. However, I am know receiving the following error:

Fatal error: Call to undefined function: getparameter() in /home/hascupd/public_html/members/tournaments.php on line 9

 
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Is getParameter your own function or is it just what you think it should be. I see no reference of that function in the manual. Have you tried?
Code: 
$pname = $_POST["name"];
 
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Please forgive me if there is something I'm not seeing here..
(Very likely(Sincerely)).

But I think USADarts is trying to parse data in an .htm file.
If you want values there you have to parse it thru php
Like:

tournamentsa.php
Code: 
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<html>
<head>
  <title></title>
  
</head>
<body>
<?php
$name = "Fred";
$nameid = 15;

echo "<form action=\"tournamentsb.php?name={$name}&nameid={$nameid}\" method=\"post\">\n";
echo "<input type=\"hidden\" name=\"name\" value=\"{$name}\">\n";
echo "<input type=\"hidden\" name=\"nameid\" value=\"{$nameid}\">\n";
?>

<input type="submit">
</form>
</body>
</html>

tournamentsb.php
Code: 
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<html>
<head>
  <title></title>
</head>
<body>
<?php
echo "Hello ". $_GET['name']. "<br>\n";
echo "Your id is ". $_GET['nameid']. "<br>\n";

//But you can use the hidden form fields also that were "posted"
echo "Hello ". $_POST['name']. "<br>\n";
echo "Your id is ". $_POST['nameid']. "<br>\n";

?>
</body>
</html>

You use $_GET["name"] and $_GET["nameid"] to retreive the url parameters.
For "posted" parameters $_POST["name"] and $_POST["nameid"].

Thanks,
 
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Ok, this is what I am trying to do going back a few steps.
name = David Hascup
nameid = usadarts

profile.php - link passing 2 fields from an sql database
Code: 
<a href="[URL unfurl="true"]http://www.usadarts.com/members/tournaments.htm?$name&$nameid">[/URL]

tournaments.htm
Code: 
<html>
<head><title></title>
</head>
<body>
<center>
<form action="tournaments.php?htmname=$htmname&nameid=$nameid" method="POST"> 
......

tournaments.php -
Code: 
<?php 
$phpname = $_GET['htmname'];
$phpnameid = $_GET['htmnameid'];
this then updates a seperate sql database
Code: 
$insert = "INSERT INTO results (name, username, points) 
VALUES ('$phpname', '$phpnameid', '$pts')";

I am getting no more erro message, however the following is ending up in the SQL database:
name field = $htmname instead of David Hascup
username field = $htmnameid instead of usadarts
 
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