Passing a form entry

Sapper

Hi I wrote a script in php3 and it worked fine I have now transfered this script to php4 and it doesnt work.

Basically it is for a transfer list so the first page for example is called transfer.php which has a form on it like this:

<select name="position2">
<option value="Select From List">Select From List</option>
<option value="Goalie">Goalies</option>
<option value="Defender">Defenders</option>
<option value="Winger">Wingers</option>
<option value="Midfield">Midfielders</option>
<option value="Striker">Striker</option>
</select>

I havent put the whole page in but that form is then posted to transfer2.php which used to take the form input name position2 and post it to the transfers2.php page which would then convert that position2 into $position2

Then in the transfers2.php page the following line would check for all entries that matched the form input:

$sql = "select * from players where Position = '$position2' and listed ='1'";
$result = mysql_query($sql);

Why doesnt this work now? and is there an easier way of passing a selection from
a form onto the next form which checks the database and displays all matching records.

Sorry if ive confused you but I can see no reason why it is no longer changing the input name into a variable.

Any Ideas

Thanx.....

mogster

If you have upgraded php, your globals may be set to off, in case you need to specify the variables being sent into the script with:

$position2 = $_POST['position2'];

Every variable being sent to the script must be defined. Or, you may turn your globals to on in php.ini 😉

knutm

stolzyboy

instead of $position2, use $_POST['position2'], since with php4, register_globals is probably off in your php.ini

Sapper

Thanks for your responses.

I tried the above suggestion and I am now getting the following error:

PHP Notice Undefined index position2 in filename line 49

here is the entire code I have written maybe I just put something in the wrong place...

<?php
/ This program gets members from the database /
$db = mysql_connect("localhost", "username" , "password");
mysql_select_db("database", $db);
$position2 = $_POST['position2'];
$sql = "select * from players where Position = '$position2' and listed ='1'";
$result = mysql_query($sql);

while ($row = mysql_fetch_array($result)) {
print("<tr><td bgcolor=\"#003399\">");
printf(" %s</td></tr>\n",
$row["Name"]);
printf("<td>Current Owner: %s\n",
$row["TeamName"]);
printf("&nbsp&nbsp&nbsp&nbsp&nbspAge: %s\n",
$row["Age"]);
printf("&nbsp&nbsp&nbsp&nbsp&nbspPosition: %s\n",
$row["Position"]);
printf("&nbsp&nbsp&nbsp&nbsp&nbspSkill Level:%s<hr>\n",
$row["Skill"]);
}
?>

stolzyboy

are you using method="get" in your form, if so replace

$_POST['variable']

with

$_GET['variable']

other than that i will look closer later and answer back

stolzyboy

or you can try

$HTTP_POST_VARS["varname"]

Sapper

I am indeed using post not get.....

Sapper

I stripped it right back down to print out a single field and found a typo, it seems to work now with the single field so I will build it back up agian see if it still works.

Cheers for all the help though...