[RESOLVED] Passing variable to multiple pages

[tzuriel]

Hi,

I hope I can ask this correctly.

I have a page with a drop-down (<select name="years">) containing years. The user selects a year, clicks submit and that choice is passed using $POST["years"]. This works fine and the script returns the correct data from the mySQL db that used $POST["years"] in the SELECT query.

What I want to do is have another submit button on this 2nd page that would pull data using the next year in the SELECT query by incrementing the value of $POST["years"] by 1. However, the value of $POST["years"] is gone, it seems, after it posts to the 2nd page. $POST["years"] used the <select name="years"> from page 1 and there is no such element found on page 2. When I echo the value of $POST["years"]+1 I get 1. Imagine that! :bemused:

How do I get the initial choice made on page 1 available to the script on page 3?

Thanks for your help

[NogDog]

You could add it to a hidden field in your form...

echo "<input type='hidden' name='years' value='{$_POST['years']}'>";

...or if you're using sessions, you could just add it to the $_SESSION array.

[tzuriel]

Please forgive my near-complete ignorance...

On which page would this hidden field go? Here's the relevant code from page 1:

<form name="form2" method="POST" action="snlquery2.php">

$res2=mysql_query("select * from years");


echo'<select name="years">';

if(mysql_num_rows($res2)==0) echo "there is no data in table..";
else
for($i=0;$i<mysql_num_rows($res2);$i++) {
$row2=mysql_fetch_assoc($res2);
echo"<option>$row2[years]</option>";


}
echo'	</select>';

<input type="submit" name="Submit" value="Submit Query">
    </div>
</form>

Page 2 creates a SELECT query using the choice made on page 1 that's stored in $POST["years"]. I've assigned it to $yearchoice and $nextyear is the incremented value of $POST["years"]:

$sql_query3="SELECT casts . memberID , members . memberName , casts . yearID "
        . " FROM members INNER JOIN casts ON members . memberID = casts . memberID "
        . " WHERE casts . yearID = '$_POST[years]' ORDER by members.memberName";

$nextyear=$_POST["years"]+1;        

$result3=mysql_query($sql_query3);

$num3=mysql_numrows($result3);
//echo $num;
echo "<br><font face=verdana size=4 color=#9966CC >Cast data for: <b>$yearchoice</b></font><br>";
echo "<table width=200 border=0>";
mysql_data_seek ($result3, 0); 
while ($row3 = mysql_fetch_assoc($result3)) {
	echo '<tr><td><b><font face=arial size=3 color=white>' . $row3['memberName']. '</td>';
}
echo "</table><br>\n";

So far, this all works. $nextyear holds the incremented value as expected because this script has access to $_POST["years"]. I added the following to try and get this value to page 3:

<form name="form1" method="POST" action="nextyear.php">


<input type="submit" name="Submit" value="See Next Year"></form>

And here is what's in nextyear.php

$nextyear=$yearchoice+1;

$sql_query3="SELECT casts . memberID , members . memberName , casts . yearID "
        . " FROM members INNER JOIN casts ON members . memberID = casts . memberID "
        . " WHERE casts . yearID = $_POST["years"]+1 ORDER by members.memberName";


$result3=mysql_query($sql_query3);

$num3=mysql_numrows($result3);
//echo $num;
echo "<br><font face=verdana size=4 color=#9966CC >Cast data for: <b>$nextyear</b></font><br>";
echo "<table width=200 border=0>";
mysql_data_seek ($result3, 0); 
while ($row3 = mysql_fetch_assoc($result3)) {
	echo '<tr><td><b><font face=arial size=3 color=white>' . $row3['memberName']. '</td>';
}
echo "</table><br>\n";

But $_POST["years"] is empty so no data is returned and i get ugly errors:

Warning: mysql_data_seek() [function.mysql-data-seek]: Offset 0 is invalid for MySQL result index 3 (or the query data is unbuffered
😕
What am I doing wrong? Thanks for taking a look at this mess!

[NogDog]

You have some quoting problems on that last query. It should be something like:

$sql_query3="SELECT casts . memberID , members . memberName , casts . yearID "
        . " FROM members INNER JOIN casts ON members . memberID = casts . memberID "
        . " WHERE casts . yearID = {$_POST['years']}+1 ORDER by members.memberName";

[tzuriel]

Thanks for the suggestion. I was able to understand (finally) your previous comment about the hidden value. I changed the Request Method to GET instead of POST, and the incremented year was passed through to page 3. Works perfectly! Thanks!

[NogDog]

<form name="form1" method="POST" action="nextyear.php">
<input type='hidden' name='years' value='<?php echo $_POST['years']; ?>'>
<input type="submit" name="Submit" value="See Next Year"></form>

I'm not 100% sure which value you want to pass, so adjust the variable being echo'd if you want something else.