[RESOLVED] Passing values to next page

MitchEvans

I have a page that allows users to choose from checkboxes, each with text next to the three checkbox options. If users don't choose from one of the checkboxes then i don't want the text to appear on the next page. But if others have been chosen i want them to show. Maybe i need another elseif in the statement.

Any help cheers?!?!

Code:

$res= mysql_query("SELECT * FROM survey WHERE rtid=1 ORDER BY id");
$num= mysql_num_rows($res);
echo"<tr>";
echo"<td>1 - Operational<td>";
echo"</tr>";
for ($i=0; $i<$num; $i++) {
// Fetch data from survey table here
$row= mysql_fetch_array ($res);

$s='';
$t='';
$u='';
$array_name= 'myradio'.$row['id'];

if (isset($POST[$array_name])) {
foreach ($POST[$array_name] as $key => $val) {
if ($val=='Currently In Place') $s='X';
else if ($val=='Required Essential') $t='X';
else if ($val=='Required-Recommended') $u='X';
}
}

// echo table cells with requirement_txt, $s, $t, and $u, here

echo "<tr>";
echo "<td>{$row['requirement_txt']}</td>";
echo "<td><center>$s</center></td>";
echo "<td><center>$t</center></td>";
echo "<td><center>$u</center></td>";
echo "<td>Module 1</td>";
echo "</tr>";
}

Tezread

here is something to maybe get you moving. Im showing yuo some code that calls checkboxes and respective field names from a database- the condition of if it is checked or not and whether it shows is at the bottom

<?php 
    include ("connect.php"); 

// Get the activities (checkboxes and names)
$sql_activity = "SELECT activity FROM activity ORDER by activity"; 
$activity_result = mysql_query($sql_activity) or die("Error fetching activities: ".mysql_error()); 
$activity_array = mysql_fetch_assoc($activity_result); 


?> 
//display them on the screen
<?php 
      echo '<form name="searchform" method="GET" action="page2.php">'; 
echo ' 
<table> 
<tr> 
<td>Select the activities:</td> 
</tr> 
<tr> 
<td>'; 

foreach($activity_array as $activity){ 
echo '<input type="checkbox" name="activity_checkbox[]" value="'.$activity['activity_participant_id'].'">'.$activity['activity'].''; 
} 

echo ('</td> 
</tr> 


echo ('</td> 
</tr> 
<tr> 
<td> 
  <input type="submit" name="search_submit" value="Search!"> 
</td> 
</tr> 
</table>'); 

echo '</form>'; 
      ?> 

///next page show checkboxes that are checked with the name next to them:

//insert values into a table to run a select ;ater
foreach ($_POST["activity_checkbox"] as $activity) { 
$ary = "INSERT INTO activityuser VALUES (" . $username . ", '" . $activity . "')"; 
// execute query 
$result = mysql_query($ary) or die("Error in search query: ".mysql_error()); 
	} 


//show the checkboxes that have been ticked
foreach ($_POST["activity_checkbox"] as $activity) { 
$ary = "SELECT * FROM activityuser where .  VALUE = '1' ....// execute query 
$activity_result = mysql_query($sql_activity) or die("Error fetching activities: ".mysql_error()); 
$activity_array = mysql_fetch_assoc($activity_result); 


//show the checkboxes that are left

foreach($activity_array as $activity){ 
echo '<input type="checkbox" name="activity_checkbox[]" value="'.$activity['activity_participant_id'].'">'.$activity['activity'].''; 
} 

echo ('</td> 
</tr> 


echo ('</td> 
</tr> 
<tr> 
<td> 
  <input type="submit" name="search_submit" value="Search!"> 
</td> 
</tr> 
</table>'); 

echo '</form>'; 
      ?>

MitchEvans

I'm afraid this doesn't really help my problem I don't think.
Maybe I need another SELECT statement that checks whether $s='X' or $t='X' or $u='X'.

What do you think, how would I do this.

I only want results to show if $s,$t or $u =X.

MitchEvans

The problem is solved yippeeeee!

I had to echo the values within the if statement for this to work!