Passing hidden form field values to other pages

peteraub

I hope someone can help me as I am baffled. I have a script which sets up a number of check boxes for people to choose the fields of selected table which they wish to have in report. There are two hidden form variables to be passed to next script. One is the $table variable which comes from query string on previous page. The other is $sql variable, the value of which is generated by the selections made by user.

The problem is that no matter what I do I cant get the $sql value to be passed by the hidden form field. If I just echo the value to the screen by submitting form to self it appears ok so there is a value in the variable. I can even get it to appear in a query string.
Here is an example of my code.

function list_fields($table) {

$fields = mysql_list_fields('brahm001', $table);
$columns = mysql_num_fields($fields);

?><form method=post action="display_report.php" target="_blank"><?php        
for ($i = 0; $i < $columns; $i++) { ?>
    <input type="checkbox" name="<?php echo 'list'.$i; ?>" value="<?php echo mysql_field_name($fields, $i); ?>"><?php
    echo mysql_field_name($fields, $i) . "\n<br />";

}

for ($i = 0; $i < $columns; $i++) {
    $check="list".$i;
    if($_POST[$check])
    $sql = $sql.$_POST[$check].' , ';

}

$sql = substr($sql, 0, -2);
?>
<input type="hidden" name="sql" value="<?php echo $sql; ?>">
<input type="hidden" name="table" value="<?php echo $table; ?>">
<input type="submit" value="Generate Report">
</form>
[/COLOR]
This generates the check boxes for user to make selections. If I echo the value to the screen the value appears like retreat_id , booking_status , subheading2 or similar depending on what check boxes have been ticked.

The next page appears
$sql = $_POST['sql'];
echo $sql;

If I change the statements to read

$sql = $_POST['table'];
echo $sql;
the $table variable value is passed no problems.

If someone could tell me why the $sql value isn't being passed I would be most appreciative.

Thanks
Peter

Tj_C

Looks to me like you have a bit of an error in the below loop.


for ($i = 0; $i < $columns; $i++) { 
$check="list".$i; 
if($_POST[$check]) 
$sql = $sql.$_POST[$check].' , '; 

}

Try makin $POST[$check] $POST['check'] and see what happens.

peteraub

thanks tj
I have tried that but it still doesn't make any difference. There is a value generated for the $sql variable but for some reason doesn't pass via the hidden form field.

Tj_C

How do you know there is a value there? Doen't look to me like there would be.

Have you tried just making input a text instead of hidden and see if there is a value there?

It seems like you are trying to resolve a var inside a var name and I don't think you can do that.

Never tried this but it should work...


$i=0;

while(list($key,$item)=each($_POST)) {

  $check="list".$i;

 if($key==$check and $item!="")  {

        $sql.= $item
 } 
 $i++;
}

Let me know how it goes. Never tried to loop with a $_Array but it should work in theory.

ahundiak

I think you may have a concept problem.

Generate a form with select boxes for the fields.
User presses the Generate Report button.
display_report reads the select box values and generates the $sql variable. It's not going to get passed as input data.

And by the way, consider using an array for the select boxes.

<select name="field[]" value="the field name" />

if (isset($_POST['field']) $fields = $_POST['field'];
else $fields = array();

foreach($fields as $field) echo $field;

That saves you from having to build the select names dynamically.

peteraub

Thanks for all your help guys

After all your input and suggestions I finally found a solution that works. I generate the hidden form fields using a foreach statement on the $_POST array.

foreach($_POST as $key => $value) {
?><input type="hidden" name="<?php echo $key; ?>" value="<?php echo $value; ?>"><?php
}

This seems to work fine. It is strange when things in theory should work but don't work in a practical sense, and the answer always seems so simple when you find it.

Thanks again for your help.