Passing variables from page to page

[ausgezeichnete]

I made this form in the "index page"

<form action="search_results.php" method="get" name="search">
<label for="days">
Choose days:
<select name="days" iid="days">
<option value="7">+/-7 days</option>
<option value="14">+/-14 days</option>
<option value="21">+/-21 days</option>
</select>
<input name="search" type="submit" value="search">
</form>

and when i did : echo $ days it in the "search_results.php" page ,it was ok

Now i have made another form in the "search_results.php"

echo "<form action='calculate.php' method='post'>";
while($record=mysqli_fetch_assoc($result)){
$days_db=$record['days'];
echo "<input type='checkbox' name='choose[]' value='{$days_db}' />";
echo "<input type='submit' name='submit' value='Continue'>";
echo "</form>";

and in the "continute.php"
i calculated the sum of the checkboxes selection
but my problem now is:
i want to compare between the $days in the "index.php " and the $sum of the checkboxs selection
but each time i try to echo $days in the "contniue.php" it deosnt seem to see it
why???

[kairno]

an easy way to solve this issue is to create a session and store the information in it, then call it whenever you want

example :

<file1.php>
session_start();
$_SESSION['test']=value;

<file2.php>
session_start();
echo $_SESSION['test'];

<file3.php>
session_start();
echo $_SESSION['test'];

this is the way most sites pass user credentials, you might want to check the manual help on sessions, but this way you can pass info to the pages without the use of the "<form>" tag...

for passing arrays and other objects through sessions use the serialise functions

hope the info helped you

[ausgezeichnete]

thnxxx it solved
by doing this

echo "<input type='hidden' name='days' value='$days' />";

[ausgezeichnete]

and ur way was so good too by restoring it in the session thnxx again

[ausgezeichnete]

but please i have another problem in mysql query

$query="select tourname,description ,days from tours where
days >='".$days."'";

my select is

<select name="days" iid="days">
<option value="7">+/-7 days</option>
<option value="14">+/-14 days</option>
<option value="21">+/-21 days</option>
</select>

i have records conatins :5,10,15 days
every time i choose +/-7 days which has the values 7
it gives me 0
and when i select +/14 which has the value 14
its gives me the days 5,15
and when i select +/-21
its gives me the days 5
why????????
i think mysql query contains something wrong ???
Edit/Delete Message

[kairno]

do an echo on the $days ... as far as i see it, the query is ok so the problem might be in the values passed...

for your help : edit your php.ini file to show all errors, and when executing queries, use the "or die" function to show if there's a problem with the sql code...

but as i said, for this problem, check the value of $days ...

[ausgezeichnete]

i checked the value of the $days
i did echo $days in the top of the page
and it gave me the expecting values,either 7 or 14 or 21
and there no sql errors

[kairno]

could you show me the result for

$query='select tourname,description ,days from tours where days >="'.$days.'"';
$res=mysql_query($query);
$data=mysql_fetch_array($res);

echo "-= ".$data['tourname']." =--= ".$data['description'].'" =--= ".$data['days'];

i have no idea why the code is wrong if the value sent is ok, and the query is ok...
(i assume the echo on $days was done in the page where you use them in the database..)

[ausgezeichnete]

this when i chooses -/+14
= Golden Triangle with Taj Mahal =--= Delhi – Agra – Jaipur.more" =--= "5
and the same when i choosed +/-21
and NOne when i choosed +/-7
-= =--= " =--= "

[kairno]

😐
... never happened to me before... i'll check the script when i get home... and if i get anything i'll post back... hope someone knows what the problem is...

[ausgezeichnete]

ok thnxxx dear