how to store values in session properly

Question

I have these pages that asks user for values. when I try to use the links to previous pages, the session values are gone and it requires user to go back to the first and repeat the process.

Here are sample codes based of my original code:

page1.php

<?php
 session_start();
 echo "<form method='POST' action='page2.php'>";
    echo "<input type='text' name='date1'>";
    echo "<input type='text' name='date2'>";
    echo "<input type='submit'>"
 echo "</form>";
?>

page2.php

<?php
 session_start();
 $_SESSION['date1'] = $_POST['date1'];
 $_SESSION['date2'] = $_POST['date2'];
      echo "<form method='POST' action='page3.php'>";
        echo "<input type='text' name='info1'>";
        echo "<input type='text' name='info2'>";
        echo "<input type='submit'>"
     echo "</form>";
echo "<a href='page1.php'>Change value in page 1</a>";
?>

page3.php

<?php
 session_start();
 $_SESSION['info1'] = $_POST['info1'];
 $_SESSION['info2'] = $_POST['info2'];
 if(isset($_POST['confirm'])){
$mysql_host = 'localhost';
$mysql_user = 'root';
$mysql_pass = '';
$db = 'sampDB';

$conn = mysqli_connect($mysql_host, $mysql_user, $mysql_pass, $db);

$query = mysqli_query($conn, 'INSERT INTO info(date1, date2, info1, info2)
                               VALUE ('$_SESSION[date1]', '$_SESSION[date2]', '$_SESSION[info1]', '$_SESSION[info2]')');

}
      echo "<form method='POST'>";
        echo $_SESSION['date1'];
        echo $_SESSION['date2'];
        echo $_SESSION['info1'];
        echo $_SESSION['info2'];
        echo "<input type='submit' name='confirm'>"
     echo "</form>";
echo "<a href='page1.php'>Change value in page 1</a>";
echo "<a href='page2.php'>Change value in page 2</a>";
?>

How do I go back to previos pages without destroying the values from the session?

Ben · Accepted Answer · 2016-02-08 11:42:31Z

3

As long as you're not $_POSTing values back to the other pages, you can simply check if $_POST is set before overwriting values, like so:

session_start();
if(isset($_POST["date1"])) $_SESSION['date1'] = $_POST['date1'];
if(isset($_POST["date2"])) $_SESSION['date2'] = $_POST['date2'];

Reference: PHP isset

edited Feb 8, 2016 at 11:42

answered Feb 8, 2016 at 11:37

Ben

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2 Comments

chemical_elii Over a year ago

I tried $_SESSION['date1'] = isset($_POSTS['date1']) ? $_POSTS['date1'] : ''; not sure why i did not work

Ben Over a year ago

@chemical_elii It won't work because if the $_POST value isn't set, it'll make the $_SESSION value blank. You don't want to affect the $_SESSION value if $_POST is not set!

Pupil · Accepted Answer · 2016-02-08 11:39:07Z

0

Everytime, you are overwriting values of SESSION.

You need to check whether session is already set.

If already set, then there should be not change.

Also, posted values should be applied only if form is posted.

Add checks if sessions already exist like this:

$_SESSION['date1'] = !isset($_SESSION['date1']) ? $_POST['date1'] : 
$_SESSION['date1'];

$_SESSION['date2'] = !isset($_SESSION['date2']) ? $_POST['date2'] : $_SESSION['date2'];

Same for other session values.

answered Feb 8, 2016 at 11:39

Pupil

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Comments

Machavity · Accepted Answer · 2016-02-08 19:27:57Z

0

You should write your query this way

$query = mysqli_query($conn, "INSERT INTO info(date1, date2, info1, info2) 
VALUE ('{$_SESSION['date1']}', '{$_SESSION['date2']}', 
    '{$_SESSION['info1']}', '{$_SESSION['info2']}')");

edited Feb 8, 2016 at 19:27

Machavity♦

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answered Feb 8, 2016 at 11:42

makmesh

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