PHP: Isset with if else statement

Question

iso.obj.php

$from = null;
$too = null;

if (isset($_POST['search'])) {
    if ($from = null) {
        $from = date("Y-m-d");
    } else {
        $from = $_POST['FromDateTime'];
    }

    if ($too = null) {
        $too = date('Y-m-d', strtotime("+30 days")) . " 23:59:59"; 
    } else {
        $too = $_POST['ToDateTime']." 23:59:59";
    }
}
$sql ="//statement here with
WHERE capDateTime >='{$from}' AND capDateTime <='{$too}'";

The code is working fine but i got problem when person request the page and its return an empty table. But if i select date and time manualy its will display according to the date and time.

My question is: How do i put current date if not set and if set follow the $_POST?

if($from = null){ ??? and if ($too = null) { you are assigning the values not comparing — devpro
– devpro, Commented Oct 7, 2016 at 8:00
if ($too = null) this ---> = is an assignment operator. If you want to compare it use == comparison operator. See here. — Slatyoo
– Slatyoo, Commented Oct 7, 2016 at 8:03
Sidenote: By directly using variables in your $sql, you're exposing yourself to SQL injection. Please use PDO or Mysqli see owasp.org/index.php/SQL_Injection — Neil Yoga Crypto
– Neil Yoga Crypto, Commented Oct 7, 2016 at 8:03

danopz · Accepted Answer · 2016-10-07 08:05:39Z

2

$from = date("Y-m-d");
$too = date('Y-m-d', strtotime("+30 days")) . " 23:59:59";

if (isset($_POST['search'])) {
    if (!empty($_POST['FromDateTime'])) {
        $from = $_POST['FromDateTime'];
    }

    if (!empty($_POST['ToDateTime'])) {
        $from = $_POST['ToDateTime']." 23:59:59";
    }
}

answered Oct 7, 2016 at 8:05

danopz

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