elseif not working in php as expected

Question

<?php

$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "test_db";


// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}

//check radio button and update sql accordingly

if (isset($_GET['calc'])) {

$calc = $_GET['calc'];
}

if ($calc='average'){
$sql1 = "UPDATE `config` SET `calc_category`=1 WHERE 1";
$conn->query($sql1);
} elseif ($calc='high'){
$sql1 = "UPDATE `config` SET `calc_category`=2 WHERE 1";
$conn->query($sql1);
} else {
$sql1 = "UPDATE `config` SET `calc_category`=3 WHERE 1";
$conn->query($sql1);
}


?>

In the above code even if the $calc is equal to 'high' or something other than 'average', it still updates the database with 'calc_category'=1 which is only supposed to happen when first 'if' condition is true.

I have echoed the value of $calc and it changes based on which radio button is selected but I am not sure why always first 'if' condition code is getting executed.

I am new to PHP,SQL please help me where am I going wrong.

in the query where 1 means all rows will be updated.. check where clause and also use `==` (equal operator) instead of `=` (assignment operator) — Bilal Ahmed, Aug 11 '18 at 06:14
`$calc='average'` is an assignment, not a test for equal to. `$calc=='average'` — Nigel Ren, Aug 11 '18 at 06:15

Dr Manish Lataa-Manohar Joshi · Accepted Answer · 2018-08-11T06:23:02.733

2

double equal to needed...

 if ($calc=='average'){ /****// == instead of = //*****/
     $sql1 = "UPDATE `config` SET `calc_category`=1 WHERE 1";
     $conn->query($sql1);
 } elseif ($calc=='high'){ /****// == instead of = //***/
     $sql1 = "UPDATE `config` SET `calc_category`=2 WHERE 1";
     $conn->query($sql1);
 } else {
     $sql1 = "UPDATE `config` SET `calc_category`=3 WHERE 1";
     $conn->query($sql1);
 }

also check your query near Where clause.... it is where 1 ?????

edited Aug 11 '18 at 06:23

answered Aug 11 '18 at 06:16

Dr Manish Lataa-Manohar Joshi

2,380
5
17
38

Thanks I now feel that was a stupid question. Love how quickly you responded :) – Pie Aug 11 '18 at 06:30
Welcome... and I am happy that your problem is sorted out... – Dr Manish Lataa-Manohar Joshi Aug 11 '18 at 06:33
Also there is only one row in that column so maybe "where 1" doesnt matter. I actually copied it from phpmyadmin query so it exists. – Pie Aug 11 '18 at 06:35

score 1 · Answer 2 · answered Aug 11 '18 at 06:16

1

In your elseif statement you are setting $calc = 'high'. That is not a comparison. In order for you to evaluate if the terms are equal you need to use == or === depending on how strict you need the comparison to be.

if ($calc == 'average'){
$sql1 = "UPDATE `config` SET `calc_category`=1 WHERE 1";
$conn->query($sql1);
} elseif ($calc == 'high'){
$sql1 = "UPDATE `config` SET `calc_category`=2 WHERE 1";
$conn->query($sql1);
} else {
$sql1 = "UPDATE `config` SET `calc_category`=3 WHERE 1";
$conn->query($sql1);
}

answered Aug 11 '18 at 06:16

Joseph_J

3,654
2
13
22

Thanks for the answer and detailed explanation, I couldnt upvote being a new user – Pie Aug 11 '18 at 06:31
No problem and no worries! – Joseph_J Aug 11 '18 at 06:33

score 0 · Answer 3 · answered Aug 11 '18 at 06:33

0

for Comparison two values you must use '==' instead of '='. appending value to a variable return true value so elseif section not work. i hope it be useful.

false usage : if ($calc='average') true useage : if ($calc == 'average')

answered Aug 11 '18 at 06:33

MHashem Hashemi

1

elseif not working in php as expected

3 Answers3