elseif不能按预期在php中工作[重复]

Question

This question already has an answer here:

• The 3 different equals 5 answers

<?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.

</div>

Answer 1

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 ?????