Error in PHP code

Question

Code:

$db = mysqli_connect("localhost","root","54825482","MyO"); $result = mysqli_query("SELECT id FROM users WHERE login = '$login'" ,$db) or die('Cann`t query '.mysql_error);

Mistake:

mysqli_query () expects parameter 1 to be mysqli, string given in C: \ www ... Cann`t query mysql_error

I can not understand what the problem

Possible duplicate question: mysql_fetch_array () expects parameter 1 to be resource (or mysqli_result), boolean given
Documentation read php.net/manual/ru/mysqli.query.php Procedural style

Vladimir Stepanov Vladimir Stepanov 20 1 bronze sign · Accepted Answer · 2017-08-19T10:47:31

 $db = mysqli_connect("localhost","root","54825482","MyO"); $result = mysqli_query($db,"SELECT id FROM users WHERE login = '$login'");

This answer, like all the others, at the moment, contains an alleged (with a high probability) vulnerability such as SQL injection .

user242433 · Answer 2 · 2017-08-19T10:49:04

 $db = mysqli_connect("localhost","root","54825482","MyO"); $sql = "SELECT id FROM users WHERE login = '$login'"; $result = mysqli_query($db, $sql); if (!$result) die('Can`t query '. mysql_error);

Procedural style
mysqli_query ( mysqli $link , string $query [, int $resultmode MYSQLI_STORE_RESULT ] )

mysqli_query

Answer 3 · 2017-08-19T10:32:39

 $mysqli = new mysqli("localhost","root","54825482","MyO"); $result = $mysqli->query("SELECT id FROM users WHERE login = '$login'");

In " " you can simply write a variable. echo "Hello $world"; - user242433

Error in PHP code

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