Mysqli statement returning warnings

Question

Here is my code:

include "db_conx.php";
$sql = mysqli_query("INSERT INTO table(column) VALUES('" . mysqli_real_escape_string($var) . "')");
if ($sql) {echo "connection successful";
} else {
echo "failure";
}

It returns these errors:

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given
Warning: mysqli_query() expects at least 2 parameters, 1 given

I tried using PDO but that didn't work either...

simply pass connection link identifier in the function – Ankur Tiwari Apr 25 '16 at 05:47 — Ankur Tiwari, Apr 25 '16 at 05:47

score 0 · Accepted Answer · answered Apr 25 '16 at 05:49

First Parameter is mysql connection link identifier, and second is string For more details, you can visit this link : http://in2.php.net/manual/en/mysqli.real-escape-string.php.

include "db_conx.php";
$sql = mysqli_query(pass_your_connection_identifier_here ,"INSERT INTO table(column) VALUES('" . mysqli_real_escape_string(pass_your_connection_identifier_here, $var) . "')");
if ($sql) {echo "connection successful";
} else {
echo "failure";
}

score 0 · Answer 2 · answered Apr 25 '16 at 05:58

0

You are missing connection identifier in both mysqli_real_escape_string() and mysqli_query() change your code to mysqli_query($connection,$sql) and mysqli_real_escape_string($connection,$string)

answered Apr 25 '16 at 05:58

Mujahidh

428
4
19

score 0 · Answer 3 · answered Apr 25 '16 at 06:18

You are missing connection variable at both the lines that's why you are facing issues :

Try to replace your code with this :

//$con // it is your connection variable
include "db_conx.php";
$sql = mysqli_query($con,"INSERT INTO table(column) VALUES('" . mysqli_real_escape_string($con,$var) . "')");
if ($sql) {echo "connection successful";
} else {
echo "failure";
}

Mysqli statement returning warnings

3 Answers3