I am making a system in which at the time of entering the data, this error marks me.
Error: Warning: mysqli_query () expects at least 2 parameters, 1 given in
<?php
require("connect_db.php");
if (isset($_POST['descripcion']) && !empty($_POST['descripcion'])) {
$descripcionTurno = $_POST['descripcion'];
mysqli_query("INSERT INTO ctg_turno (descripcion) values ('$descripcionTurno'");
echo '<script>alert("Datos Ingresados Correctamente")</script> ';
}
First you need to pass the connection as the first parameter to mysqli_query (style procedure) ,
mysqli_query($con, "INSERT....") { ...}
But the even bigger problem is the concatenation of variables in a statement, thus it is prone to receive attacks from Injection SQL . You should use prepared sentences. (Ejm base)
$mysqli = new mysqli("localhost", "my_user", "my_password", "midatabase");
if (isset($_POST['descripcion']) && !empty($_POST['descripcion'])) {
$descripcionTurno = $_POST['descripcion'];
if ($stmt = $mysqli->prepare("INSERT INTO ctg_turno (descripcion) values (?)")) {
$descripcionTurno = $_POST['descripcion'];
/* añadir los valores para los marcadores con la variable descripcionTurno*/
$stmt->bind_param("s", $descripcionTurno);
/* ejecutamos la consulta */
if($stmt->execute()){
echo "INSERT CORRECTO";
}
}
}
The function requires two arguments, the first is the connection and the second the SQL.
You have to put the connection to the function as an argument to mysqli_query , for example:
mysqli_query($con,"SELECT * FROM Persons");
$ with - > It is the first parameter, the connection you have established.
"SELECT * FROM Persons" - > It's the SQL you want to run
The error is because you are using the procedural style of the mysqli function as described in the manual. The first parameter must be
A link identifier returned by mysqli_connect () or mysqli_init ()
Applies to style by procedures.
I recommend that you also review your work style, it is not advisable to directly pass the user data to the query directly, for that there are the parameterized queries.