Uncaught Error: Call to undefined function mysql_query() , PLEASE HELP ME

Question

When i run my code, the navegator show me that:

Notice: Use of undefined constant datos1 - assumed 'datos1' in C:\xampp\htdocs\formulario\conexion1.php on line 8

Warning: mysqli_select_db() expects exactly 2 parameters, 1 given in C:\xampp\htdocs\formulario\conexion1.php on line 8 No se pudo conectar a la base de datos Notice: Undefined index: Teachers_Name in C:\xampp\htdocs\formulario\conexion1.php on line 13

Notice: Undefined index: School_Name in C:\xampp\htdocs\formulario\conexion1.php on line 14

Notice: Undefined variable: Implementation_Quality in C:\xampp\htdocs\formulario\conexion1.php on line 15

Notice: Undefined variable: Implementation_Quality in C:\xampp\htdocs\formulario\conexion1.php on line 17

Fatal error: Uncaught Error: Call to undefined function mysql_query() in C:\xampp\htdocs\formulario\conexion1.php:19 Stack trace: #0 {main} thrown in C:\xampp\htdocs\formulario\conexion1.php on line 19

What i do?

if(!$conexion){
echo "Conexión no exitosa";
} else {

$base= mysqli_select_db(datos1);
    if(!$base){
        echo "No se pudo conectar a la base de datos";
    }
 }
$Teachers_Name=$_POST['Teachers_Name'];
$School_Name=$_POST['School_Name'];
$Implementation_Quality['Implementation_Quality'];

$sql= "INSERT INTO datos_1 VALUES('$Teachers_Name', 'School_Name', 
'$Implementation_Quality')";

$ejecutar = mysql_query($sql);

if(!$ejecutar){
echo "Hubo algun error";
} else {
 echo "Datos guardados correctamente<br><a href='index.html'>Volver</a>";
}
?>

Answer 1

Replace this:

$base= mysqli_select_db(datos1);

With this:

$base= mysqli_select_db($conexion,'datos1');

---

replace this:

$sql= "INSERT INTO datos_1 VALUES('$Teachers_Name', 'School_Name', '$Implementation_Quality')";

With this:

$sql= 'INSERT INTO datos_1 (col1, col2, col3) VALUES ( ?, ?, ? )';

Replace col1,col2 and col3 with the names of columns in the datos_1 table.
e.g. ... datos_1 (teacher_name, school_name, implementation_quality) VALUES ...

Note that this is a static string literal. There's no variable substitution. The question marks are bind placeholders which we will reference later.

---

Replace this:

$ejecutar = mysql_query($sql);

With this:

$sth = mysqli_prepare($conexion,$sql);
mysqli_stmt_bind_param($sth,'sss', $Teachers_Name, $School_Name, $Implementation_Quality);
$ejecutar = mysqli_stmt_execute($sth);

Don't mix mysql_ functions with mysqli_ functions. That won't work. (We shouldn't be using any mysql_ functions in new development; the mysql_ interface functions have been deprecated a long time ago, and are finally removed in newest versions of PHP.

---

Do NOT incorporate potentially unsafe values in SQL text. Use prepared statements with bind placeholders. (Or less optimally, properly escape any values that are incorporated into the text.)

Recommended:

Little Bobby Tables - Exploits of a Mom https://xkcd.com/327/

OWASP SQL Injection https://www.owasp.org/index.php/SQL_Injection