How to insert variable value in SQL

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Hello, I am newbies in PHP and Bootstrap. Currently, I am developing an online examination system. I would try to store the score of students in database but it's a failure. The score was being declared in variable named $per. Could you help me to figure out the coding for doing it ?

What I have tried:

I've try to code. Here are the sample coding.

users.php

PHP

<?php
session_start();
class users{

public $per;

public function __construct()
	{
		$this->conn=new mysqli($this->host,$this->username,$this->pass,$this->db_name);
		if($this->conn->connect_errno)
		{
			die ("database connection failed".$this->conn->connect_errno);
		}
	}

public function store_result($per)
	{
		$this->conn->query($per);
		return true;
	}
?>

store.php

<?php 
include("class/users.php");
$score=new users('per');
$query = "SELECT * FROM table PERCENTAGE $per";

if ($score->store_result($query))
{
	echo $per;
}

?>

And here is the piece of code in answer.php where percentage of score was being declared here. 

        Your result        <?php 
		$per=($answer['right']*100)/($total_qus);
		
		echo $per."%";
		?>

Posted 5-Dec-17 7:00am

Azifatilah Ramli

Updated 5-Dec-17 7:54am

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W∴ Balboos, *GHB* · Answer 1 · 2017-12-05T07:11:00

Solution 1

"SELECT * FROM table PERCENTAGE $per"

Since you stipulate that $per is a variable name I can safely say that your SQL query is failing because it's invalid.

Perhaps you mean something like:

SELECT * FROM table PERCENTAGE WHERE per=$per

Not enough in your code sample to be sure and assuming per is an int

Resolve that and let's go on from there, if necessary.

Posted 5-Dec-17 7:11am

W∴ Balboos, GHB

Updated 5-Dec-17 7:13am

Comments

Azifatilah Ramli 5-Dec-17 22:12pm

Thank you sir Balboos, however, how can I solve this error ?
Undefined variable: per
I am screwed up.

W∴ Balboos, GHB 6-Dec-17 6:36am

If you are within a class and use a member variable in the class ($per), you must tell the system where that variable is located:
If withing the class methods, you need to use: $this->per
If in an outside instance of the class ($score): $score->per

This way you can tell (if more than one class contains $per) which one you are referring to. In your case, there is no $per that is not either within your class or part of an instance of the class.

Here's an example of a class, instance, and it's useage