How to fix a undefined variable in PHP

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hi
I am trying to insert and display images in my database but in the insert file I am getting this error

is there anyway that i can fix this. can some one point me in the direction. i am not asking no one to do this assignment just looking for some explaining and some resolution to it. this is the error

PHP

Notice: Undefined variable: img_id in C:\wamp64\www\luana_itec244\php\upload.php on line 15

this is what I have tried.

What I have tried:

PHP

<pre><?php

		session_start();
		
	if(isset($_POST['submit']))
	{

		$db_host='localhost';
		$db_username='root';
		$db_password="";
		
		$con=mysqli_connect($db_host,$db_username,$db_password) or die(mysqli_connect_error());
		
			mysqli_select_db($con, 'food') or die(mysqli_error($con));
			$sql="SELECT * FROM tbl_images WHERE img_id='$img_id'";
			$result=mysqli_query($con, $sql) or die("Error:" .mysqli_error($con));
			$rowcount=mysqli_num_rows($result);
			
			if($rowcount >=1)
			{
				echo"<script type=\"text/javascript\";
						alert('files not uploaded');
						window.location=\"login.html\";
					</script>";
			}
				else
				{
					//insert images into table
					
					$sql = "INSERT INTO tbl_images
					VALUES('img_id','name','image')";
				
					if(mysqli_query($con, $sql))
					{
						mysqli_close($con);
						header("location:dashboard.php");
					}
					else
					{
						echo "Error inserting images";
					}
				}
	}
	
	
?>

Posted 24-May-18 15:38pm

divinity02

Updated 16-Oct-21 3:46am

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Patrice T · Answer 1 · 2018-05-24T17:54:00

Looks like the variable $img_id do not exist. You have to create a variable befire using it.

PHP

$sql="SELECT * FROM tbl_images WHERE img_id='$img_id'";

Another problem you have on same line.
Never build an SQL query by concatenating strings. Sooner or later, you will do it with user inputs, and this opens door to a vulnerability named "SQL injection", it is dangerous for your database and error prone.
A single quote in a name and your program crash. If a user input a name like "Brian O'Conner" can crash your app, it is an SQL injection vulnerability, and the crash is the least of the problems, a malicious user input and it is promoted to SQL commands with all credentials.
SQL injection - Wikipedia[^]
SQL Injection[^]
SQL Injection Attacks by Example[^]
PHP: SQL Injection - Manual[^]
SQL Injection Prevention Cheat Sheet - OWASP[^]

This one is weird too:

PHP

$sql = "INSERT INTO tbl_images VALUES('img_id','name','image')";

Mukesh Jakhar · Answer 2 · 2018-05-30T20:20:00

"$img_id" variable does not exist. You have to declare a variable before using it. using it.
You have to store your image id into $img_id variable like: $img_id=5;
Below updated code you can use, i think it will be help of you.

PHP

<?php
	session_start();		
	if(isset($_POST['submit']))
	{

		$db_host='localhost';
		$db_username='root';
		$db_password="";
		$img_id=5;
		
		$con=mysqli_connect($db_host,$db_username,$db_password) or die(mysqli_connect_error());
		
			mysqli_select_db($con, 'food') or die(mysqli_error($con));
			$sql="SELECT * FROM tbl_images WHERE img_id='$img_id'";
			$result=mysqli_query($con, $sql) or die("Error:" .mysqli_error($con));
			$rowcount=mysqli_num_rows($result);
			
			if($rowcount >=1)
			{
				echo"<script type=\"text/javascript\";
						alert('files not uploaded');
						window.location=\"login.html\";
					</script>";
			}
				else
				{
					//insert images into table
					
					$sql = "INSERT INTO tbl_images
					VALUES('img_id','name','image')";
				
					if(mysqli_query($con, $sql))
					{
						mysqli_close($con);
						header("location:dashboard.php");
					}
					else
					{
						echo "Error inserting images";
					}
				}
	}	
?>

How to fix a undefined variable in PHP

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