Warning: use of undefined constant search_query - assumed 'search_query' (this will throw an error in a future version of PHP) in C:\xampp\htdocs\CMS\all_post.php on line 721

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Hello all
I am try to make search engine on my post page but this message show on page (
Warning: Use of undefined constant search_query - assumed 'search_query' (this will throw an Error in a future version of PHP) in C:\xampp\htdocs\CMS\All_post.php on line 721) in echo "Some RESULT"; i am try to print something but not work also

What I have tried:

 <?php
            
            
        if(isset($_POST['submit'])) {
            
        $search = $_POST['search'];
            
        $query = "SELECT * FROM posts WHERE post_title LIKE '%$search%' ";   
        $query_search = mysqli_query($connection,$query); 
        }
//        if(!query_search) {
//        die("QUERY FAILED" . mysqli_error($connection));
//        }
//            $count = mysqli_num_rows($query_search);
//            if($count == 0){
//            echo "<h1> no result</h1>";
//               
//            }else {
//                
//                echo "Some RESULT";
//            }
//            }   
           
        ?>

Posted 19-May-18 22:35pm

MuhammadNaamh

Updated 19-May-18 22:53pm

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Patrice T · Accepted Answer · 2018-05-19T22:54:00

Quote:
Warning: Use of undefined constant search_query - assumed 'search_query' (this will throw an Error in a future version of PHP)

PHP

<?php


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

	$search = $_POST['search'];

	$query = "SELECT * FROM posts WHERE post_title LIKE '%$search%' ";
	$query_search = mysqli_query($connection,$query);
	// $query_search is defined inside a conditional structure
}
//  if(!query_search) { // thus it may not exist at this point
//      die("QUERY FAILED" . mysqli_error($connection));
//  }
//  $count = mysqli_num_rows($query_search);
//  if($count == 0){
//      echo "<h1> no result</h1>";
//  }else {
//      echo "Some RESULT";
//  }
//}

?>

PHP

$query = "SELECT * FROM posts WHERE post_title LIKE '%$search%' ";

Not a solution to your question, but another problem you have.
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[^]