John Davidson

MYSQL | PHP-MySQLi-Database-Class | Subquery

0 comments
Message:


I want help in forming this query using class object wrapper(MysqliDb ). You can find here (MysqliDb )


This is the query i needed to convert


SELECT * FROM (SELECT * FROM users WHERE eventId=34) as t2 WHERE name like "alex" or email like "alex";

For Example


    $db = getDbInstance();
$db->where('eventId', 34);
$status = $db->get('users');
//SELECT * FROM USERS WHERE eventId=34



My Answer:
Here is how you can form the query using the MysqliDb class object wrapper:

php
// Include the MysqliDb class
require_once 'path/to/MysqliDb.php';

// Create a new instance of MysqliDb
$db = new MysqliDb('hostname', 'username', 'password', 'database');

// Form the subquery
$subQuery = $db->subQuery();
$subQuery->where('eventId', 34);
$subQuery->get('users');

// Form the main query
$db->join($subQuery, 't2', 'INNER');
$db->where('name', '%your_search_term%', 'like');
$results = $db->get('t2', null, '*');

// Execute the query
if($db->count > 0) {
foreach($results as $row) {
// Process the results
}
} else {
echo 'No results found';
}


Make sure to replace `'hostname'`, `'username'`, `'password'`, `'database'`, and `'%your_search_term%'` with your actual database connection details and search term.

Rate this post

4 of 5 based on 5308 votes

Comments




© 2024 Hayatsk.info - Personal Blogs Platform. All Rights Reserved.
Create blog  |  Privacy Policy  |  Terms & Conditions  |  Contact Us