I want to select a table which holds all the issues a specific project id has. For that I am planning to do a LEFT JOIN and get the projects_id in Projects table to the column 'issues_projects_id' in issues table. Isn't it a JOIN
that has to be used here?
// $this->db->join('issues', 'issues.issues_projects_id = projects.projects_id', 'left');
// $query1 = $this->db->get();
$squery="SELECT * FROM projects LEFT JOIN issues ON issues.issues_projects_id=projects.projects_id";
$query1= $this->db->query($squery);
$query= $this->db->query($squery);
return $query->result();
return $query1->result();
The above code shows below error & displays no output.
A Database Error Occurred
Error Number: 1064
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'LEFT JOIN `issues` ON `issues`.`issues_projects_id` = `projects`.`projects_id`' at line 2
SELECT * LEFT JOIN `issues` ON `issues`.`issues_projects_id` = `projects`.`projects_id`
Filename: C:/wamp64/www/projects/issue_tracker/system/database/DB_driver.php
Line Number: 691
Projects Table
projects_id | projects_name |
---|---|
1 | abc |
2 | xyz |
Issues Table
issues_id | issues_description | issues_projects_id |
---|---|---|
1 | gfj | 1 |
2 | fghgj | 1 |
you are missing from statement
:
SELECT *
FROM `projects` --< here
LEFT JOIN `issues`
ON `issues`.`issues_projects_id` = `projects`.`projects_id`
WHERE projects_id = ?