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+--- Thread: MySQL commend to CodeIgniter (/showthread.php?tid=70790)
MySQL commend to CodeIgniter - azmath - 05-30-2018
SELECT COUNT(*) FROM job_progress WHERE status='Runing';
This is MySQL commend please help me for CodeIgniter commend.
RE: MySQL commend to CodeIgniter - Krycek - 05-30-2018
PHP Code:
$this->db->select('COUNT(*) AS total');
$this->db->from('job_progress');
$this->db->where('status', 'running');
$get = $this->db->get();
$result = $get->result_object();
This should do the trick, $result contains the rows from the database
RE: MySQL commend to CodeIgniter - dave friend - 05-30-2018
The previous answer shows how to build the statement using Query Builder (QB). QB is a great tool but a lot of the time it is not necessary. Consider what QB does - it builds query statements, which are strings used to pass commands to a database.
When the required query statement is simple and won't need to be dynamically altered then QB is serious overkill. The previous answer, while completely correct, will result in a lot of code (hundreds of lines?) being executed in order to produce the string "SELECT COUNT(*) FROM job_progress WHERE status='Runing'". It also required a lot of typing to code all those QB calls.
QB will eventually use the string it builds as an argument to the method query(). Why not remove the middle-man and go straight to that method?
PHP Code:
return $this->db
->query("SELECT COUNT(*) FROM job_progress WHERE status='Runing'")
->result();