CODEWITHSUNDEEP
phpmysqlsqlcodeigniter
user3622745
user3622745

Reputation: 31

passing variable into SELECT clause with codeigniter

This is my model, which works when I query one table only:

function cari_username()
    {
        $username = $this->input->post('username');
        $this->db->like('username',$username);
        $data = $this->db->get('user');
        return $data;
    }

What I need is to join data from several tables, but now my code fails:

function cari_username()
    {
        $this->db->from('user');
        $username = $this->input->post('username');
        $data = $this->db->query("SELECT * FROM (SELECT user.id_user,mahasiswa.nama,user.password,user.username 
           FROM (`mahasiswa`) 
           LEFT JOIN `pendaftaran_anggota` ON `pendaftaran_anggota`.`nim` = `mahasiswa`.`nim`
           LEFT JOIN `anggota` ON `pendaftaran_anggota`.`id_anggota` = `anggota`.`id_anggota` 
           LEFT JOIN `user` ON `user`.`id_user` = `anggota`.`id_user` UNION 
           SELECT pelatih.id_user, pelatih.nama,user.password,user.username FROM (`pelatih`)
           LEFT JOIN `user` ON `user`.`id_user` = `pelatih`.`id_user` UNION 
           SELECT admin.id_user, admin.nama, user.password,user.username FROM (`admin`)
           LEFT JOIN `user` ON `user`.`id_user` = `admin`.`id_user`) userdata WHERE `username` LIKE username
            ORDER BY `userdata`.`id_user`
             ");

        return $data;
    }

Please tell me how to change the query to solve this.

Thank you.

Upvotes: 1

Views: 1597

Answers (3)

Palash Nag
Palash Nag

Reputation: 1

$result = $this ->db -> query("select * from tablename where fname ='$string'");

Don't forget to pass the $string in the function where the model is created.

Upvotes: 0

Vickel
Vickel

Reputation: 7997

You have 2 errors in your script:

first:

WHERE username LIKE username will not work, as username like any php variable should be $username

second:

mysql doesn't know which kind of variable $username is, but we know it's a string. Therefore you need to place it within single quotes: 

WHERE `username` LIKE '$username'

Upvotes: 2

newBee
newBee

Reputation: 1319

I dont know if I understood you right to be honest: So your Selection from the admin-table by username is not working?

Well I guess it should be username LIKE $username. So you forgot the $.

LEFT JOIN `user` ON `user`.`id_user` = `admin`.`id_user` userdata WHERE `username` LIKE $username ORDER BY `userdata`.`id_user`

Upvotes: 0

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