douzhechi2435
2014-05-18 15:18
浏览 48
已采纳

使用codeigniter将变量传递给SELECT子句

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.

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3条回答 默认 最新

  • drrvnbwle80177811 2014-05-18 20:15
    已采纳

    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'
    
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