Question

我有一个查询，可以在phpmyadmin上尝试它，并且可以正常工作。

SELECT  * FROM `penyakit` LEFT JOIN puskesmas ON penyakit.id_puskesmas=puskesmas.id_puskesmas WHERE (YEAR(waktu_upload) = 2018 AND MONTH(waktu_upload) = 12)  AND penyakit.id_puskesmas = 19

我将此查询用于我的过滤器功能，如下所示：

function status_data($year,$month){
    $this->db->select('*');
    $this->db->from('puskesmas');
    $this->db->group_by('puskesmas.nama_puskesmas');
    $this->db->join('penyakit', 'puskesmas.id_puskesmas=penyakit.id_puskesmas', 'left');

    $data= $this->db->get()->result();

    foreach ($data as $value) {

        $query = $this->db->query("SELECT * FROM `penyakit` LEFT JOIN puskesmas ON penyakit.id_puskesmas=puskesmas.id_puskesmas WHERE YEAR(waktu_upload) = '$year' AND MONTH(waktu_upload) = '$month' 
            AND puseksmas.id_puskesmas=".$value->id_puskesmas);

        $num = $query->num_rows();

         $value->num = $num;
        if($num < 19 ){
            $value->status = "<button type='button' class='btn btn-danger btn-rounded'> Uncomplete</button>";
        }else{
            $value->status = "<button type='button' class='btn btn-success btn-rounded'> Complete </button>";
        }

    }
    //die(print_r($data));
    return $data;
}

但是此查询无法获取ID并显示这样的错误。

SELECT *从penyakit左起，在penyakit.id_puskesmas = puskesmas.id_puskesmas上加入puskesmas，其中YEAR（waktu_upload）='2018'AND MONTH（waktu_upload）='12'和puseksmas.id_puskesmas =

此ID未显示。我该如何解决？

我如何从Codeigniter的表中获取ID？

0 个答案: