mysql - 如何实现“SHOW COLUMN FROM (SELECT Result) WHERE Column_name = 'column_name'”？

Question

晚安，我的兄弟/姐妹。

我想检查选择结果中是否存在列。

我已经尝试了如下的一些努力，但它给了我一个错误。

SHOW COLUMNS 
    FROM (SELECT
                    hasil.skpd, hasil.kode_skpd, SUM(hasil.sudah_isi) AS sudah_isi, SUM(hasil.belum_isi) AS belum_isi
                FROM
                    (
                        SELECT
                            a.id_pegawai, c.skpd, c.kode_skpd, a.id_satuan_organisasi, a.nama_pegawai, a.nip,
                            CASE
                                WHEN i.id_pegawai IS NULL THEN 0
                                ELSE 1
                            END AS sudah_isi,
                            CASE
                                WHEN i.id_pegawai IS NULL THEN 1
                                ELSE 0
                            END AS belum_isi
                        FROM
                            tbl_pegawai a
                            LEFT JOIN ref_skpd c ON a.id_satuan_organisasi = c.id_skpd
                            LEFT JOIN tbl_bapertarum i ON (a.id_pegawai = i.id_pegawai AND YEAR(i.tgl_lapor)='2015')
                        GROUP BY
                            a.id_pegawai
                    ) hasil
                WHERE
                    1 AND hasil.kode_skpd LIKE 'Badan Kepegawaian Daerah%'
                GROUP BY
                    hasil.id_satuan_organisasi
                ORDER BY
                    hasil.kode_skpd) WHERE Field = "kode_skpd"

这是完整的错误结果：

#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 '(SELECT
                    hasil.skpd, hasil.kode_skpd, SUM(hasil.sudah_isi) AS s' at line 2

我如何从 SELECT 结果中找到一列或多列的存在？

在此先感谢兄弟/姐妹。

标签： mysqlsqlxampp