[英]Show data record using id in codeigniter
Column 'id_siswa' in where clause is ambiguous
where子句中的“ id_siswa”列不明确
SELECT * FROM `siswa` `a`
LEFT JOIN `pembayaran_spp` `b` ON `b`.`id_siswa`=`a`.`id_siswa`
WHERE `id_siswa` = '7%E2%80%8B'
I have 2 table. 我有2张桌子。
1.table 'siswa' structure(id_siswa,nama_siswa,id_tahun_masuk) 1.表'siswa'结构(id_siswa,nama_siswa,id_tahun_masuk)
2.table 'pembayaran_spp'-> structure(id_pembayaran,id_siswa,jml_pembayaran,id_tahun,date) 2.表'pembayaran_spp'->结构(id_pembayaran,id_siswa,jml_pembayaran,id_tahun,日期)
I want to show data 'pembayaran_spp' by id_siswa. 我想显示id_siswa的数据“ pembayaran_spp”。 so when i click detail on 'siswa', data 'pembayaran' is showed by id_siswa.
因此,当我单击“ siswa”的详细信息时,id_siswa会显示数据“ pembayaran”。
function detailtagihan($id_siswa)
{
$data['siswa'] = $this->M_keuangan->tagihansiswa($id_siswa);
$this->load->view('template/header');
$this->load->view('template/sidebar');
$this->load->view('keuangan/v_detailtagihan',$data);
$this->load->view('template/footer');
}
function tagihansiswa($id_siswa)
{
//$data = array('id_siswa' => $id_siswa );
$this->db->select('*');
$this->db->from('siswa a');
$this->db->join('pembayaran_spp b','b.id_siswa=a.id_siswa', 'left');
$this->db->where('id_siswa',$id_siswa);
$query = $this->db->get();
if($query->num_rows()>0)
return $query->result();
}
You should mention of which table you want to use the column id_siswa
in your where clause. 您应该在where子句中提到要使用哪个表
id_siswa
表。 As both of the tables are having a column with same name, you are getting this error. 由于两个表都具有相同名称的列,因此会出现此错误。
If you want to use siswa
then in your where condition write a.id_siswa
如果要使用
siswa
,请在您的where条件中编写a.id_siswa
And if you want to use pembayaran_spp
then in your where condition b.id_siswa
. 如果要使用
pembayaran_spp
则在您的where条件b.id_siswa
。
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