[英]Making mysql query to tables based on the values from another table
我有以下查询并且工作正常。 如果您看到查询,您会注意到有一个名为$code
的变量,我曾经像这样手动设置它的值$code="1001";
function check1() {
$code="1001";
$getData = $this->db->query("SELECT * FROM vouchers
LEFT JOIN details on vouchers.voucher_no = details.voucher_no
LEFT JOIN accounts on accounts.code = vouchers.account_code
WHERE (voucher_type='1' AND t_code=$code)
UNION ALL
SELECT * FROM vouchers
LEFT JOIN details on vouchers.voucher_no = details.voucher_no
LEFT JOIN accounts on accounts.code = details.t_code
WHERE (voucher_type='0' AND account_code=$code)
");
if($getData->num_rows() > 0)
return $getData->result_array();
else
return null;
}
当我在我的视图文件中显示结果时,我使用以下脚本:
<?php if(count($records) > 0) { ?>
<?php $i = $this->uri->segment(3) + 0; foreach ($records as $row){ $i++; ?>
<?php echo $row['voucher_date']; ?>
<?php echo $row['name']; ?>
<?php echo $row['amount']; ?> <br>
<?php } ?>
<?php } else { echo "No Record Found";} ?>
<?php
$sum = 0;
foreach ( $records as $row ) {
$sum += str_replace(",", "", $row['amount']);
}
?>
Total: <?php echo number_format( $sum, 2 );?>
现在我想要做的是从下面的 mysql 表中获取$code
的值并获取其所有值的结果,这样当我在视图文件中显示结果时它不会显示$code="1001"
的结果$code="1001"
不仅适用于$code="1002"
, $code="1003"
。
为了实现我想要的,我必须更改上面的 mysql 查询。但我不知道该怎么做,请你告诉我。
提前致谢:)
表名:accounts
code name
1001 Cash Account
1002 Advertising Expense
1003 Accounts Receivable
SELECT * FROM vouchers
LEFT JOIN details on vouchers.voucher_no = details.voucher_no
LEFT JOIN accounts on accounts.code = vouchers.account_code
WHERE voucher_type='1' AND t_code IN (SELECT * FROM code_table)
UNION ALL
SELECT * FROM vouchers
LEFT JOIN details on vouchers.voucher_no = details.voucher_no
LEFT JOIN accounts on accounts.code = details.t_code
WHERE voucher_type='0' AND account_code IN (SELECT * FROM code_table)
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.