Question

我有以下选择查询。

    $this->db->select("$order_table.user_id, $order_table.total_price,
                       $order_table.shipping, $order_table.tax, 
                       $order_table.filled, $order_table.shipped, 
                        $order_table.tracking");
    $this->db->select("$user_table.first_name as user_first_name, 
                       $user_table.last_name as user_last_name, 
                       $user_table.email as user_email");
    $this->db->select("$mailing_address_table.first_name as mailing_first_name, 
                       $mailing_address_table.last_name as mailing_last_name, 
                       $mailing_address_table.address as mailing_address, 
                       $mailing_address_table.address_2 as mailing_address_2, 
                       $mailing_address_table.city as mailing_city, 
                       $mailing_address_table.state as mailing_state, 
                       $mailing_address_table.zip_code as mailing_zip_code, 
                       $mailing_address_table.country as mailing_country");

    $this->db->select("$billing_address_table.first_name as billing_first_name, 
                       $billing_address_table.last_name as billing_last_name, 
                       $billing_address_table.address as billing_address, 
                       $billing_address_table.address_2 as billing_address_2, 
                       $billing_address_table.city as billing_city, 
                       $billing_address_table.state as billing_state,              
                       $billing_address_table.zip_code as billing_zip_code, 
                       $billing_address_table.country as billing_country");

    $this->db->join($user_table, 
                   "$user_table.id = $order_table.user_id", "left");
    $this->db->join($mailing_address_table, 
                   "$mailing_address_table.user_id = $user_table.id", "left");
    $this->db->join($billing_address_table, 
                   "$billing_address_table.user_id = $user_table.id", "left");

无论如何，我想是使用CONCAT，我可以使用前缀＆＃34;计费_＆＃34;到我在结算中选择的每一列。当我没有选择所有内容，而只是选择特定的列时，我似乎无法解决这个问题。

我喜欢从$ user_table中选择的每一列都是＆＃34; user _＆＃34;，$ mailing_table＆＃34; mailing _＆＃34;等等。

注意：

1）我没有选择这些选择查询中每列的所有字段，因此我无法进行SELECT * FROM ...

2）我不想更新数据库中的列名，只是为了这个特定的SELECT使用。

Answer 1

您可以在此处创建字段列表：

SET @SCHEMA = 'toimport';

SET @PREFIX = 'billing_';
SET @POSTFIX = 'address';

SELECT GROUP_CONCAT(
 CONCAT(@SCHEMA,'.',@PREFIX,'.',@POSTFIX,'.',COLUMN_NAME,
 ' AS ',@PREFIX,COLUMN_NAME) SEPARATOR ',\n')
 FROM information_schema.COLUMNS
 WHERE TABLE_SCHEMA = @SCHEMA AND TABLE_NAME = CONCAT( @PREFIX,  @POSTFIX );

结果看起来像：

toimport.billing_.address.id AS billing_id,
toimport.billing_.address.name AS billing_name,
toimport.billing_.address.street AS billing_street,
toimport.billing_.address.a1 AS billing_a1,
toimport.billing_.address.a2 AS billing_a2

Answer 2

在这里您的答案更多表：

查询：

SET @SCHEMA = 'toimport';

SELECT
  GROUP_CONCAT(
    CONCAT(
      @SCHEMA,'.',
      SUBSTRING_INDEX(TABLE_NAME, '_', 1),'.',
      SUBSTRING_INDEX(TABLE_NAME, '_', -1),'.',
      COLUMN_NAME, ' AS ',
      SUBSTRING_INDEX(TABLE_NAME, '_', 1),'_',COLUMN_NAME
    ) SEPARATOR ',\n'
  )
  FROM information_schema.COLUMNS
  WHERE TABLE_SCHEMA = @SCHEMA
  AND TABLE_NAME IN ('billing_address', 'email_post');

结果如下：

toimport.billing.address.id AS billing_id,
toimport.billing.address.name AS billing_name,
toimport.billing.address.street AS billing_street,
toimport.billing.address.a1 AS billing_a1,
toimport.billing.address.a2 AS billing_a2,
toimport.email.post.id AS email_id,
toimport.email.post.name AS email_name,
toimport.email.post.street AS email_street,
toimport.email.post.a1 AS email_a1,
toimport.email.post.a2 AS email_a2

Codeigniter：选择所有指定列的前缀字符串

2 个答案: