别名左联接上的列名

本文介绍了别名左联接上的列名的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

假设我有两个表，并且两个表的主要标识符都使用名称"id".如果要对这两个表执行联接，该如何别名要与前一个表联接的表的id?

Let's say I have two tables, and both their primary identifiers use the name 'id'. If I want to perform a join with these two tables, how would I alias the id of the table that I want to join with the former table?

例如:

SELECT * FROM `sites_indexed` LEFT JOIN `individual_data` ON `sites_indexed`.`id` = `individual_data`.`site_id` WHERE `url` LIKE :url

现在，site_id应该与sites_indexed.id链接.代表individual_data行的实际id与sites_indexed具有相同的 title .

Now, site_id is supposed to link up with sites_indexed.id. The actual id which represents the row for individual_data however has the same title as sites_indexed.

就个人而言，我喜欢为所有内容使用名称id，因为它可以使内容保持一致.但是，在对服务器端进行脚本编写时，这会使事情变得混乱.

Personally, I like to just use the name id for everything, as it keeps things consistent. When scripting server-side however, it can make things confusing.

例如

$var = $result['id'];

鉴于上述查询，这不会使解释器感到困惑吗?

Given the aforementioned query, wouldn't this confuse the interpreter?

无论如何，这是如何完成的?

Anyway, how is this accomplished?

推荐答案

而不是使用"SELECT *"选择所有字段，而应显式命名所需的每个字段，并根据需要使用AS对其进行别名.例如:

Instead of selecting all fields with "SELECT *" you should explicitly name each field you need, aliasing them with AS as required. For example:

SELECT si.field1 as si_field1, si.field2 as si_field2, ind_data.field1 as ind_data_field1 FROM sites_indexed as si LEFT JOIN individual_data as ind_data ON si.id = ind_data.site_id WHERE `url` LIKE :url

然后您可以在结果集中引用别名.

And then you can reference the aliased names in your result set.