PHP/SQL:仅使用一个查询，如果两个表中都有数据，则从两个表中选择行；否则，仅从一个表中进行SELECT [英] PHP/SQL: Using only one query, SELECT rows from two tables if data is in both tables, or just SELECT from one table if not

问题描述

我有两个评分/投票表.一个用于用户投票和统计，另一个用于外部投票. 外部投票表始终具有数据，因为它的DEAFULT = 0，但用户投票仅包含任何用户为该特定ID投票的数据.

I have two rating/votes tables. One for user votes and stats and another for external votes. The external votes table always have data because it has a DEAFULT = 0, but the users votes only contains data if any user have voted for that specific ID.

所以我做这样的事情:

    $sql = 'SELECT  ratings_stats.votes, ratings_stats.total_value,
                   ratings_stats.view, ratings_stats.fav, ratings_stats.wish,
                   ratings_external.votes, ratings_external.total_value
            FROM ratings_stats, ratings_external
            WHERE ratings_stats.imdbID = ?
            AND ratings_stats.imdbID = ratings_external.imdbID
            LIMIT 1';

我想从两个表中选择数据(如果可用)，或者如果没有，则仅形成第二个(外部选票)表.

I want to select data from both tables if available OR only form the second (external votes) table if not.

如何在不进行新查询的情况下做到这一点?

How to can I do it without making a new query?

推荐答案

SELECT  ratings_stats.votes, 
        ratings_stats.total_value,
        ratings_stats.view, 
        ratings_stats.fav, 
        ratings_stats.wish,
        ratings_external.votes, 
        ratings_external.total_value
FROM  ratings_external
LEFT JOIN ratings_stats ON ratings_stats.imdbID = ratings_external.imdbID
WHERE ratings_external.imdbID = ?
LIMIT 1

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