Question

我有3个表 - 用户，期刊，期刊。我使用选择的标签从3个表中选择数据。

$sqltheme="SELECT users.id as uid, users.name, users.surname, users.avatar, journals.id, journals.author_id, journals.title, journals.text, journals.create_date, journaltags.name as jname FROM users
INNER JOIN journals ON users.id=journals.author_id
INNER JOIN journaltags ON journaltags.journal_id = journals.id WHERE journals.create_date LIKE ? AND journals.author_id=? AND (".$expression.") ORDER BY journals.id DESC LIMIT 10";
$stmtheme=$conn->prepare($sqltheme);
$stmtheme->execute($array);

但是如果一个期刊的两个标签相同，那么它将被选择两次相同的期刊。我怎样才能制作DISTINCT journals.id。我试过GROUP BY journals.id，但没有帮助。

Answer 1

如果我理解正确，您的问题是journaltags表可能有一行或多行，其中包含重复的journal_id和name列值，对吧？您只需在单词distinct后面的select语句中添加SELECT子句：

SELECT DISTINCT users.id as uid, users.name, users.surname, users.avatar, journals.id, journals.author_id, journals.title, journals.text, journals.create_date, journaltags.name as jname FROM users
INNER JOIN journals ON users.id=journals.author_id
INNER JOIN journaltags ON journaltags.journal_id = journals.id WHERE journals.create_date LIKE ? AND journals.author_id=? AND (".$expression.") ORDER BY journals.id DESC LIMIT 10

您的GROUP BY journals.id无效的原因是因为您还需要将其他列包含在分组中。添加distinct本质上是撰写group by [all selected columns]

的简短方法

只选择一列不同的mysql

1 个答案: