未验证 提交 f2e2086d 编写于 作者: T Tim Wade

[ci skip] Use spaceship operator to indicate precedence.

上级 2ec1fe43
......@@ -590,7 +590,7 @@ If you create an association some time after you build the underlying model, you
If you create a `has_and_belongs_to_many` association, you need to explicitly create the joining table. Unless the name of the join table is explicitly specified by using the `:join_table` option, Active Record creates the name by using the lexical order of the class names. So a join between customer and order models will give the default join table name of "customers_orders" because "c" outranks "o" in lexical ordering.
WARNING: The precedence between model names is calculated using the `<` operator for `String`. This means that if the strings are of different lengths, and the strings are equal when compared up to the shortest length, then the longer string is considered of higher lexical precedence than the shorter one. For example, one would expect the tables "paper_boxes" and "papers" to generate a join table name of "papers_paper_boxes" because of the length of the name "paper_boxes", but it in fact generates a join table name of "paper_boxes_papers" (because the underscore '_' is lexicographically _less_ than 's' in common encodings).
WARNING: The precedence between model names is calculated using the `<=>` operator for `String`. This means that if the strings are of different lengths, and the strings are equal when compared up to the shortest length, then the longer string is considered of higher lexical precedence than the shorter one. For example, one would expect the tables "paper_boxes" and "papers" to generate a join table name of "papers_paper_boxes" because of the length of the name "paper_boxes", but it in fact generates a join table name of "paper_boxes_papers" (because the underscore '_' is lexicographically _less_ than 's' in common encodings).
Whatever the name, you must manually generate the join table with an appropriate migration. For example, consider these associations:
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