Postgresql - 使用 LIKE 创建表语法错误 -

Postgresql - CREATE TABLE SYNTAX ERROR USING LIKE -

提问by jzafrilla

I'm trying to create a temp table from a parent table:

我正在尝试从父表创建临时表：

This is the code that I execute with pgAdmin III (or by JDBC in Java):

这是我使用 pgAdmin III（或 Java 中的 JDBC）执行的代码：

CREATE TEMP TABLE table1_tmp LIKE table1 INCLUDING DEFAULTS;

And the error I received is:

我收到的错误是：

[WARNING  ] CREATE TEMP TABLE table1_tmp LIKE table1 INCLUDING DEFAULTS
        ERROR:  syntax error at or near ?LIKE?
        LíNEA 1: CREATE TEMP TABLE table1_tmp LIKE table1 INCLUDING DEFAULTS
                                              ^

Reading postgresql 8.4 documentation, create tables using this, its very easy, but I don't understand where is the syntax problem.

阅读 postgresql 8.4 文档，使用它创建表，非常简单，但我不明白语法问题在哪里。

回答by Heiko Rupp

You need to put the like in to parens like

你需要把喜欢放在括号里

CREATE TEMP TABLE table1_tmp ( LIKE table1 INCLUDING DEFAULTS ) ;

This is not obvious from the docsif you don't count parens 1:1

如果您不计算括号 1:1，这在文档中并不明显

回答by fluminis

I'm not a Postgresql user but the manualsay that there is ( ) around the like setence.

我不是 Postgresql 用户，但手册上说类似 setence 有 ( ) 。

CREATE TEMP TABLE table1_tmp (LIKE table1 INCLUDING DEFAULTS);