Postgres 从带有 CSV 数据的变量中复制

问题描述

我知道您可以从文件中运行这样的复制命令:

I know you can run a copy command like this from a file:

"COPY zip_codes FROM '/path/to/csv/ZIP_CODES.txt' DELIMITER ',' CSV;"

我想从 ruby​​ 变量中复制 csv 数据，这样我就可以这样做

I'd like to copy csv data from a ruby variable so I can do like so

"COPY zip_codes FROM '#{csv_data}' DELIMITER ',' CSV;"

推荐答案

这是不可能的 SQL COPY 命令.COPY 仅从 file 或 STDIN 复制.

That's not possible with the SQL COPY command. COPY only copies from a file or STDIN.

您可以将变量的内容写入文件或通过 STDIN 管道传输.仅对多于几行才有意义.

You can either write the content of the variable to a file or pipe it via STDIN. Only makes sense for more than a couple of rows.

我想我在更新之前误解了你的问题，你可能不需要这个:

I think I misunderstood your question before the update, you probably don't need this:

文件路径不能像其他数据项一样交换，并且您不能为此使用准备好的语句.在执行或使用服务器端函数使用动态 SQL 之前构建整个语句:

The file path can not be exchanged like other data items, and you can't use a prepared statement for that. Build the whole statement before executing or resort to dynamic SQL with a server-side function like:

CREATE OR REPLACE FUNCTION f_cp(_file text)
  RETURNS void AS
$BODY$
BEGIN
EXECUTE format($$COPY zip_codes FROM %L DELIMITER ',' CSV$$, $1);
END
$BODY$
  LANGUAGE plpgsql;

调用:

SELECT f_cp('/var/lib/postgres/sync/myfile.csv')

format()需要 Postgres 9.1 或更高版本.

format() requires Postgres 9.1 or later.

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