如何在双引号内转义双引号? [英] How can I escape a double quote inside double quotes?

问题描述

如何在Bash中的双字符串中转义双引号?

How can I escape double quotes inside a double string in Bash?

例如，在我的shell脚本中

For example, in my shell script

#!/bin/bash

dbload="load data local infile \"'gfpoint.csv'\" into table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '\"' LINES TERMINATED BY \"'\n'\" IGNORE 1 LINES"

我无法获得带双引号的ENCLOSED BY '\"'才能正确转义.我不能对变量使用单引号，因为我想使用变量$dbtable.

I can't get the ENCLOSED BY '\"' with double quote to escape correctly. I can't use single quotes for my variable, because I want to use variable $dbtable.

推荐答案

使用反斜杠:

echo "\""     # Prints one " character.

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