bash命令替换期间的空白变量 [英] Blank variable during bash command substitution

问题描述

我正在执行命令替换并将结果保存到变量中.但是，命令的结果包含双引号，这导致变量为空.

I am doing command substitution and saving the result to a variable. However, the results of the command contain double quotes and this is causing the variable to be empty.

运行 test ="$(java -version)" 时，我得到以下结果:

When running test="$(java -version)" I get the following result:

openjdk version "1.8.0_65"
OpenJDK Runtime Environment (build 1.8.0_65-b17)
OpenJDK 64-Bit Server VM (build 25.65-b01, mixed mode)

但是运行 echo $ test 会产生空白行.

However running echo $test yields a blank line.

推荐答案

test ="$(java -version)" 直接将结果打印到终端的原因是 java -version 输出到标准错误(stderr)，而不是标准输出(stdout).

The reason that test="$(java -version)" prints the result to the terminal directly is that java -version outputs to standard error (stderr), not standard output (stdout).

因为没有stdout输出(这是 $(...)捕获的内容)，所以为 $ test 分配了一个空字符串.

Because there is no stdout output (which is what $(...) captures), $test is assigned an empty string.

解决方案是将标准错误(stderr)重定向到标准输出(stdout).

The solution is to redirect standard error (stderr) to standard output (stdout).

version=$(java -version 2>&1)

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