bash - bash: can read be used as a command or expression?

Question

I want to make a oneliner loop that reads and checks what it read.

This surely won't work:

while [[ read line != "q" ]]; do; echo "enter q to quit: "; done

Zsh here tells me condition expected: read.

Perhaps read does not even have this concept of a return value.

Answer 1

Bash 陷阱#9：

如果接受命令。[ 是一个命令，而不是 if 语句的语法标记。它等价于 test 命令，只是最后一个参数必须是 ]。

[[不是if条件的特殊语法。[[本身就是一个命令。尝试这样的循环：

while read -r line; do
  [[ $line = "q" ]] && break
  # more code
done

read当它无法读取表达式并且[[是命令时返回失败，所以这有效。有关详细信息，请参阅http://mywiki.wooledge.org/BashFAQ/001。

注意：将-r开关添加到read是一个好主意 - 它可以防止输入中的反斜杠发生有趣的事情。

Answer 2

while read line; do; [[ "$line" == 'q' ]] && break; done;

似乎工作。

但我不确定为什么我不能用 替换[[ ]]，[ ]它奇怪地失败了：

$ bash
bash-3.2$ while true; do; read line; [ "$line" == 'q' ] && break; done;
bash: syntax error near unexpected token ';'                                       

$ while true; do; read line; [ "$line" == 'q' ] && break; done;
ajlfkj
zsh: = not found