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- How do I preserve newlines in a quoted string in Bash? [duplicate] 1 answer
如何在Bash中用引号字符串保留换行符? [重复] 1回答
How can I get cat to add a literal line break after each line? (for echo -e to read it) in bash
如何让cat在每行后添加文字换行符? (用于回显-e来读取它)在bash中
root@111[~]# cat names
Joe Smith
John St. John
Jeff Jefferson
root@111 [~]# var=`cat names`;echo -e $var
Joe Smith John St. John Jeff Jefferson
I want the second command to produce output identical to the first.
我希望第二个命令产生与第一个命令相同的输出。
1 个解决方案
#1
First, don't use echo -e; it adds no value here, and the POSIX standard for echo explicitly disallows the behavior you're expecting it to provide (a POSIX-compliant echo with XSI extensions will honor backslash escapes without -e; a POSIX-compliant echo without XSI extensions will just echo the literal string -e and still ignore backslash escapes; the GNU implementation that changes its behavior based on whether -e breaks both XSI and baseline versions of the standard).
首先,不要使用echo -e;它在这里没有增加任何价值,并且echo的POSIX标准明确禁止你期望它提供的行为(符合POSIX的回声与XSI扩展将在没有-e的情况下支持反斜杠转义;没有XSI扩展的符合POSIX的回声将只是回显文字字符串-e并仍然忽略反斜杠转义; GNU实现根据-e是否同时破坏标准的XSI和基线版本来改变其行为。
Second, use quotes:
二,使用引号:
echo "$var"
...or, better, skip echo altogether in favor of printf:
...或者,更好的是,完全跳过echo以支持printf:
printf '%s\n' "$var"
#1
First, don't use echo -e; it adds no value here, and the POSIX standard for echo explicitly disallows the behavior you're expecting it to provide (a POSIX-compliant echo with XSI extensions will honor backslash escapes without -e; a POSIX-compliant echo without XSI extensions will just echo the literal string -e and still ignore backslash escapes; the GNU implementation that changes its behavior based on whether -e breaks both XSI and baseline versions of the standard).
首先,不要使用echo -e;它在这里没有增加任何价值,并且echo的POSIX标准明确禁止你期望它提供的行为(符合POSIX的回声与XSI扩展将在没有-e的情况下支持反斜杠转义;没有XSI扩展的符合POSIX的回声将只是回显文字字符串-e并仍然忽略反斜杠转义; GNU实现根据-e是否同时破坏标准的XSI和基线版本来改变其行为。
Second, use quotes:
二,使用引号:
echo "$var"
...or, better, skip echo altogether in favor of printf:
...或者,更好的是,完全跳过echo以支持printf:
printf '%s\n' "$var"