bash sed：如何打印到文件末尾的行范围
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/32889645/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
sed: how to print a range of line to end of file
提问by JosephSmith47
I can print a range of lines from a file using this cmd:
我可以使用此 cmd 从文件中打印一系列行：
sed -n 267975,1000000p < dump-gq1-sample > dump267975
but how to print to the end? I tried
但如何打印到最后？我试过
sed -n 267975,$p < dump-gq1-sample > dump267975
and it gives me:
它给了我：
sed: 1: "267975,": expected context address
sed: 1: "267975,": 预期的上下文地址
回答by edi9999
You're the victim of Shell Parameter Expansion
你是Shell 参数扩展的受害者
sed -n 267975,$p < dump-gq1-sample > dump267975
is received by sedas
被接收sed为
sed -n 267975, < dump-gq1-sample > dump267975
because the pvariable is undefined.
因为p变量未定义。
You should quote your parameter with single quotes '
你应该用单引号引用你的参数 '
sed -n '267975,$p' < dump-gq1-sample > dump267975
See https://www.gnu.org/software/bash/manual/html_node/Shell-Expansions.htmlFor the full list of existing shell expansions.
有关现有 shell 扩展的完整列表，请参阅https://www.gnu.org/software/bash/manual/html_node/Shell-Expansions.html。
回答by jessup Hymanson
the single-quote response does not work in cases where the start/end of range are passed to sed as a variable.
在范围的开始/结束作为变量传递给 sed 的情况下，单引号响应不起作用。
including a space between the end-symbol $ and p in single quotes works on my systems to prevent unintended expansion of $p
在单引号中的结束符号 $ 和 p 之间包含一个空格适用于我的系统，以防止 $p 意外扩展
sed -n '267975,$ p' ${INPUT_FILE} > ${OUTPUT_FILE} #double quotes work here too
If you need to pass the initial line as a variable, double quotes are required for the expansion of $L1 so the space is generally a good idea:
如果您需要将初始行作为变量传递，则 $L1 的扩展需要双引号，因此空格通常是一个好主意：
L1=267975
sed -n "${L1},$ p" ${INPUT_FILE} > dump${L1}
