如何从Unix文件删除空行 [英] How to remove blank lines from a Unix file

问题描述

我需要从一个输入文件中删除所有的空行，并写入到输出文件。这里是我的数据如下。

  11216,33,1032747,64310,1,0,0,1.878,0,0,0,1,1,1.087,5,1,1,18-JAN- 13,00060322132111216,33,1033196,31300,1,0,0,1.5391,0,0,0,1,1,1.054,5,1,1,18-JAN-13,05976215300311216,33,1033246,31300,1,0,0,1.5391,0,0,0,1,1,1.054,5,1,1,18-JAN-13,00060321103211216,33,1033280,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,05511103400111216,33,1033287,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,00037868970111216,33,1033358,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,00009373730111216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,04580204192611216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,04580204195411216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,04580204932611216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,04580204938311216,33,1036985,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,00009341558011216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,00078120200111216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,00078126130511216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,00078160395511216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781615746
 

解决方案

  SED -i'/ ^ $ / d的'富
 

这告诉 SED 删除每一行相匹配的正则表达式 ^ $ 即每一个空行。在 -i 标记编辑就地该文件，如果你的 SED 不支持，你可以写输出到一个临时文件，并替换原来的：

  sed的'/ ^ $ / d的'富＆GT; foo.tmp
MV foo.tmp富
 

如果您也想删除只包含空格的行（不只是空行），然后使用：

  SED -i/ ^ [[：空间：]] * $ / d的'富
 

编辑：还行的末尾删除空格，因为显然你已经决定你需要它：

  SED -i/ ^ [[：空间：]] * $ /天; S / [[：空间：]] * $ //'富
 

I need to remove all the blank lines from an input file and write into an output file. Here is my data as below.

11216,33,1032747,64310,1,0,0,1.878,0,0,0,1,1,1.087,5,1,1,18-JAN-13,000603221321

11216,33,1033196,31300,1,0,0,1.5391,0,0,0,1,1,1.054,5,1,1,18-JAN-13,059762153003

11216,33,1033246,31300,1,0,0,1.5391,0,0,0,1,1,1.054,5,1,1,18-JAN-13,000603211032

11216,33,1033280,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,055111034001

11216,33,1033287,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,000378689701

11216,33,1033358,31118,1,0,0,1.5513,0,0,0,1,1,1.115,5,1,1,18-JAN-13,000093737301

11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802041926

11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802041954

11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802049326

11216,33,1035476,37340,1,0,0,1.7046,0,0,0,1,1,1.123,5,1,1,18-JAN-13,045802049383

11216,33,1036985,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000093415580

11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781202001

11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781261305

11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781603955

11216,33,1037003,15151,1,0,0,1.4436,0,0,0,1,1,1.065,5,1,1,18-JAN-13,000781615746

解决方案

sed -i '/^$/d' foo

This tells sed to delete every line matching the regex ^$ i.e. every empty line. The -i flag edits the file in-place, if your sed doesn't support that you can write the output to a temporary file and replace the original:

sed '/^$/d' foo > foo.tmp
mv foo.tmp foo

If you also want to remove lines consisting only of whitespace (not just empty lines) then use:

sed -i '/^[[:space:]]*$/d' foo

Edit: also remove whitespace at the end of lines, because apparently you've decided you need that too:

sed -i '/^[[:space:]]*$/d;s/[[:space:]]*$//' foo

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