bash 如何在Unix中计算文件中的所有空格
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How to count all spaces in a file in Unix
提问by Baba
I want to count all the spaces from my file in Unix and I have tried the following command:
我想在 Unix 中计算我文件中的所有空格，我尝试了以下命令：
grep " " file1 | wc
It is giving me the following output:
它给了我以下输出：
3 6 34
There are three spaces in my file so is it accurate command and further more how can I filter this to get exactly the spaces so only '3' should come as output and also how can I remove it
我的文件中有三个空格，所以它是准确的命令，更进一步，我如何过滤它以准确获取空格，所以只有“3”应该作为输出出现，以及如何删除它
回答by hek2mgl
Use grepand wcin a way like this to count the occurrences of spaces:
以这样的方式使用grep和wc来计算空格的出现次数：
grep -o ' ' | wc -l
grep -owill print every match in a separate line. The number of those lines can afterwards counted easily using wc -l
grep -o将在单独的行中打印每个匹配项。这些行的数量之后可以很容易地使用wc -l
回答by Gordon Davisson
Use trto remove everything but the spaces, then wc -cto count the remaining (space) characters:
使用tr去除一切，但空格，然后wc -c计算剩余的（空间）的字符：
tr -cd ' ' <file1 | wc -c
回答by David Elliman
This sed (stream editor) command will remove all the white space in a text file:
这个 sed（流编辑器）命令将删除文本文件中的所有空白：
sed -e 's/^[ \t]*//' -e 's/[ \t]*$//' yourFile
