从 Bash 中的行中删除中间的 n 个字符
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/18186481/
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
Remove the middle n characters from lines in Bash
提问by Felixia
I am trying to cut out the middle of each line in a file. All the lines are like this:
我试图剪掉文件中每一行的中间部分。所有的行都是这样的：
79.472850 97 SILENCE
and I need to end up with:
我需要结束：
79.472850 SILENCE
As each line has the undesired portion starting at character ten and ending at character 14, I was trying to use sed in this way:
由于每一行都有不想要的部分，从第 10 个字符开始到第 14 个字符结束，我试图以这种方式使用 sed：
sed "s/\(.\{9\}\).\{6\}//"but I just end up with everything after character 14. The numbers following the tab space change in every file. What can I do to make sed just cut out the tab and two digits?
sed "s/\(.\{9\}\).\{6\}//"但我最终得到了字符 14 之后的所有内容。每个文件中制表符空间后面的数字都会发生变化。我该怎么做才能让 sed 只剪掉标签和两位数字？
Thanks for your help.
谢谢你的帮助。
回答by fedorqui 'SO stop harming'
As per your input and expected output, this can be a way:
根据您的输入和预期输出，这可以是一种方式：
$ echo "79.472850 97 SILENCE" | tr -s " " | cut -d" " -f1,3
79.472850 SILENCE
tr -s " "deletes repeated spaces.cut -d" " -f1,3prints 1st and 3rd field based on splitting by spaces.
tr -s " "删除重复的空格。cut -d" " -f1,3根据空格分割打印第一个和第三个字段。
With sed:
与sed：
$ sed 's#\([^ ]*\)[ ]*\([^ ]*\)[ ]*\([^ ]*\)# #g' <<< "79.472850 97 SILENCE"
79.472850 SILENCE
回答by devnull
Looks like you need the first and the third fieldsfrom the input:
看起来您需要输入中的第一个和第三个字段：
$ echo "79.472850 97 SILENCE" | awk '{print , }'
79.472850 SILENCE
回答by cdarke
No need to call external programs like sedor cut, or other languages like awk, bashcan do it for you:
不需要调用像外部程序sed或cut类似，或其他语言awk，庆典能为你做：
var="79.472850 97 SILENCE"
echo ${var:0:9}${var:14}
79.472850 SILENCE
${var:0:9}copies 9 characters starting at position 0 (start of text).
${var:0:9}从位置 0（文本开头）开始复制 9 个字符。
${var:14}copies from character position 14 to the end of text.
${var:14}从字符位置 14 复制到文本末尾。
Alternatively, if it is space-delimited fields you need:
或者，如果是空格分隔的字段，您需要：
read one two three <<< "$var"
echo "$one $three"
79.472850 SILENCE
Again, that uses pure bash.
同样，它使用纯 bash。
回答by svante
Here is an awk solution that works even if you have spaces in your third column :
这是一个 awk 解决方案，即使您的第三列中有空格也能正常工作：
awk '{=""; print}' file
Where $2=""empties the second column and
printoutputs all columns.
其中$2=""清空第二列并
print输出所有列。
回答by Kent
if you want to extract exactly like what you described with sed:
如果你想完全像你用 sed 描述的那样提取：
kent$ echo "79.472850 97 SILENCE"|sed -r 's/(^.{9})(.{4})(.*)/ /'
79.472850 SILENCE
if you want to extract the 1st and 3rd/last columns, from your space separated input, you could use awk: awk '$2="";7'
如果要从空格分隔的输入中提取第一列和第三列/最后一列，可以使用 awk： awk '$2="";7'
kent$ echo "79.472850 97 SILENCE"|awk '="";7'
79.472850 SILENCE
