请问用sed怎么删除行尾的空格！

waker

cat -A或od看看你那个空格究竟是啥

lydongkill

13987669934,9017151398766993450132,-HYWLDZ 0 ^M$
13431799076,9017151343179907650132,-HYWLDZ CB:0005^M$
13431799076,9017151343179907650132,-HYWLDZ CB:0005^M$
13686463099,9017151368646309950132,-HYWLDZ 0 ^M$
13424759500,9017151342475950050132,-HYWLDZ 0 ^M$
13424759500,9017151342475950050132,-HYWLD 0 ^M$
13694237031,9017151369423703150132,-HYWLD CB:0047^M$
13694237031,9017151369423703150132,-HYWLDZ CB:0047^M$
13686463099,9017151368646309950132,-HYWLDZ 0 ^M$
13424759500,9017151342475950050132,-HYWLDZ 0 ^M$
13424759500,9017151342475950050132,-HYWLD 0 ^M$
13694237031,9017151369423703150132,-HYWLD CB:0047^M$
13694237031,9017151369423703150132,-HYWLDZ CB:0047^M$
13686014060,9017151368601406050132,-HYWLDZ CB:0005^M$
13686014060,9017151368601406050132,-HYWLDZ CB:0005^M$
13414259248,9017151341425924850132,-HYWLDZ CB:0005^M$
13420202140,9017151342020214050132,-HYWLD CB:0047^M$
13420202140,9017151342020214050132,-HYWLDZ CB:0047^M$
13414259248,9017151341425924850132,-HYWLDZ CB:0005^M$
13420202140,9017151342020214050132,-HYWLD CB:0047^M$
13420202140,9017151342020214050132,-HYWLDZ CB:0047^M$
15992745538,9017151599274553850132,-HYWLDZ 0 ^M$
13432465741,9017151343246574150132,-HYWLDZ CB:0047^M$
15992745538,9017151599274553850132,-HYWLDZ 0 ^M$
13432465741,9017151343246574150132,-HYWLDZ CB:0047^M$
13420248855,9017151342024885550132,-HYWLDZ CB:0047^M$
13420248855,9017151342024885550132,-HYWLDZ CB:0047^M$
后面全是这些东西，怎么作才能清楚干净。

海天

dos2unix ufile

ly5066113

sed 's/ *.$//' urfile

9er

原帖由 ly5066113 于 2007-9-14 15:36 发表
sed 's/ *.$//' urfile

不适用于12楼

ly5066113

原帖由 9er 于 2007-9-14 16:35 发表



不适用于12楼

哪里不适用？

SimpleLittle

我也是如10楼一样，都不管用啊
cat -A test.txt | sed -e 's/[\t]*$//g'
  test$
sjdflsjfl$
     sdffsdf$
sfsdlfij  $
ljijijdf   $
[test]delete inner test$
[ ] delete delete inner kong$
[] delete innter no kong$

davidbeckham921

vi 然后
%s/^M//g
这里的^M需要用ctrl+v+m打出来才行。

试试！

用bin传asc的东西就有这个^M。有米高手给解释一下！

alan9101

sed 's/[ \t]*$//g' text3