bash sed 返回“sed：命令乱码”
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sed returns "sed: command garbled"
提问by user2877175
I have this data in file.txt:
我在 file.txt 中有这些数据：
1234-abca-dgdsf-kds-2;abc dfsfds 2
123-abcdegfs-sdsd;dsfdsf dfd f
12523-cvjbsvndv-dvd-dvdv;dsfdsfpage
I want to replace the string after "-" and up to ";" with just ";", so that I get:
我想替换“-”之后的字符串，直到“;” 只用“;”，所以我得到：
1234;abc dfsfds 2
123;dsfdsf dfd f
12523;dsfdsfpage
I tried with the command:
我尝试使用以下命令：
sed -e "s/-.*;/;" file.txt
But it gives me the following error:
但它给了我以下错误：
sed command garbled
sed 命令乱码
Why is this happening?
为什么会这样？
采纳答案by iamauser
This should work :
这应该工作：
sed 's/-.*;/;/g' file > newFile
回答by fedorqui 'SO stop harming'
sedreplacement commands are defined as (source):
sed替换命令定义为（源）：
's/REGEXP/REPLACEMENT/[FLAGS]'(substitute) Match the regular-expression against the content of the pattern space. If found, replace matched string with REPLACEMENT.
's/REGEXP/REPLACEMENT/[FLAGS]'（替代）将正则表达式与模式空间的内容进行匹配。如果找到，用 REPLACEMENT 替换匹配的字符串。
However, you are saying:
然而，你是说：
sed "s/-.*;/;"
That is:
那是：
sed "s/REGEXP/REPLACEMENT"
And hence missing a "/" at the end of the expression. Just add it to have:
因此在表达式末尾缺少“/”。只需添加它即可：
sed "s/-.*;/;/"
# ^
回答by damienfrancois
You are missing a slash at the end of the sedcommand:
您在sed命令末尾缺少斜线：
Should be "s/-.*;/;/"
应该 "s/-.*;/;/"
回答by Jotne
-.*here the *greedy, so this would fail if there are more than one ;
-.*这里是*贪婪的，所以如果有多个，这将失败;
echo "12523-cvjbsvndv-dvd-dvdv;dsfdsfpage;test" | sed -e "s/-.*;/;/"
12523;test
Change to -[^;]*
改成 -[^;]*
echo "12523-cvjbsvndv-dvd-dvdv;dsfdsfpage;test" | sed -e "s/-[^;]*;/;/"
12523;dsfdsfpage;test
