我对此感到非常困惑，并且它弄乱了我的代码： 我正在寻找一个在一行中包含空格的模式，我想在该模式之后打印单词。 虽然我可以在line变量中使用引号转引号（使用\“），但是在我的实际数据中却不能这样做（直接从给出格式的文件中读取该行...）
我看到了我不理解的行为：
lineone='#include "COMPLEX_Other_chain_X.itp"'
linetwo='Other_chain_X 2'
pattern='Other_chain_X '
#pattern='Other_chain_X\ \ \ \ \ \ ' #alternative pattern, same issues
问题1：尽管该行中不存在该模式，但sed还是打印了sth（代表$ lineone）
#this prints the entire value of $lineone, but should empty...
echo "$lineone" |sed -e 's/.*Other_chain_X //; s/}.*//'
#this works
echo "$linetwo" |sed -e 's/.*Other_chain_X //; s/}.*//'
#this give the output I want - because there are no spaces in the pattern I am looking for ?
echo "$lineone" |sed -e 's/.*Other_chain_X//; s/}.*//'
echo "$linetwo" |sed -e 's/.*Other_chain_X//; s/}.*//'
问题2：同样，即使模式不匹配（与问题1相同，但不使用变量lineone和linetwo），仍会打印sth。
#why is the whole line printed here?
echo '#include "COMPLEX_Other_chain_X.itp"' |sed -e 's/.*Other_chain_X //; s/}.*//'
#my guess is that the syntax of the pattern with the spaces is wrong?
echo "Other_chain_X 2" |sed -e 's/.*Other_chain_X //; s/}.*//' #prints what I want
echo "#include "COMPLEX_Other_chain_X.itp"" |sed -e "s/.*$pattern//; s/}.*//" #why is the whole line printed here?
#my guess is that the syntax of the pattern with the spaces is not properly put into the variable?
echo "Other_chain_X 2" |sed -e "s/.*$pattern//; s/}.*//" #prints what I want
#why is the whole line printed here?
echo "$lineone" |sed -e "s/.*$pattern//; s/}.*//"
echo "$linetwo" |sed -e "s/.*$pattern//; s/}.*//" #works fine
我意识到，当人们不得不逃脱空格或引号时，这与其他问题类似-尽管那里的解释都没有帮助我。问题似乎在于如何将字符串保存在变量中。 有人对此行为有解释吗？
这是行得通的，但我不明白为什么...
echo "$linetwo" |sed -e "s/.*$pattern//; s/}.*//"
问题1和问题2：无论是否使用模式MISMATCH（lineone），sed都可以打印，无论使用变量还是字符串