bash 在bash中提取两个引号之间的字符串
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extracting a string between two quotes in bash
提问by Vik
I have example string like
Test "checkin_resumestorevisit - Online_V2.mt" Run
我有像这样的示例字符串
Test "checkin_resumestorevisit - Online_V2.mt" Run
and i want to extract the text between the two quotes in bash. I tried using command like
我想在 bash 中提取两个引号之间的文本。我尝试使用类似的命令
SUBSTRING=${echo $SUBSTRING| cut -d'"' -f 1 }
but it fails with error: bad substitution.
但它失败了error: bad substitution。
回答by Arton Dorneles
In order to extract the substring between quotes you can use one of these alternatives:
为了提取引号之间的子字符串，您可以使用以下替代方法之一：
Alternative 1:
备选方案 1：
SUBSTRING=`echo "$SUBSTRING" | cut -d'"' -f 2`
Alternative 2:
备选方案 2：
SUBSTRING=`echo "$SUBSTRING" | awk -F'"' '{print }'`
Alternative 3:
备选方案 3：
set -f;IFS='"'; SUBSTRING=($SUBSTRING); SUBSTRING=${SUBSTRING[1]};set +f
回答by Charles Duffy
If you want to extract content between the first "and the last ":
如果要提取第一个"和最后一个之间的内容"：
s='Test "checkin_resumestorevisit - Online_V2.mt" Run'
s=${s#*'"'}; s=${s%'"'*}
echo "$s"
回答by bufh
Other weird ideas to not let @arton-dornelesalone :^)
其他不让@arton-dorneles一个人呆着的奇怪想法:^)
But my favourite is @charles-duffyanswer.
但我最喜欢的是@charles-duffy 的回答。
[[ "$SUBSTRING" =~ \"(.*)\" ]] && SUBSTRING="${BASH_REMATCH[1]}"
IFS=\" read -r _ SUBSTRING _ <<<"$SUBSTRING"
# bonus, match either ' or " by pair
SUBSTRING=$(echo "$SUBSTRING" | sed -r "s/.*?([\"'])(.*).*//")
Note that bash create a temporary file for "here string" (<<<) as it does for here documents.
请注意，bash 为“here string” ( <<<)创建一个临时文件，就像它为 here 文档所做的那样。
Edit
编辑
It took me way too long to fully understand Charles Duffy critics; for readers, this could be when:
我花了很长时间才完全理解查尔斯·达菲 (Charles Duffy) 的批评者；对于读者来说，这可能是：
SUBSTRING='*"inside text" outside text'
In this scenario, the following would be unsafe because the star would be expanded, listing files and moving the desired result index:
在这种情况下，以下内容将是不安全的，因为星号将被扩展，列出文件并移动所需的结果索引：
IFS=\" eval '_=($SUBSTRING);SUBSTRING="${_[1]}"'
IFS=\" read -ra _ <<<"$SUBSTRING";SUBSTRING=${_[1]}
![[: bash 脚本中缺少‘]’](/res/img/loading.gif)