如何在backticks中传递bash变量[duplicate]

这与你的倒数无关,完全是由于你的单引号。见 this question 了解如何在单引号字符串中使用变量。下面是根据这个问题修复的示例:

for d in $(cat file_with_databasesnames.txt)
do
  echo $d;
  LC_ALL=C sed -n '/^-- Current Database: `'"$d"'`/,/^-- Current Database: `/p' mysqldump.sql > $d.sql
done

下面是一个示例,说明它的工作原理:

$ cat file_with_databasesnames.txt
mytarget

$ cat mysqldump.sql
-- Current Database: `users`
Some dump here
-- Current Database: `mytarget`
This is what you want
-- Current Database: `etc`
And so on

$ for d in $(cat file_with_databasesnames.txt)
do
  echo $d;
  LC_ALL=C sed -n '/^-- Current Database: `'"$d"'`/,/^-- Current Database: `/p' mysqldump.sql > $d.sql
done
mytarget

$ cat mytarget.sql
-- Current Database: `mytarget`
This is what you want
-- Current Database: `etc`

这里有一个版本,对阅读行进行了一般性的改进 while read 循环并以明确的形式显示数据库名称:

while IFS= read -r db
do
  printf 'Processing: %q\n' "$db"
  LC_ALL=C sed -n '/^-- Current Database: `'"$db"'`/,/^-- Current Database: `/p' mysqldump.sql > "$db.sql"
done < file_with_databasesnames.txt