mysql - 如何将当前的 mysql 变量干净地输出到 my.cnf？

Question

我正在尝试运行以下命令来生成 my.cnf:

{
  # Print version, user, host and time
  echo -e "# MYSQL VARIABLES {{{1\n##\n# MYSQL `
      mysql -V | sed 's,^.*\(V.*\)\, for.*,\1,'
      ` - By: `logname`@`hostname -f` on `date +%c`\n##"
  for l in {a..z}; do
      # Get mysql global variables starting with $l
      echo '#'; mysql -NBe "SHOW GLOBAL VARIABLES LIKE '${l}%'" |
      # Transorm it
      sed 's,\t,^= ,' |
      column -ts^ |
      tr "\n" '@' |
      eval $(echo "sed '" "s,@\("{a..u}{a..z}"\),\n\n\1,;" "'") |
      eval $(echo "sed '" "s,@\(innodb_"{a..z}{a..z}"\),\n\n\1,;" "'") |
      tr '@' "\n" |
      sed 's,^,# ,g'
  done
  echo -e "#\n##\n# MYSQL VARIABLES }}}1";
} | tee ~/mysql-variables.log

(命令的原始来源:http://www.askapache.com/mysql/view-mysql-variables-my-cnf.html)

但是 sed 不断抛出 未终止的替代模式 错误:

sed: 3: " s,@\(aa\),\n\n\1,; s,@ ...": unterminated substitute pattern
sed: 2: " s,@\(innodb_aa\),\n\n\ ...": unterminated substitute pattern
#
sed: 3: " s,@\(aa\),\n\n\1,; s,@ ...": unterminated substitute pattern
sed: 2: " s,@\(innodb_aa\),\n\n\ ...": unterminated substitute pattern
#

如何干净地输出我当前的 mysql 变量？

Answer 1

正如在这个other post中讨论的那样，FreeBSD sed 似乎对脚本中各行的大小有限制，{a..u}{a..z} bash 扩展。

这是一个 sed 脚本，它应该具有与合并的两个 eval … 行完全相同的效果 (及其周围的 tr 命令):

sed 'N;P;s/^innodb_//;tl;/.*\ninnodb_/{x;p;x;D};:l /^\(.\)\(.\).*\n\(innodb_\)\?\1\2/D;{x;p;x;D}' | …

---

编辑:
mklement0 更正版本，使用 FreeBSD sed 语法(见下面的评论):

sed 'N;P;s/^innodb_//;tl'$'\n''/.*\ninnodb_/{x;p;x;D;};:l'$'\n''/^\(.\)\(.\).*\n\(in‌​nodb_\)\?\1\2/D;{x;p;x;D;}' | …
# Changes:              ^^^^^^^                     ^    ^^^^^^^                                            ^

---

(应该替换所有这些:)

  tr "\n" '@' |
  eval $(echo "sed '" "s,@\("{a..u}{a..z}"\),\n\n\1,;" "'") |
  eval $(echo "sed '" "s,@\(innodb_"{a..z}{a..z}"\),\n\n\1,;" "'") |
  tr '@' "\n" |

现在详细说明:我们想要用这两行获得的是一种节，第一行包含以aa，第二行以ab等开头

OP 脚本中的

Bash 扩展 应该会产生类似的结果

sed ' s,@\(aa\),\n\n\1,;
      s,@\(ab\),\n\n\1,;
      ...
      s,@\(uz\),\n\n\1,;' | # without line breaks

然后将被评估。为了获得节，我们必须执行每个子命令恰好一次(并且按照这个顺序，因为输入是按字母顺序排序的)，所以使用

sed 's/@\([a-u][a-z]\)/\n\n\1/' | ...

相反不是解决方案。

回到我的脚本:

sed 'N;                              # Append next input line to pattern space
                                     # Pattern space is "input1 \n input2"

     P;                              # Print first line of pattern space

     s/^innodb_//;                   # Delete (already printed) suffix if present

         tl;                         # If substitution succeeded goto label l

         /.*\ninnodb_/{x;p;x;D};     # Else: if we\'re on first ^innodb_ line,
                                     # then print a new line before (see below)

  :l /^\(.\)\(.\).*\n\(innodb_\)\?\1\2/D;   # If the two lines in pattern space
                                     # begins with the same two chars, remove 1st
                                     # line and go on with next input line

     x;                              # We have to print a blank line: swap space
                                     # and hold patterns

     p;                              # Print space pattern (now empty), i.e.
                                     # print a new line

     x;                              # Swap again patterns

     D                               # Delete "input1 \n" from space pattern and
                                     # loop 
  ' | ...

除非我弄错了，否则它应该不包含 GNU 扩展。