编写最简单的SQL解析程序（原理演示）

如何写一个SQL解析器呢？这里先抛出第一步：单词切分。下面举个简单例子：

select.l 文件：

%{
/* nothing */
  enum{
    BEGIN_OF_INPUT = 0,
    SELECT,
    COLUMN,
    FROM,
    TABLE,
    SEP,
    END_OF_INPUT
  };
  int state = BEGIN_OF_INPUT;
%}

%%

\n {
  if (SEP == state)
  {
    printf("=========   Cong!!! found a valid sql   ==========\n");
  }
  else if (BEGIN_OF_INPUT < state)
  {
    printf("=========   :-(  not a valid sql   ==========\n");
  }
  state = BEGIN_OF_INPUT;
  printf("please input a sql and press Enter\n");
}
SELECT {
  if (state == (SELECT - 1))
    state++;
  printf("select state=%d\n", state);
}

FROM {
    if (state == (FROM - 1))
          state++;
  printf("from state=%d\n", state);
}


[a-zA-Z_]+ {
  if (state == (COLUMN - 1) || state == (TABLE - 1))
    state++;
  printf("any word state=%d\n", state);
}

; {
  if (state == (SEP - 1))
  {
    state++;
  }
  printf("seperator. state=%d\n", state);
}

. ; /* ignore others */

%%

main()
{
  yylex();
}

编译：

lex select.l     
gcc lex.yy.c  -ll

运行：

./a.out

效果截图：

注意：

写lex文件规则的时候，严格的规则必须写在松散的规则前面。下面是一个错误的例子：

[a-zA-Z_]+ {
  if (state == (COLUMN - 1) || state == (TABLE - 1))
    state++;
  printf("any word state=%d\n", state);
}

FROM {
    if (state == (FROM - 1))
          state++;
  printf("from state=%d\n", state);
}

因为FROM这个永远都走不到，全部都被[a-zA-Z_]拦住了。