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oracle instr函数详解
阅读量:793 次
发布时间:2023-02-24

本文共 1036 字,大约阅读时间需要 3 分钟。

Oracle instr???PLSQL????????????????????????????????instr??????????????????????????????

instr??????????

instr(string1, string2 [, start_position [, nth_appearance ] ])

?????

  • string1???????????
  • string2???????????
  • start_position???????????1?
  • nth_appearance???????n??????????1?
  • ?????

    SELECT instr('syranmo','s') FROM dual; -- ??1SELECT instr('syranmo','ra') FROM dual; -- ??3SELECT instr('syran mo','a',1,2) FROM dual; -- ??0SELECT instr('syranmo','an',-1,1) FROM dual; -- ??4SELECT instr('abc','d') FROM dual; -- ??0

    ??????

    • ????0????????????????????
    • ??????????????????????1??????
    • ??start_position????????????????

    ???????

    SELECT code, name, dept, occupation FROM staff WHERE instr('A10001,A10002', code) > 0;

    ??????????code?????'A10001,A10002'????????

    SELECT code, name, dept, occupation FROM staff WHERE instr(code, '001') > 0;

    ???????code?????'001'????

    ?????

  • instr?????????????????????????
  • ??????????????instr()????????????instrIgnoreCase()?
  • ???????????????????????start_position?nth_appearance??????
  • ??????instr?????????PLSQL???????????

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