oracle判断身份证号是否正确（分享一个oracle身份证校验函数）

有个朋友说能不能用函数来实现对身份证的校验，所以这里用Oracle的函数来实现，其他数据库异曲同工..

CREATE OR REPLACE FUNCTION Func_checkidcard (p_idcard IN VARCHAR2) RETURN INT IS v_regstr VARCHAR2 (2000); v_sum NUMBER; v_mod NUMBER; v_checkcode CHAR (11) := '10X98765432'; v_checkbit CHAR (1); v_areacode VARCHAR2 (2000) := '11,12,13,14,15,21,22,23,31,32,33,34,35,36,37,41,42,43,44,45,46,50,51,52,53,54,61,62,63,64,65,71,81,82,91,'; BEGIN CASE LENGTHB (p_idcard) WHEN 15 THEN -- 15位 IF INSTRB (v_areacode, SUBSTR (p_idcard, 1, 2) || ',') = 0 THEN RETURN 0; END IF; IF MOD (TO_NUMBER (SUBSTRB (p_idcard, 7, 2)) 1900, 400) = 0 OR ( MOD (TO_NUMBER (SUBSTRB (p_idcard, 7, 2)) 1900, 100) <> 0 AND MOD (TO_NUMBER (SUBSTRB (p_idcard, 7, 2)) 1900, 4) = 0 ) THEN -- 闰年 v_regstr := '^[1-9][0-9]{5}[0-9]{2}((01|03|05|07|08|10|12)(0[1-9]|[1-2][0-9]|3[0-1])|(04|06|09|11)(0[1-9]|[1-2][0-9]|30)|02(0[1-9]|[1-2][0-9]))[0-9]{3}$'; ELSE v_regstr := '^[1-9][0-9]{5}[0-9]{2}((01|03|05|07|08|10|12)(0[1-9]|[1-2][0-9]|3[0-1])|(04|06|09|11)(0[1-9]|[1-2][0-9]|30)|02(0[1-9]|1[0-9]|2[0-8]))[0-9]{3}$'; END IF; IF REGEXP_LIKE (p_idcard, v_regstr) THEN RETURN 1; ELSE RETURN 0; END IF; WHEN 18 THEN -- 18位 IF INSTRB (v_areacode, SUBSTRB (p_idcard, 1, 2) || ',') = 0 THEN RETURN 0; END IF; IF MOD (TO_NUMBER (SUBSTRB (p_idcard, 7, 4)), 400) = 0 OR ( MOD (TO_NUMBER (SUBSTRB (p_idcard, 7, 4)), 100) <> 0 AND MOD (TO_NUMBER (SUBSTRB (p_idcard, 7, 4)), 4) = 0 ) THEN -- 闰年 v_regstr := '^[1-9][0-9]{5}(19|20)[0-9]{2}((01|03|05|07|08|10|12)(0[1-9]|[1-2][0-9]|3[0-1])|(04|06|09|11)(0[1-9]|[1-2][0-9]|30)|02(0[1-9]|[1-2][0-9]))[0-9]{3}[0-9Xx]$'; ELSE v_regstr := '^[1-9][0-9]{5}(19|20)[0-9]{2}((01|03|05|07|08|10|12)(0[1-9]|[1-2][0-9]|3[0-1])|(04|06|09|11)(0[1-9]|[1-2][0-9]|30)|02(0[1-9]|1[0-9]|2[0-8]))[0-9]{3}[0-9Xx]$'; END IF; IF REGEXP_LIKE (p_idcard, v_regstr) THEN v_sum := ( TO_NUMBER (SUBSTRB (p_idcard, 1, 1)) TO_NUMBER (SUBSTRB (p_idcard, 11, 1)) ) * 7 ( TO_NUMBER (SUBSTRB (p_idcard, 2, 1)) TO_NUMBER (SUBSTRB (p_idcard, 12, 1)) ) * 9 ( TO_NUMBER (SUBSTRB (p_idcard, 3, 1)) TO_NUMBER (SUBSTRB (p_idcard, 13, 1)) ) * 10 ( TO_NUMBER (SUBSTRB (p_idcard, 4, 1)) TO_NUMBER (SUBSTRB (p_idcard, 14, 1)) ) * 5 ( TO_NUMBER (SUBSTRB (p_idcard, 5, 1)) TO_NUMBER (SUBSTRB (p_idcard, 15, 1)) ) * 8 ( TO_NUMBER (SUBSTRB (p_idcard, 6, 1)) TO_NUMBER (SUBSTRB (p_idcard, 16, 1)) ) * 4 ( TO_NUMBER (SUBSTRB (p_idcard, 7, 1)) TO_NUMBER (SUBSTRB (p_idcard, 17, 1)) ) * 2 TO_NUMBER (SUBSTRB (p_idcard, 8, 1)) * 1 TO_NUMBER (SUBSTRB (p_idcard, 9, 1)) * 6 TO_NUMBER (SUBSTRB (p_idcard, 10, 1)) * 3; v_mod := MOD (v_sum, 11); v_checkbit := SUBSTRB (v_checkcode, v_mod 1, 1); IF v_checkbit = upper(substrb(p_idcard,18,1)) THEN RETURN 1; ELSE RETURN 0; END IF; ELSE RETURN 0; END IF; ELSE RETURN 0; -- 身份证号码位数不对 END CASE; EXCEPTION WHEN OTHERS THEN RETURN 0; END fn_checkidcard; / Show Err;

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