关于c语言判断电话号码合法问题

Posted 2023-05-01

我要输入一个电话号码char phone[15];
输入合法数据为数字和-，类似123-123456
请问我怎么判断输入的数据是否合法，
inputTel:
printf("Please enter the engineer's telephone:");
scanf("%s",&Engineer->EngineerPhone);
if(sizeof(&Engineer->EngineerPhone)<=0||sizeof(&Engineer->EngineerPhone)>15)

printf("Error! please re-enter.\n");
goto inputTel;

for(x=0;x<sizeof(&Engineer->EngineerPhone);x++)

if(((&Engineer->EngineerPhone<'0')||(&Engineer->EngineerPhone>'9'))&&((&Engineer->EngineerPhone)!='-'))

printf("Error! please re-enter.\n");
goto rinputel;


你们看看这段代码怎么改呗 谢谢了

提取号码并验证合法性，最好使用正则表达式，你这种判断方式只能判断填入的是否为数字、判断号码长度什么的 ，但是号码本身的数字规则等等都不能进行验证。比如你输入的号码，本身就是不正确的，不会有人用123-123456这种形式。
不过正则表达式有一些复杂，很难马上上手。
可以看看这里：http://www.2cto.com/kf/201110/107088.html 参考技术A /* 检查电话合法性的代码 */
bool check_phone_num_correctness(const char* buffer)

const char* p = buffer;
for(; *p && (isdigit(*p) || *p == '-'); ++p);
return (*p == NULL) ? true : false;


bool get_phone_number(char* buffer, int max_size)

/* 使用一个更大的缓冲区保存输入, 如果字符串太长了, 不可以拷贝到你设定的 buffer[15]中 */
char big_buffer[1000];
printf("Please enter the engineer's telephone:");
scanf("%s", big_buffer);
int phone_size = strlen(big_buffer);

/* 判断是否合法, 1) 长度不超过buffer的长度 2) 字符内容符合电话号码要求(自己可以再扩展) */
bool is_size_correct = (phone_size > 0) && (phone_size < max_size);
if(!is_size_correct || !check_phone_num_correctness(big_buffer))
fprintf(stderr, "Error! please re-enter.\n");
return false;


/* copy back */
strcpy(buffer, big_buffer);
return true;


/* calling method */
char phone[15];
bool phone_correct = get_phone_number(phone, 15); 参考技术B bool bRightPhone = false;
while(!bRightPhone)

printf("Please enter the engineer's telephone:");
scanf("%s",&Engineer->EngineerPhone);
bool bEveryRight = true;
if(sizeof(&Engineer->EngineerPhone) > 0 && sizeof(&Engineer->EngineerPhone) > 15)

for(x=0;x<sizeof(&Engineer->EngineerPhone);x++)

if(((&Engineer->EngineerPhone < '0') && (&Engineer->EngineerPhone > '9')) && ((&Engineer->EngineerPhone) !='-'))

bEveryRight = false;
break;



else

bEveryRight = false;

bRightPhone = bEveryRight ;
if(!bRightPone)

printf("Wrong PhoneNumber,Again!\n");

本回答被提问者采纳 参考技术C 电话格式有几种？如果就只有上面的格式，你只需要检查前三个和后6个是否为数字，第4个是否为'-' 参考技术D 不好意思，弄说错了

oracle判断身份证号码是否合法（包含15位身份证）

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 Func_checkidcard;