从字符串中删除空格和特殊字符

Question

如何从字符串中删除空格和特殊字符？

我在Google上搜索时找不到一个答案。有很多与其他语言有关，但不是C.大多数人提到了使用正则表达式，这不是C标准（？）。

卸下易空间很简单：

char str[50] = "Remove The Spaces!!"; 

然后用if语句的简单循环：

if (str[i] != ' '); 

输出将是：

RemoveTheSpaces!! 

我该怎么加到if语句以便识别特殊字符并将其删除？

我的特殊字符的定义：

Characters not included in this list: 
A-Z a-z 0-9 

Answer 1

这是可能不是实现这一目标的最有效方式，但它可以相当快地完成工作。

注意：此代码确实需要您包括<string.h>和<ctype.h>

char str[50] = "Remove The Spaces!!"; 
char strStripped[50]; 

int i = 0, c = 0; /*I'm assuming you're not using C99+*/ 
for(; i < strlen(str); i++) 
{ 
    if (isalnum(str[i])) 
    { 
     strStripped[c] = str[i]; 
     c++; 
    } 
} 
strStripped[c] = '\0'; 

Answer 2

这是ASCII代码范围内

Char:Dec

0:48, 9:57 
A:65, Z:90 
a:97, z:122 

试试这个：

char str[50] = "Remove The Spaces!!"; 

int i =0; 
for(; i<strlen(str); i++) 
{ 
    if(str[i]>=48 && str[i]<=57 || str[i]>=65 && str[i]<=90 || str[i]>=97 && str[i]<=122) 
    //This is equivalent to 
    //if(str[i]>='0' && str[i]<='9' || str[i]>='A' && str[i]<='Z' || str[i]>='a' && str[i]<='z') 
     printf("alphaNumeric:%c\n", str[i]); 
    else 
    { 
     printf("special:%c\n", str[i]); 
     //remove that 
    } 
} 

Answer 3

使用您的if语句：

if (str[i] != ' '); 

随着一点点的逻辑（字符必须是在范围az或AZ或0-9：

If (!('a' <= str[i] && 'z' >= str[i]) && 
    !('A' <= str[i] && 'Z' >= str[i]) && 
    !('0' <= str[i] && '9' >= str[i])) then ignore character. 

Answer 4

这只是一个愚蠢的建议。

char ordinary[CHAR_MAX] = { 
    ['A']=1,['B']=1,['C']=1,['D']=1,['E']=1,['F']=1,['G']=1,['H']=1,['I']=1, 
    ['J']=1,['K']=1,['L']=1,['M']=1,['N']=1,['O']=1,['P']=1,['Q']=1,['R']=1, 
    ['S']=1,['T']=1,['U']=1,['V']=1,['W']=1,['X']=1,['Y']=1,['Z']=1, 

    ['a']=1,['b']=1,['c']=1,['d']=1,['e']=1,['f']=1,['g']=1,['h']=1,['i']=1, 
    ['j']=1,['k']=1,['l']=1,['m']=1,['n']=1,['o']=1,['p']=1,['q']=1,['r']=1, 
    ['s']=1,['t']=1,['u']=1,['v']=1,['w']=1,['x']=1,['y']=1,['z']=1, 

    ['0']=1,['1']=1,['2']=1,['3']=1,['4']=1,['5']=1,['6']=1,['7']=1,['8']=1, 
    ['9']=1, 
}; 

int is_special (int c) { 
    if (c < 0) return 1; 
    if (c >= CHAR_MAX) return 1; 
    return !ordinary[c]; 
} 

void remove_spaces_and_specials_in_place (char *str) { 
    if (str) { 
     char *p = str; 
     for (; *str; ++str) { 
      if (!is_special(*str)) *p++ = *str; 
     } 
     *p = '\0'; 
    } 
} 

Answer 5

有数百万种不同的方式可以完成。这里只是一个不使用任何额外的存储空间，并执行“就地”去除不需要的字符例如：

#include <stdlib.h> 
#include <stdio.h> 
#include <ctype.h> 

static void my_strip(char *data) 
{ 
    unsigned long i = 0; /* Scanning index */ 
    unsigned long x = 0; /* Write back index */ 
    char c; 

    /* 
    * Store every next character in `c` and make sure it is not '\0' 
    * because '\0' indicates the end of string, and we don't want 
    * to read past the end not to trigger undefined behavior. 
    * Then increment "scanning" index so that next time we read the 
    * next character. 
    */ 
    while ((c = data[i++]) != '\0') { 
     /* Check if character is either alphabetic or numeric. */ 
     if (isalnum(c)) { 
      /* 
      * OK, this is what we need. Write it back. 
      * Note that `x` will always be either the same as `i` 
      * or less. After writing, increment `x` so that next 
      * time we do not overwrite the previous result. 
      */ 
      data[x++] = c; 
     } 
     /* else — this is something we don't need — so we don't increment the 
      `x` while `i` is incremented. */ 
    } 
    /* After all is done, ensure we terminate the string with '\0'. */ 
    data[x] = '\0'; 
} 

int main() 
{ 
    /* This is array we will be operating on. */ 
    char data[512]; 

    /* Ask your customer for a string. */ 
    printf("Please enter a string: "); 

    if (fgets(data, sizeof(data), stdin) == NULL) { 
     /* Something unexpected happened. */ 
     return EXIT_FAILURE; 
    } 

    /* Show the customer what we read (just in case :-)) */ 
    printf("You have entered: %s", data); 

    /* 
    * Call the magic function that removes everything and leaves 
    * only alphabetic and numberic characters. 
    */ 
    my_strip(data); 

    /* 
    * Print the end result. Note that newline (\n) is there 
    * when we read the string 
    */ 
    printf("Stripped string: %s\n", data); 

    /* Our job is done! */ 
    return EXIT_SUCCESS; 
} 

我投入了大量的意见在里面，所以希望该代码不需要解释。希望能帮助到你。祝你好运！

Answer 6

#include <stdio.h> 
#include <string.h> 

main() 
{ 
    int i=0, j=0; 
    char c; 
    char buff[255] = "Remove The Spaces!!"; 

    for(; c=buff[i]=buff[j]; j++){ 
     if(c>='A' && c<='Z' || c>='a' && c<='z' || c>='0' && c<='9'){ 
      i++; 
     } 
    } 

    printf("char buff[255] = \"%s\"\n", buff); 
}