Question

我想检查用户的输入命令如：

add person relation person (person can be for example John, or "John Smith")

示例：add John Smith brother Jack Smith ...

我使用分隔符（空格）将字符串拆分为多个。字符串，但我必须将person的名字留作一个字符串/参数（在人的名字和姓氏之间可以是1个或更多的空格，在每种情况下我都必须将其解释为一个参数）。

// input from cmd stored in "input" variable with fgets() in main function...

char inputTerminalCommands(char *input) // takes char input from main, splits strings and then should compare type of command from user
{

int i = 0;
char *str = input;      
char *split = strtok(str, " ");  // split string into words after "space"

char *array[6];

while(split!= 0)
{
    array[i++] = split;
    split = strtok(NULL, " ");
}

return 0;

使用我的代码我按空格分割字符串，但如何忽略名称之间的空格？或者说，在一个字符串中的“add”和“relation”之间存储字符串，在一个字符串中也存储“relation”...

Answer 1

使用strchr，您可以遍历字符串，解析命令名称性别关系性。如果单词之间有多个空格，则会失败，但可以添加额外的逻辑以容纳多个空格。如果没有观察到格式command name relation name，它也会失败但是可以再次添加额外的逻辑来处理它需要添加检查以处理malloc的失败。

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

struct parse {
    char *command;
    char *name;
    char gender;
    char *relation;
    char *identity;
    char sex;
};


int main(void)
{
    //char input[] = "add John Smith III [m] brother David William Smith [m]";
    char input[] = "add John Smith III brother David William Smith";
    char *start = NULL;
    char *stop = NULL;
    int span = 0;
    struct parse item;
    //parse command - one word
    start = input;
    if ( ( stop = strchr ( start, ' '))) {
        span = stop - start;
        item.command = malloc ( span + 1);
        memmove ( item.command, start, span);
        item.command[span] = '\0';
    }
    //parse name - could be many words, each begins with upper case
    start = stop + 1;
    stop = start;;
    while ( ( stop = strchr ( stop, ' '))) {
        if ( !isupper( *(stop + 1))) {
            span = stop - start;
            item.name = malloc ( span + 1);
            memmove ( item.name, start, span);
            item.name[span] = '\0';
            break;
        }
        stop = stop + 1;
    }
    //optional - parse gender - one character in []
    start = stop;
    item.gender = '*';
    if ( *(start + 1) == '[') {
        item.gender = *(start + 2);
        stop = strchr ( start + 1, ' ');
    }
    //parse relation - one word
    start = stop + 1;
    if ( ( stop = strchr ( start, ' '))) {
        span = stop - start;
        item.relation = malloc ( span + 1);
        memmove ( item.relation, start, span);
        item.relation[span] = '\0';
    }
    //parse name - could be many words, each begins with upper case
    start = stop + 1;
    stop = start;;
    while ( ( stop = strchr ( stop, ' '))) {
        if ( !isupper( *(stop + 1))) {
            span = stop - start;
            item.identity = malloc ( span + 1);
            memmove ( item.identity, start, span);
            item.identity[span] = '\0';
            break;
        }
        stop = stop + 1;
    }
    if ( !stop) { // if nothing follows identity...strchr failed on last name
        span = strlen ( start);
        item.identity = malloc ( span + 1);
        memmove ( item.identity, start, span);
        item.identity[span] = '\0';
        stop = start + span;
    }
    //optional - parse sex - one character in []
    start = stop;
    item.sex = '*';
    if ( *(start + 1) == '[') {
        item.sex = *(start + 2);
    }

    printf ( "%s\n", item.command);
    printf ( "%s\n", item.name);
    printf ( "%c\n", item.gender);
    printf ( "%s\n", item.relation);
    printf ( "%s\n", item.identity);
    printf ( "%c\n", item.sex);

    free ( item.command);
    free ( item.name);
    free ( item.relation);
    free ( item.identity);

    return 0;
}

“C”按空格分隔字符串和分隔符，但在某些单词之间转义空格（最后一个名字）

1 个答案: