"C" 用分隔符 space 拆分字符串，但在某些单词（姓氏、名字）之间转义 space

Question

我想检查来自用户的输入命令例如：

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

示例：add John Smith brother Jack Smith ...

我使用定界符 (space) 将字符串拆分为多个。字符串，但我必须将人名保留为一个 string/parameter（人名和姓氏之间可以是 1 space 或更多，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;

在我的代码中，我按 space 拆分字符串，但是如何忽略名称之间的 space？或者说，将 "add" 和 "relation" 之间的字符串存储在一个字符串中，在 "relation" 之后也存储在一个字符串中...

Answer 1

使用strchr你可以遍历字符串，解析命令名称性别关系性别。如果单词之间有多个 space，这将失败，但可以添加额外的逻辑以容纳多个 space。如果未观察到格式 command name relation name，它也会失败，但可以再次添加额外的逻辑来处理该问题。
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] = '[=10=]';
    }
    //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] = '[=10=]';
            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] = '[=10=]';
    }
    //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] = '[=10=]';
            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] = '[=10=]';
        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" 用分隔符 space 拆分字符串，但在某些单词（姓氏、名字）之间转义 space

"C" split string by space with delimiter but escape space between certain words (last, first name)

c

delimiter