請問下 比如 char * XX = "小明衣服1|小明衣服2|小明來了|歡迎|"
怎么去分割
我的需求 是要獲取 去掉|之后 的長度 比如現在看到的 有4
然后分別取獲取 1號內容 2 號內容 3 號內容 4號內容
uj5u.com熱心網友回復：
用二維陣列讀取字串 將其保存到字符陣列 遇到 | 就停止。
接下來讀取到的保存到二維陣列的下個成員
回圈即可
uj5u.com熱心網友回復：
求教學！ 新手不會！uj5u.com熱心網友回復：
僅供參考：#include <stdio.h>
#include <string.h>
char string[80];
char seps1[3];
char seps2[3];
char *token;
char *zzstrtok (
char *string,
const char *control1,//連續出現時視為中間夾空token
const char *control2 //連續出現時視為中間無空token
)
{
unsigned char *str;
const unsigned char *ctrl1 = (const unsigned char *)control1;
const unsigned char *ctrl2 = (const unsigned char *)control2;
unsigned char map1[32],map2[32];
static char *nextoken;
static char flag=0;
unsigned char c;
int L;
memset(map1,0,32);
memset(map2,0,32);
do {
map1[*ctrl1 >> 3] |= (1 << (*ctrl1 & 7));
} while (*ctrl1++);
do {
map2[*ctrl2 >> 3] |= (1 << (*ctrl2 & 7));
} while (*ctrl2++);
if (string) {
if (control2[0]) {
L=strlen(string);
while (1) {
c=string[L-1];
if (map2[c >> 3] & (1 << (c & 7))) {
L--;
string[L]=0;
} else break;
}
}
if (control1[0]) {
L=strlen(string);
c=string[L-1];
if (map1[c >> 3] & (1 << (c & 7))) {
string[L]=control1[0];
string[L+1]=0;
}
}
str=(unsigned char *)string;
}
else str=(unsigned char *)nextoken;
string=(char *)str;
while (1) {
if (0==flag) {
if (!*str) break;
if (map1[*str >> 3] & (1 << (*str & 7))) {
*str=0;
str++;
break;
} else if (map2[*str >> 3] & (1 << (*str & 7))) {
string++;
str++;
} else {
flag=1;
str++;
}
} else if (1==flag) {
if (!*str) break;
if (map1[*str >> 3] & (1 << (*str & 7))) {
*str=0;
str++;
flag=0;
break;
} else if (map2[*str >> 3] & (1 << (*str & 7))) {
*str=0;
str++;
flag=2;
break;
} else str++;
} else {//2==flag
if (!*str) return NULL;
if (map1[*str >> 3] & (1 << (*str & 7))) {
str++;
string=(char *)str;
flag=0;
} else if (map2[*str >> 3] & (1 << (*str & 7))) {
str++;
string=(char *)str;
} else {
string=(char *)str;
str++;
flag=1;
}
}
}
nextoken=(char *)str;
if (string==(char *)str) return NULL;
else return string;
}
void main()
{
strcpy(string,"A \tstring\t\tof ,,tokens\n\nand some more tokens, ");
strcpy(seps1,",\n");strcpy(seps2," \t");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"1234| LIYI|China | 010 |201110260000|OK");
strcpy(seps1,"|");strcpy(seps2," ");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"1234|LIYI||010|201110260000|OK");
strcpy(seps1,"");strcpy(seps2,"|");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"1234|LIYI||010|201110260000|OK");
strcpy(seps1,"|");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"a");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"a,b");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"a,,b");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",a");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"a,");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",a,,b");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",,a,,b,,");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",,");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",,,");
strcpy(seps1,",");strcpy(seps2," ");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
}
//
//[A string of ,,tokens
//
//and some more tokens, ]
//Tokens:
// <A> <string> <of> <> <tokens> <> <and> <some> <more> <tokens> <>
//[1234| LIYI|China | 010 |201110260000|OK]
//Tokens:
// <1234> <LIYI> <China> <010> <201110260000> <OK>
//[1234|LIYI||010|201110260000|OK]
//Tokens:
// <1234> <LIYI> <010> <201110260000> <OK>
//[1234|LIYI||010|201110260000|OK]
//Tokens:
// <1234> <LIYI> <> <010> <201110260000> <OK>
//[a]
//Tokens:
// <a>
//[a,b]
//Tokens:
// <a> <b>
//[a,,b]
//Tokens:
// <a> <> <b>
//[,a]
//Tokens:
// <> <a>
//[a,]
//Tokens:
// <a> <>
//[,a,,b]
//Tokens:
// <> <a> <> <b>
//[,,a,,b,,]
//Tokens:
// <> <> <a> <> <b> <> <>
//[,]
//Tokens:
// <> <>
//[,,]
//Tokens:
// <> <> <>
//[,,,]
//Tokens:
// <> <> <> <>
uj5u.com熱心網友回復：
直接用split函式轉載請註明出處，本文鏈接：https://www.uj5u.com/houduan/241621.html
標籤：C++ 語言
上一篇：【C語言】內部編譯器錯誤