字符串制表符分解求助
读取文本里面的内容以后,原本打算使用 strtok 函数来分解字符串结果才发现 举例(假设用 ","来代替制表符tab) : CString st= "a1,a2,,a4"; 这种情况的时候 strtok函数会忽略掉 中间没有内容的地方 直接就被忽略了变成: a1 a2 a4 而我要的是把中间的没有内容的地方也要保留 变成 a1 a2 a4 或者空缺的地方用其他表示也行 a1 a2 NULL a4 请高手帮忙 本人是新手 麻烦高手用代码说明
[解决办法]
http://topic.csdn.net/u/20081022/20/2766c46d-6ed3-4567-b0d9-bb9c4fc79236.html
要是不行就自己写分割函数吧，也不难写
[解决办法]
#include <stdio.h>#include <string.h>char string[80];char seps1[3];char seps2[3];char *token;char *zzstrtok ( char *string, const char *control1,//连续出现时视为中间夹空token const char *control2 //连续出现时视为中间无空token ){ unsigned char *str; const unsigned char *ctrl1 = control1; const unsigned char *ctrl2 = 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=string; } else str=nextoken; string=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=str; flag=0; } else if (map2[*str >> 3] & (1 << (*str & 7))) { str++; string=str; } else { string=str; str++; flag=1; } } } nextoken=str; if (string==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:// <>, <>, <>, <>,
[解决办法]
看看这个怎么样
char *strtok_he(const char *src, char *demial,unsigned int n){ char map[32]; int count; int i = 0; char* str = strdup(src); char* ctrl = demial; char *str_org; if (src == NULL || demial == NULL) { return NULL; } for (count =0; count <32; count++) { map[count] = 0; } while (*ctrl) { map[*ctrl >> 3] |= (1 << (*ctrl & 7)); ++ctrl; } str_org = str; for (;*str; str++) { if ( map[*str >> 3] & (1 << (*str & 7))) { if (i == n) { *str++ = '\0'; break; } ++i; str_org = str+1; } } if (i<n) { return NULL; } return str_org;}int main(){ char buf[100] = "ca,bdfbwa,fef,,b,c,dfs"; char *p = NULL; int i = 0; p = strtok_he(buf,",",0); while (p != NULL) { cout << p << endl; i++; p = strtok_he(buf,",",i); } }
[解决办法]
CString str = _T("a1,a2,,a4"); LPCTSTR lpszToken = _T(","); CString strText(str); int nFlag = -1; CString tmp(_T("")); while(TRUE) { nFlag = strText.Find(lpszToken); if(-1 == nFlag) { if(!strText.IsEmpty()) tmp = strText; break; } tmp = strText.Left(nFlag); AfxMessageBox(tmp); strText = strText.Right(strText.GetLength() - nFlag - 1); } AfxMessageBox(tmp);
[解决办法]
System::String st = "a1,a2,,a4";
System::StringArray ss = st.Split(",");
ss就是分割后的结果