我正在尝试完成主题任务,但我的代码没有分裂。这是主要功能:
#define SQL_TEXT Latin_Text
#include <iostream>
#define SQL_TEXT Latin_Text
#include <sqltypes_td.h>
#include "Split.h"
#include <string>
#include <stdio.h>
#include <vector>
#include <cstring>
using namespace std;
int main ()
{
VARCHAR_LATIN *result = new VARCHAR_LATIN[512];
wchar_t *s1 = (wchar_t *)"Myýnameýisýzeeshan";
**splitstringwc s(s1);
vector<wstring> flds = s.splitwc((wchar_t)'ý');**
wstring rs = flds[1];
wcout<<rs<<endl;
for (int k = 0; k < flds.size(); k++)
cout << k << " => " << flds[k].data() << endl;
cout<<result;
return 0;
}
splitstringwc类的代码如下:
public:
splitstringwc(wchar_t *s) : wstring(s) { };
vector<wstring>& splitwc(wchar_t delim, int rep=0);
};
vector<wstring>& splitstringwc::splitwc(wchar_t delim, int rep) {
if (!flds1.empty()) flds1.clear(); // empty vector if necessary
wstring ws = data();
wcout<<ws<<endl;
//wcout<<delim<<endl;
//wstring ws;
//int j = StringToWString(ws, work);
wstring buf = (wchar_t *)"";
int i = 0;
while (i < ws.size()) {
if (ws.at(i) != delim)
buf += ws.at(i);
else if (rep == 1) {
flds1.push_back(buf);
buf = (wchar_t *)"";
} else if (buf.size() > 0) {
flds1.push_back(buf);
buf = (wchar_t *)"";
}
i++;
}
if (!buf.empty())
flds1.push_back(buf);
return flds1;
}
代码没有拆分输入字符串,当我尝试调试时,我得到分段错误:wstring ws = data();
请帮助...............
答案 0 :(得分:1)
使用strtok而不是我自己的split函数,是根据unicode分隔符拆分字符串。
代码如下:
str = "Myýnameýisýzeeshan";
char *pch;
pch = strtok(str, "ý");
while (pch != NULL)
{
printf("%s\n", pch);
pch = strtok(NULL, "ý");
}
请注意,str由UNICODE分隔符分隔的ANSI字符串组成。
答案 1 :(得分:0)
处理宽字符字符串时,不能使用普通的字符串和字符文字。他们也必须是广泛的角色,比如
const wchar_t *s1 = L"Myýnameýisýzeeshan";
注意文字前面的L
,这使得字符串成为一个宽字符串。
同样用于字符文字:
s.splitwc(L'ý')