C++ 删除字符后的所有内容（以及字符）
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Removing everything after character (and also character)
提问by James Harzs
I have a string like this:
我有一个这样的字符串：
std::string string1 = "xjdfhfakdjs%54k34k.-jk34";
I need to get only ""xjdfhfakdjs", but the string is dynamic, not hardcoded so I don't know what is it, the length etc. so I wanted to remove everything after %, and also the % char.
我只需要获得“xjdfhfakdjs”，但字符串是动态的，不是硬编码的，所以我不知道它是什么，长度等，所以我想删除 % 之后的所有内容，以及 % 字符。
How could I do this?
我怎么能这样做？
采纳答案by Shawn Johnson
I believe that will work.
我相信这会奏效。
std::string mystr = string1.substr(0, string1.find("%", 0));
回答by Edward Loper
std::string the_prefix_you_want = string1.substr(0, string1.find("%"));
See: http://www.cplusplus.com/reference/string/string/find/and http://www.cplusplus.com/reference/string/string/substr/for more details
有关更多详细信息，请参阅：http: //www.cplusplus.com/reference/string/string/find/和http://www.cplusplus.com/reference/string/string/substr/
回答by Laurie Stearn
Did a C-like thing which does work in its current form, but certainly more clunky than the methods shown above:
做了一个类似 C 的东西，它以当前的形式工作，但肯定比上面显示的方法更笨拙：
#include <windows.h>
void main ()
{
int i;
wchar_t searchString[100];
wchar_t * stringReturn;
memset(searchString, L'##代码##', sizeof(stringReturn));
wcscpy_s(searchString, 100, L"String\ to search");
stringReturn = wcschr (searchString, '\');
if (stringReturn)
{
for (i = 0; i < (int)(stringReturn - searchString); i++) stringReturn[i] = searchString[i];
stringReturn[i] = L'##代码##';
wcscpy_s(searchString, 100, stringReturn);
}
}
Can be easily modified to work for dynamic strings.
可以轻松修改以适用于动态字符串。
