C++ 如何从字符串中找到子字符串？
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How to find substring from string?
提问by CrazyCoder
How do I find a substring from the string path "/user/desktop/abc/post/" using C/C++? I want to check if folder "abc" is present or not in that path.
如何使用 C/C++ 从字符串路径“/user/desktop/abc/post/”中找到子字符串？我想检查该路径中是否存在文件夹“abc”。
Path is character pointer char *ptr = "/user/desktop/abc/post/";
路径是字符指针 char *ptr = "/user/desktop/abc/post/";
采纳答案by unwind
In C, use the strstr()standard library function:
在 C 中，使用strstr()标准库函数：
const char *str = "/user/desktop/abc/post/";
const int exists = strstr(str, "/abc/") != NULL;
Take care to not accidentally find a too-short substring (this is what the starting and ending slashes are for).
注意不要意外找到太短的子字符串（这是开始和结束斜线的用途）。
回答by Luchian Grigore
Use std::stringand find.
使用std::string和find。
std::string str = "/user/desktop/abc/post/";
bool exists = str.find("/abc/") != std::string::npos;
回答by DragonB
Example using std::stringfind method:
使用std::stringfind 方法的示例：
#include <iostream>
#include <string>
int main (){
std::string str ("There are two needles in this haystack with needles.");
std::string str2 ("needle");
size_t found = str.find(str2);
if(found!=std::string::npos){
std::cout << "first 'needle' found at: " << found << '\n';
}
return 0;
}
Result:
结果：
first 'needle' found at: 14.
回答by 1''
As user1511510 has identified, there's an unusual case when abc is at the end of the file name. We need to look for either /abc/or /abcfollowed by a string-terminator '\0'. A naive way to do this would be to check if either /abc/or /abc\0are substrings:
正如 user1511510 所确定的那样，当 abc 位于文件名末尾时，这是一种不寻常的情况。我们需要查找/abc/或/abc后跟一个字符串终止符'\0'。一种天真的方法是检查/abc/或/abc\0是否是子字符串：
#include <stdio.h>
#include <string.h>
int main() {
const char *str = "/user/desktop/abc";
const int exists = strstr(str, "/abc/") || strstr(str, "/abc#include <stdio.h>
#include <string.h>
int main() {
const char *str = "/user/desktop/abc", *substr;
const int exists = (substr = strstr(str, "/abc")) && (substr[4] == 'char * str_str(const char *s, const char *t)
{
int i, j, k;
for (i = 0; s[i] != '##代码##'; i++)
{
for (j=i, k=0; t[k]!='##代码##' && s[j]==t[k]; j++, k++);
if (k > 0 && t[k] == '##代码##')
return (&s[i]);
}
return NULL;
}
' || substr[4] == '/');
printf("%d\n",exists);
return 0;
}
");
printf("%d\n",exists);
return 0;
}
but existswill be 1 even if abc is not followed by a null-terminator. This is because the string literal "/abc\0"is equivalent to "/abc". A better approach is to test if /abcis a substring, and then see if the character after this substring (indexed using the pointer returned by strstr()) is either a /or a '\0':
但exists即使 abc 后面没有空终止符，也会为 1。这是因为字符串文字"/abc\0"等效于"/abc". 更好的方法是测试 if/abc是子字符串，然后查看此子字符串后面的字符（使用 返回的指针索引strstr()）是 a/还是 a '\0'：
This should work in all cases.
这应该适用于所有情况。
回答by Omkant
Use strstr(const char *s , const char *t)and include<string.h>
使用strstr(const char *s , const char *t)和include<string.h>
You can write your own function which behaves same as strstrand you can modify according to your requirement also
您可以编写自己的功能，其行为与strstr您的要求相同，也可以根据您的要求进行修改
回答by NatiSpark
If you are utilizing arrays too much then you should include cstring.hbecause it has too many functions including finding substrings.
如果您过多地使用数组，那么您应该包括在内，cstring.h因为它具有太多功能，包括查找子字符串。
