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/”找到子字符串?我想检查一下“abc”文件夹是否存在于此路径中。
Path is character pointer char *ptr = "/user/desktop/abc/post/";
Path是字符指针char *ptr = "/user/desktop/abc/post/";
6 个解决方案
#1
17
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).
注意不要偶然地发现一个太短的子字符串(这是开始和结束斜线的目的)。
#2
35
Use std::string
and find
.
使用std::string和发现。
std::string str = "/user/desktop/abc/post/";
bool exists = str.find("/abc/") != std::string::npos;
#3
8
Example using std::string
find method:
示例使用std::string find方法:
#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.
#4
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 /abc
followed by a string-terminator '\0'
. A naive way to do this would be to check if either /abc/
or /abc\0
are substrings:
正如user1511510所指出的,当abc位于文件名末尾时,会出现一个不寻常的情况。我们需要找到/abc/或/abc后面跟着一个字符串终结者'\0'。一种简单的方法是检查/abc/或/abc\0是否为子字符串:
#include <stdio.h>
#include <string.h>
int main() {
const char *str = "/user/desktop/abc";
const int exists = strstr(str, "/abc/") || strstr(str, "/abc\0");
printf("%d\n",exists);
return 0;
}
but exists
will 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 /abc
is a substring, and then see if the character after this substring (indexed using the pointer returned by strstr()
) is either a /
or a '\0'
:
但是,即使abc没有被一个空终止符跟踪,它仍然是1。这是因为字符串文字“/abc\0”等同于“/abc”。更好的方法是测试/abc是否是子字符串,然后查看这个子字符串(使用strstr()返回的指针进行索引)后面的字符是A /还是'\0':
#include <stdio.h>
#include <string.h>
int main() {
const char *str = "/user/desktop/abc", *substr;
const int exists = (substr = strstr(str, "/abc")) && (substr[4] == '\0' || substr[4] == '/');
printf("%d\n",exists);
return 0;
}
This should work in all cases.
这应该适用于所有情况。
#5
0
Use strstr(const char *s , const char *t)
and include<string.h>
使用strstr(const char *s, const char *t)并包含
You can write your own function which behaves same as strstr
and you can modify according to your requirement also
您可以编写自己的函数,其行为与strstr相同,还可以根据您的需求进行修改
char * str_str(const char *s, const char *t)
{
int i, j, k;
for (i = 0; s[i] != '\0'; i++)
{
for (j=i, k=0; t[k]!='\0' && s[j]==t[k]; j++, k++);
if (k > 0 && t[k] == '\0')
return (&s[i]);
}
return NULL;
}
#6
0
If you are utilizing arrays too much then you should include cstring.h
because it has too many functions including finding substrings.
如果您过多地使用数组,那么应该包含cstring。因为它有太多的函数,包括查找子字符串。
#1
17
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).
注意不要偶然地发现一个太短的子字符串(这是开始和结束斜线的目的)。
#2
35
Use std::string
and find
.
使用std::string和发现。
std::string str = "/user/desktop/abc/post/";
bool exists = str.find("/abc/") != std::string::npos;
#3
8
Example using std::string
find method:
示例使用std::string find方法:
#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.
#4
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 /abc
followed by a string-terminator '\0'
. A naive way to do this would be to check if either /abc/
or /abc\0
are substrings:
正如user1511510所指出的,当abc位于文件名末尾时,会出现一个不寻常的情况。我们需要找到/abc/或/abc后面跟着一个字符串终结者'\0'。一种简单的方法是检查/abc/或/abc\0是否为子字符串:
#include <stdio.h>
#include <string.h>
int main() {
const char *str = "/user/desktop/abc";
const int exists = strstr(str, "/abc/") || strstr(str, "/abc\0");
printf("%d\n",exists);
return 0;
}
but exists
will 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 /abc
is a substring, and then see if the character after this substring (indexed using the pointer returned by strstr()
) is either a /
or a '\0'
:
但是,即使abc没有被一个空终止符跟踪,它仍然是1。这是因为字符串文字“/abc\0”等同于“/abc”。更好的方法是测试/abc是否是子字符串,然后查看这个子字符串(使用strstr()返回的指针进行索引)后面的字符是A /还是'\0':
#include <stdio.h>
#include <string.h>
int main() {
const char *str = "/user/desktop/abc", *substr;
const int exists = (substr = strstr(str, "/abc")) && (substr[4] == '\0' || substr[4] == '/');
printf("%d\n",exists);
return 0;
}
This should work in all cases.
这应该适用于所有情况。
#5
0
Use strstr(const char *s , const char *t)
and include<string.h>
使用strstr(const char *s, const char *t)并包含
You can write your own function which behaves same as strstr
and you can modify according to your requirement also
您可以编写自己的函数,其行为与strstr相同,还可以根据您的需求进行修改
char * str_str(const char *s, const char *t)
{
int i, j, k;
for (i = 0; s[i] != '\0'; i++)
{
for (j=i, k=0; t[k]!='\0' && s[j]==t[k]; j++, k++);
if (k > 0 && t[k] == '\0')
return (&s[i]);
}
return NULL;
}
#6
0
If you are utilizing arrays too much then you should include cstring.h
because it has too many functions including finding substrings.
如果您过多地使用数组,那么应该包含cstring。因为它有太多的函数,包括查找子字符串。