如何按文件名查找文件

Question

How can I find the a file by its part of name in c++/linux? The file full name maybe like:

foo.db.1

or

foo.db.2

How can I find the file by part of its name "foo.db", Thanks

Answer 1

The Posix standard function for your use case is glob() . Earlier somebody posted an easy to use C++ wrapper around it here on Stackoverflow .

Answer 2

For example with:

find /your/path -name "foo.db.*"

It will match all files in /your/path structure whose name is like foo.db.<something> .

If the text after db. is always a number, you can do:

find /your/path -name "foo.db.[0-9]*"

You can also use ls :

ls -l /your/path/ls.db.[0-9]

to get files whose name is like foo.db.<number> or even

ls -l /your/path/ls.db.[0-9]*

to get files whose name is like foo.db.<different numbers>

Answer 3

Do like this:

find /filepath -name "foo.db*"

In C/C++, you can do like this:

void findFile(const char *path)
{
    DIR *dir;
    struct dirent *entry;
    char fileName[] = "foo.db";

    if (!(dir = opendir(path)))
        return;
    if (!(entry = readdir(dir)))
        return;

    do {
        if (entry->d_type == DT_DIR) {
            char path[1024];
            int len = snprintf(path, sizeof(path)-1, "%s/%s", path, entry->d_name);
            path[len] = 0;
            if (strcmp(entry->d_name, ".") == 0 || strcmp(entry->d_name, "..") == 0)
                continue;


           if(strstr(fileName, entry->d_name) != NULL) {
                printf("%s\n",  "", entry->d_name);
            }
            findFile(path);
        }
        else
            printf(" %s\n", "", entry->d_name);
    } while (entry = readdir(dir));
    closedir(dir);
}