如何匹配 c 中的用戶名和密碼？

Question

我想寫一個登錄程序。 我希望每個用戶名都與其特定的密碼相匹配。 例如，用戶名“Asma”只能使用密碼“Hello123”而不是“Welcome”登錄。

但是不知何故，當我輸入字符串數組中的每個用戶名和密碼時，它會顯示“已成功登錄”。 如何讓每個用戶名都匹配自己的密碼？ 另外，當用戶名和密碼都不正確時，如何再次要求有效輸入？

這是我的代碼：

#include <stdio.h>
#include <string.h>

int main(void)
{
    char username[10];
    char password[10];

    printf("Enter your username: ");
    scanf("%s", username);

    printf("Enter your password: ");
    scanf("%s", password);

    if(strcmp(username,"Asma") == 0 || strcmp(username,"Iffat") == 0 || strcmp(username,"Kamal") == 0 ){
    if(strcmp(password,"Hello123") == 0 || strcmp(password,"Welcome") == 0  || strcmp(password,"fr1324") == 0 ){

        printf("Logged In Successfully");   

        }else{
    printf("Incorrect username or password, please try again! ");
}

}else{

printf("Incorrect username or password, please try again! ");    
}

Answer 1

如果你想匹配用戶名和密碼，你應該使用&&來確保兩個檢查都成功：

#include <stdio.h>
#include <string.h>

int main(void)
{
    char username[10];
    char password[10];

    printf("Enter your username: ");
    scanf("%s", username);

    printf("Enter your password: ");
    scanf("%s", password);

    if (strcmp(username, "Asma") == 0 && strcmp(password, "Hello123") == 0
        || strcmp(username, "Iffat") == 0 && strcmp(password, "Welcome") == 0
        || strcmp(username, "Kamal") == 0 && strcmp(password, "fr1324") == 0) {
            printf("Logged In Successfully");
    } else {
        printf("Incorrect username or password, please try again! ");
    }
}

Answer 2

如何讓每個用戶名都匹配自己的密碼？

您的代碼當前編寫的方式，您所做的就是確保它們匹配任何用戶名和任何密碼 - 您沒有將特定密碼與特定用戶名匹配。

雖然作為第一次嘗試明確檢查每個名稱和密碼是可以的，但它不會擴展——超過 5 個左右的用戶名很快就會變得難以管理。 使用struct類型創建名稱和密碼的查找表可能會更好：

struct usertable {
  char *name;
  char *pass;
};

...

struct usertable users[] = { // array size is taken from the number of initializers
  {"Asma", "Hello123"},
  {"Iffat", "Welcome"},
  {"Kamal", "fr1234"},
  ...                        // any other entries
  {NULL, NULL}               // explicitly mark the last entry in the table
};

然后在您輸入用戶名和密碼后，您可以像這樣循環瀏覽表格：

/**
 * Cycle through all the entries in users until logged_in is true
 * or we hit the end of the table.  
 */
int logged_in = 0;
for ( size_t i = 0; users[i].name != NULL && !logged_in; i++ )
{
    /**
     * Perform a logical AND against the results of strcmp for both
     * the username and password, assign the result (0 or 1) to logged_in.
     */
  logged_in = (strcmp( username, users[i].name ) == 0 &&
               strcmp( password, users[i].pass ) == 0 );
}

if ( !logged_in )
  printf( "Username or password not found, try again\n" );

另外，當用戶名和密碼都不正確時，如何再次要求有效輸入？

將以上所有內容放入一個循環中，該循環僅在您匹配密碼或在輸入時發出EOF信號時才退出（使用 Ctrl-D 或 Ctrl-Z，具體取決於您的系統）：

int logged_in = 0;

/**
 * Loop until the logged_in flag is true or we see a break statement
 */
while ( !logged_in ) 
{
  printf( "Enter username: " );         // scanf is *not* the right tool for
  if ( scanf( "%s", username ) == EOF ) // this, but we'll keep it in place
    break; // exit while loop           // because otherwise this example
                                        // would be three times as big
  printf( "Enter password: " );         // and I don't to bury the main point
  if ( scanf( "%s", password ) == EOF ) // of the exercise.
    break; // exit while loop

  for( size_t i = 0; users[i].name != NULL && !logged_in; i++ )
  {
    logged_in = (strcmp( users[i].name, username ) == 0 && 
                 strcmp( users[i].pass, password ) == 0 );
  }

  if ( !logged_in )
  {
    fprintf( stderr, "Username or password not found, try again.\n" );
    fprintf( stderr, "Type Ctrl-Z to quit.\n" );
  }
}

/**
 * If logged_in is still 0 at this point, that means the user entered
 * Ctrl-Z to quit, and we exit the program.
 */
if ( !logged_in )
{
  fprintf( stderr, "Exiting...\n" );
  exit( EXIT_SUCCESS );              // "Normal" exit
}

printf( "Logged in successfully\n" );