Question

我想在类中创建一个函数，创建用户名，函数将检查用户名是否存在，然后它将增加用户名如username_1。并检查此用户名是否存在（如果存在）再次将其增加到username_2，直到创建新用户名。我创建了这个函数，但它什么都没有给我。请帮我解决我的代码中的错误。

class a{

function check_username($username){


  if($usernameexist){

    return true;
  }
  else
  {

   return false;

  }

}

function create_username($username) {

        $__name = __FUNCTION__;

        if ($this->check_username($username)) {                
            $n++;
            $username = $username . "_" . $n;
            //return $__name($username);  this return fatal error.
            return call_user_func('create_username', $username);

        } else {
            return $username;               
        }
    }
}

Answer 1

不需要为此使用递归，一个简单的while(){}循环将执行：

Plain-Jane Interator方法

// your original function
function create_username($username){
  // check if the username (as-is) already exists
  if ($this->check_username($username)){
    // use $n to keep a counter
    $n = 1;
    // while {username}_{n} exists, keep incrementing the counter
    while ($this->check_username($username.'_'.$n)){
      $n++;

      /* If you don't want this to check to infinity, uncomment
       * the below portion. the 100 is an arbitrary number, but use
       * whatever you want as a limitation (could even make it a
       * parameter in the method). Also, returning FALSE allows you to
       * gracefully catch when max attempts are reached.
       *
       * e.g.
       *   if (($new_user = $obj->create_username('BradChristie')) !== FALSE){
       *     // user was successfully created within the max allowed attempts
       *   }
       */
      //if ($n > 100) return FALSE
    }
    // return the result
    return $username.'_'.$n;
  }
  // username was fine, return it back
  return $username;
}

递归方法

// recursive username check
public function create_username($username, $n = 0)
{
  /* Same as above function, this is a check to prevent counting
   * to infinity. uncomment to apply it
   */
  //if ($n > 100) return FALSE;

  // establish the username we're testing. if $n is 0,
  // it's the original call to the function (don't add _0)
  // if it's >0, it's part of the search so include it
  $_username = $username . ($n > 0 ? '_'.$n : '');

  // check if the username exists.
  if ($this->check_username($_username))
  {
    // it exists, so make a call to this same function passing
    // the original username and the value of n + 1 (move to next
    // possibility)
    return $this->create_username($username, $n+1);
  }

  // the name, as-is, was fine. return it
  return $_username;
}

Example

Answer 2

在这种情况下不需要递归...一个简单的循环可以完美地完成：

function create_username($username) {

    $original_username = $username;
    $i=1;
    while(! $this->check_username($username) ) {
        $username = $original_username . '_' .$i++;
    }

    return $username;
}

Answer 3

你的代码在几个方面都是错误的，正如其他地方所指出的那样，你想要的函数可以更好地迭代编写。

您的代码存在以下问题：

check_username成功后，您正在进行递归检查。因此，如果您找不到原始$username，则永远不会修改它，所以永远不要检查修改后的值。
您正在修改传递给create_username的名称，方法是附加_n（适用于n）。由于您在递归调用中传递了修改后的名称，因此您最终会在名称上添加多个_n部分。
由于你没有限制你的递归调用，即使这是正确编写的，你最终也会嵌套太深。

php类中的递归函数

3 个答案:

Plain-Jane Interator方法

递归方法