2

This function returns NULL while $alias having value in second recursion. In first call it return the required value but when first if not matched and it recurse first the required value in avilable in $alias variable but it does not return anything.

public function checkAlias($fname='',$lname=''){

        if(!empty($fname)){
        $fname = mysql_real_escape_string($fname);
        }
        if(!empty($lname)){
        $lname = mysql_real_escape_string($lname);
        }

    $alias = strtolower($fname).strtolower($lname);
    $sql = "Select ALIAS from table where ALIAS = '$alias'";
    $query = mysql_query($sql);
    $row = mysql_fetch_row($query);
    $string_length = strlen($alias) - 1;
    $result_string = substr($alias,0,$string_length);

    if(!$row){
            print $alias;   // is printing value 
        return $alias;  // but here it returns null
    }else{
        $this->checkAlias($result_string);
        } 
    }
4

2 回答 2

12

您忘记返回递归调用的结果：

return $this->checkAlias($result_string);
于 2013-09-16T12:22:00.347 回答
3

您之前忘记了 return 关键字$this->checkAlias($result_string);

将您的代码更改为：

public function checkAlias($fname='',$lname=''){

        if(!empty($fname)){
        $fname = mysql_real_escape_string($fname);
        }
        if(!empty($lname)){
        $lname = mysql_real_escape_string($lname);
        }

    $alias = strtolower($fname).strtolower($lname);
    $sql = "Select ALIAS from table where ALIAS = '$alias'";
    $query = mysql_query($sql);
    $row = mysql_fetch_row($query);
    $string_length = strlen($alias) - 1;
    $result_string = substr($alias,0,$string_length);

    if(!$row){
            print $alias;   // is printing value 
        return $alias;  // but here it returns null
    }else{
        return $this->checkAlias($result_string);
        } 
    }

因为第一次代码会到达 else 语句，而第二次运行它会进入 if 语句。if 返回值，但在 else 中你不做任何事情，所以返回它，你得到你的值。

斯特凡

于 2013-09-16T12:22:50.367 回答