php - 将函数返回值设置为公共(public)/静态变量

Question

学生表包含:ID名姓氏

函数get_db_fields获取学生表字段。

class Student{

public function get_db_fields(){
        $sql = "SELECT * FROM students";
        $result = mysql_query($sql);
        if(!$result){
         die("Failed :" . mysql_error());
        }
        $db_fields = array();
        while($field = mysql_fetch_field($result)){
           $db_fields[] = "'".$field->name."'";
        }
        return join(',',$db_fields);
    }

    public $db_fields = array($this->get_db_fields()); <--- This is not working

}

return join(',',$db_fields) = 'id','firstname','lastname'

我的问题是如何将返回值设置为公共(public) $db_fields。所以 public $db_fields 将是“public $db_fields = array('id','firstname','lastname')”

到目前为止我已经尝试过

1.)

public $db_fields = array($this->get_db_fields()); <--- This is not working

错误:解析错误:语法错误，第 33 行 C:\Program Files\xampp\htdocs\ieti\includes\student.php 中出现意外的 T_VARIABLE

2.) $database_fields = $this->get_db_fields();

错误:解析错误:语法错误，第 33 行 C:\Program Files\xampp\htdocs\ieti\includes\student.php 中出现意外的 T_VARIABLE

Answer 1

创建一个构造函数，并将 get_db_fields 的返回值赋给成员变量 db_fields。成员变量只能使用编译时计算的值进行初始化。

class Student{

  public $db_fields;

  public function __construct() {
    $this->db_fields = $this->get_db_fields();
  }

// Rest of your code