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HomeBackend DevelopmentPHP Tutorial求问以下方法为何不能得到返回值

  求问以下方法为何不能得到返回值?

由解,以下疑问!

public static function query($sql, $unbuffered = false) {
    $ret = self::$db->query($sql, $unbuffered);//该句成功时$ret为1  注::self::$db->query为mysql_query
    if ($unbuffered===true) {
      $cmd = trim(strtoupper(substr($sql, 0, strpos($sql, ' '))));
      if ($cmd === 'SELECT') { } 
      elseif ($cmd === 'UPDATE' || $cmd === 'DELETE') {
          $ret = self::$db->affected_rows();
      } 
      elseif ($cmd === 'INSERT') {
          $ret = self::$db->insert_id();//该处运行时证明执行了,返回的是int类型id号
      }
   }
   return $ret;//最终返回不是int类型id号
}

                           


------解决方案--------------------
那你得到的是什么?
------解决方案--------------------
无论是否有后续操作,至少会返回 $ret = self::$db->query($sql, $unbuffered); 的值
既然是:该句成功时$ret为1,而你说返回为空,显然是查询失败了
------解决方案--------------------
由 public function query($sql, $unbuffered = false) { ... 可知
要么 query 返回资源,要么 query 返回逻辑值
------解决方案--------------------
你执行的sql是什么?执行query时,$unbuffered是什么?
------解决方案--------------------
你用elseif  如果某个条件断了下面的程序就不会再走

 以上就是求问以下方法为何不能得到返回值的内容,更多相关内容请关注PHP中文网(www.php.cn)!


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