求问以下方法为何不能得到返回值?
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 }
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那你得到的是什么?
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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, $unbuffered = false) { ... 可知
要么 query 返回资源,要么 query 返回逻辑值
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你执行的sql是什么?执行query时,$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
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