我如何“回声"? “资源ID#6"从PHP的MySql响应?

问题描述

一个非常简单的问题，我有这个问题:

very simple question, i have this:

$datos1=mysql_query("SELECT TIMEDIFF(NOW(), '" . $row['fecha'] . "');");
echo($datos1);

但是结果是:资源ID#6

我如何按照预期的方式打印结果?谢谢！

How can i print the result in the way is meant to be? Thanks!

推荐答案

您需要使用提取功能.例如:

You need to use a fetch function. for example:

$result = mysql_query(sprintf("SELECT TIMEDIFF(NOW(), '%s') as time_delta", $row['fecha']));
if($result){
  $data = mysql_fetch_assoc($result);
  echo $data['time_delta'];
}

但是，除非绝对必要，否则我不会使用mysql函数.不建议将mysql扩展名用于新项目.相反，您应该将 PDO与PDO_mysql 或.

However, i wouldnt use the mysql functions unless absolutely necessary. the mysql extension is NOT recommended for use in new projects. Instead you should use PDO with PDO_mysql or mysqli.

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