php - 将日期时间输出更改为今天并中断时间

Question

如果我有以下代码，我会从我的数据库中输出项目。我想知道如何检查某些日期是否与今天相同，然后输出的数据将回显“今天”而不是 7-23-15，如果我正在查看发布的内容，“昨天”也是如此昨天。

然后我想看看是否可以断开数据库中日期时间条目的时间部分，只回显时间？

我的数据库中有 DATETIME 作为我的数据类型。

$stmt2 = $con->prepare("SELECT `id`, `category_id`, `topic_id`, `post_creator`, `post_content`, `post_date` FROM forum_posts WHERE `category_id`=? AND `topic_id`=?");
    if($stmt2===false) {
    die();
    } else {
        //var_dump($stmt2);

            $stmt2->bind_param("ii", $cid, $tid);
            $stmt2->execute();
            $stmt2->store_result();
            $stmt2->bind_result($post_id, $post_category_id, $post_topic_id, $post_creator, $post_content, $post_date);
            //$result2 = $stmt2->get_result();
            if (!$stmt2) {
            throw new Exception($con->error);
            }
        }
    $num_rows2 = $stmt2->num_rows;
    if($num_rows2) {
        while($stmt2->fetch()) {
        $post_id;
        $post_category_id;
        $post_topic_id;
        $post_creator;
        $post_content;
        $post_date;

Answer 1

替换:

$stmt2 = $con->prepare("SELECT `id`, `category_id`, `topic_id`, `post_creator`, `post_content`, `post_date` FROM forum_posts WHERE `category_id`=? AND `topic_id`=?");

与:

$stmt2 = $con->prepare("SELECT `id`, `category_id`, `topic_id`, `post_creator`, `post_content`, IF(DATE(`post_date`) = CURDATE(), 'Today', DATE(`post_date`)) FROM forum_posts WHERE `category_id`=? AND `topic_id`=?");

这将只返回日期部分。如果是当前日期，它将返回“今天”。