php - 如何根据另一个字符串的值获取表值？

Question

我不确定我是否使用了上面的正确行话，但我想做的是遵循...我如何让 $author 根据帖子的 $author_id 与用户 ID 匹配来输出用户名(来自 mySQL 用户表)？

<?php 
include "inc/mysql-connect.php"; 
$DynamicFeed = "";
$sql = mysql_query("SELECT * FROM posts ORDER BY id DESC LIMIT 20");
$postCount = mysql_num_rows($sql); // count the output amount
if ($postCount > 0) {
  while($row = mysql_fetch_array($sql)){ 
    $id = $row["id"];
    $title = $row["title"];
    $post = $row["post"];
    $author_id = $row["author_id"];
    $author = $row["author_id"];
    $type = $row["type"];
    $date_posted = $row["date_posted"];
    $DynamicFeed .= "<div class=\"item\"><img class=\"img-type\" align=\"left\"       src=\"img/$type.jpg\"> <p><a href=\"droplet.php?id=$id\"><b>$title</b></a><br />by <a    href=\"profile.php?id=$author_id\">$author</a> on $date_posted<br /><br /> $post</p>    </div>";
}
} else {
$DynamicFeed = "It would appear that there ar'nt any Droplets here...";
}
?> 

Answer 1

您可以按如下方式使用 JOIN 语句:

$sql = mysql_query("SELECT posts.*,users.name FROM posts LEFT JOIN users on posts.author_id=users.id ORDER BY posts.id DESC LIMIT 20");