php - 如果 nodeid 不是来自具有相同内容 id 的 nodeid 的标题副本，则 nodeid 是最新的

Question

我写了一个 php 文件来显示我数据库中的最后一篇文章。我找到最后一篇文章，但最后一篇文章没有标题我想显示最后一篇文章和其他没有标题的文章 复制自有标题且内容相同的文章。我的数据库很大。我准备了一张图片作为我的数据库的示例

the example pic

这是我的 php 代码:

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "vbtest";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT title , nodeid FROM node where contenttypeid='30' order by nodeid desc limit 10";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "title: " . $row["nodeid"]. " title: " . $row["title"]. "<br>";
    }
} else {
    echo "0 results";
}
$conn->close();
?>

谁能帮帮我？

Answer 1

那么，如何只留下一个唯一的标题和第一个 nodeid 进行排序。

SELECT * FROM
(SELECT T.title,
       MIN(nodeid) AS nodeid
FROM
    (SELECT
           (SELECT N2.title FROM node AS N2
            WHERE N2.contentid = N.contentid
                  AND N2.starter = N.starter
            ORDER BY N2.nodeid ASC
            LIMIT 1) AS title,
            nodeid
    FROM node AS N
    WHERE contentid='30' 
    ORDER BY nodeid DESC LIMIT 10) AS T
GROUP BY title 
) AS T2
ORDER BY nodeid DESC