其中nodeid是最新的，如果不是标题的话，从具有相同内容id的nodeid复制。

Question

我编写了一个php文件来显示我的数据库中的最后一篇文章。我发现最后一篇文章，但有些最后一篇文章没有标题，我想显示最后一篇文章和技术的帖子，没有标题副本，从帖子有标题，是相同的contentid.my数据库是多大。我准备了一张图片，这是我的数据库的例子。

示例图

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();
?>

有人能帮我吗？

Answer 1

因此，如何只保留一个唯一的标题和第一个节点与排序。

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

Answer 2

我认为子查询将有助于存档您的目标。在下面的SQL查询中，子查询将从外部查询中找到contentid的第一个标题。

$sql = "
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,
        N.nodeid
FROM node AS N
WHERE contentid='30' 
ORDER BY nodeid DESC LIMIT 10";