为什么这个 SQL 查询返回一篇文章的作者姓名为 NULL，如果作者写了不止一篇文章？

Question

以下是我的创作陈述：

            String createFeedTable = "CREATE TABLE  rss_feed ("
                + "channel_ID INTEGER NOT NULL GENERATED ALWAYS AS IDENTITY (START WITH 1, INCREMENT BY 1), "
                + "channelTitle VARCHAR(255), "
                + "channelLink VARCHAR(255), "
                + "channelDescription VARCHAR(255), "
                + "channelPubDate TIMESTAMP, "
                + "channelLastBuildDate TIMESTAMP,"
                + "channelIcon BLOB(1M),"
                + "PRIMARY KEY (channel_ID))";

        String createFeedItemTable = "CREATE TABLE  feed_item ("
                + "  item_ID INTEGER NOT NULL GENERATED ALWAYS AS IDENTITY (START WITH 1, INCREMENT BY 1),"
                + "  itemTitle VARCHAR(255),"
                + "  itemLink VARCHAR(255),"
                + "  itemDescription VARCHAR(255),"
                + "  itemGuid VARCHAR(255),"
                + "  itemPubDate TIMESTAMP,"
                + "  channel_ID INTEGER REFERENCES rss_feed(channel_ID),"
                + "  haveRead BOOLEAN DEFAULT FALSE,"
                + "  PRIMARY KEY (item_ID))";

        String createCategoryTable = "CREATE TABLE  category ("
                + "category_ID INTEGER NOT NULL GENERATED ALWAYS AS IDENTITY (START WITH 1, INCREMENT BY 1),"
                + "categoryName VARCHAR(255), "
                + "CONSTRAINT category_constraint UNIQUE (categoryName),"
                + "PRIMARY KEY (category_ID))";

        String createFeedItemCategoryTable = "CREATE TABLE  feed_item_category ("
                + "  item_ID INTEGER REFERENCES feed_item(item_ID),"
                + "  category_ID INTEGER REFERENCES category(category_ID))";

        String createFeedCategoryTable = "CREATE TABLE  rss_feed_category ("
                + "feed_ID INTEGER REFERENCES rss_feed(channel_ID),"
                + "category_ID INTEGER REFERENCES category(category_ID))";

        String createAuthorTable = "CREATE TABLE  author ("
                + "author_ID INTEGER NOT NULL GENERATED ALWAYS AS IDENTITY (START WITH 1, INCREMENT BY 1),"
                + "authorName VARCHAR(255),"
                + "CONSTRAINT author_constraint UNIQUE (authorName),"
                + "PRIMARY KEY (author_ID))";

        String createFeedItemAuthorTable = "CREATE TABLE  feed_item_author ("
                + "item_ID INTEGER REFERENCES feed_item(item_ID),"
                + "author_ID INTEGER REFERENCES author(author_ID))";

我的查询：

SELECT i.ITEMTITLE, i.ITEMLINK, i.ITEMDESCRIPTION, i.ITEMPUBDATE, i.CHANNEL_ID,
  i.ITEM_ID, a.AUTHOR_ID, a.AUTHORNAME

    FROM FEED_ITEM i
    LEFT JOIN FEED_ITEM_AUTHOR fia
         ON i.ITEM_ID = fia.ITEM_ID
    LEFT JOIN AUTHOR a
         ON fia.ITEM_ID = a.AUTHOR_ID;

结果：

<table border =1>
<tr>  
<th>ITEMTITLE</th>
<th>ITEM_ID</th>
<th>AUTHOR_ID</th>
<th>AUTHORNAME</th>
</tr>
<tr>
<td align="center">Lawyers...</td>
<td align="center">1</td>
<td align="center">1</td>
<td align="center">Joe Mullin</td>
</tr>
<tr>
<td align="center">AT&T/Time Warner...</td>
<td align="center">2</td>
<td align="center">2</td>
<td align="center">Jon Brodkin</td>
</tr>
<tr>
<td align="center">The “technosphere” ...</td>
<td align="center">19</td>
<td align="center">19</td>
<td align="center">Annalee Newitz</td>
</tr>
  <tr>
<td align="center">Decrypted: ...</td>
<td align="center">20</td>
<td align="center">&lt;null&gt;</td>
<td align="center">&lt;null&gt;</td>
</tr>


</table>

table、feed_item_author 具有正确的作者和条目数 - 15 位作者和 20 条条目并且逐行正确匹配数据.

tables、author 和 feed_items 也有正确的行数 - 15 位作者, 和 20 feed_items.

我的代码中也有验证输入是否正确的代码：

  try {

        feedItem_AuthorStatement = conn.prepareStatement(feed_item_author_SQL, PreparedStatement.RETURN_GENERATED_KEYS);

        feedItem_AuthorStatement.setInt(1, item_ID);
        feedItem_AuthorStatement.setInt(2, author_ID);
        int entrySuccess = feedItem_AuthorStatement.executeUpdate();

        System.out.println("SUCCESS? :" + entrySuccess);
    } catch (SQLException ex) {
        Logger.getLogger(FeedItemCategoryHelper.class.getName()).log(Level.SEVERE, null, ex);
    }

我在这里错过了什么？如果作者写了不止一篇文章，为什么我的查询 return 作者姓名和 ID 为空？

Answer 1

你不应该加入 ON fia.AUTHOR_ID = a.AUTHOR_ID;

您希望作者 ID 相同。除非我遗漏了什么，否则让 itemID 匹配 authorID

是没有意义的

为什么这个 SQL 查询返回一篇文章的作者姓名为 NULL，如果作者写了不止一篇文章？

Why is this SQL query returning the author name for an article as NULL, if the author has written more than one article?

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