Question

查询如下：

  SELECT v2.video_id
    FROM `VideoTags` AS v1
    JOIN `VideoTags` AS v2 USING ( tag_id )
   WHERE v1.video_id =1
     AND v1.video_id <> v2.video_id
GROUP BY v2.video_id
ORDER BY COUNT( * ) DESC

我正在尝试获取相关视频，当我在phpmyadmin中运行此查询时，它返回正确的结果集但是当我尝试使用php编写此内容并回显结果时，我得不到返回的值，我使用的查询代码是：

$result = mysql_query("SELECT v2.video_id FROM `VideoTags` AS v1 JOIN `VideoTags` AS v2 USING ( tag_id ) WHERE v1.video_id =1 AND v1.video_id <> v2.video_id GROUP BY v2.video_id ORDER BY COUNT( * ) DESC"); 

$row = mysql_fetch_array( $result );

echo "ID ".$row['video_id'];

我确定它简单的东西我似乎无法弄明白？如果有人可以帮忙...

Answer 1

尝试使用此查询： SELECT v2.video_id as v2id FROM VideoTags AS v1 JOIN VideoTags AS v2 USING ( tag_id ) WHERE v1.video_id =1 AND v1.video_id <> v2.video_id GROUP BY v2.video_id ORDER BY COUNT( * ) DESC

然后

echo "ID ".$row['v2id'];

Answer 2

它返回一个数组，以便你可以将它循环出去

$result = mysql_query("SELECT v2.video_id FROM `VideoTags` AS v1 JOIN `VideoTags` AS v2 USING ( tag_id ) WHERE v1.video_id =1 AND v1.video_id <> v2.video_id GROUP BY v2.video_id ORDER BY COUNT( * ) DESC"); 

while ($row = mysql_fetch_array( $result )) {
    echo "ID " . $row[0] . "<br />";
}

Answer 3

我不确定为什么它不会让我对Trikks的答案发表评论，但他基本上是正确的，除了一件事，所以我会把它给他。而不是$ row [0]，你会想要$ row ['video_id']; （根据提供的例子）。我将回声更改为单引号，因为你没有在回声中使用变量，单引号比双引号更快。

$result = mysql_query("SELECT v2.video_id FROM `VideoTags` AS v1 JOIN `VideoTags` AS v2 USING ( tag_id ) WHERE v1.video_id =1 AND v1.video_id <> v2.video_id GROUP BY v2.video_id ORDER BY COUNT( * ) DESC"); 

while ($row = mysql_fetch_array( $result ))
{
    echo 'ID ' . $row['video_id'] . '<br />';
}

编辑：此外，如果您没有收到任何值，我会检查您是否使用print_r(array_key($row));使用了正确的密钥：

$result = mysql_query("SELECT v2.video_id FROM `VideoTags` AS v1 JOIN `VideoTags` AS v2 USING ( tag_id ) WHERE v1.video_id =1 AND v1.video_id <> v2.video_id GROUP BY v2.video_id ORDER BY COUNT( * ) DESC"); 

while ($row = mysql_fetch_array( $result ))
{
    print_r(array_key($row));
    echo 'ID ' . $row['video_id'] . '<br />';
}

我如何使用PHP返回加入查询的结果？

3 个答案: