从中间行包含变量的查询中显示11个连续行

时间:2011-12-09 00:50:13

标签: php sql pagination

在我的网页上,有一个名为$submission的变量。我想从下面的查询中精确显示11行:$submission等于$row["title"]的行,高于它的5行,以及它下面的5行。全部按points降序排列。

我该怎么做?

$sqlStr = "SELECT title, points, submissionid
             FROM submission 
         ORDER BY points DESC"; 


$result = mysql_query($sqlStr);

$arr = array();

$count=1;

echo "<table class=\"samplesrec\">";

while ($row = mysql_fetch_array($result)) { 


    echo '<tr >';

    echo '<td>'.$count++.'.</td>';

    echo '<td class="sitename1">'.$row["title"].'</td>';

    echo '<td class="sitename2"><div class="pointlink2">'.number_format($row["points"]).'</div></td>';
    echo '</tr>';

}

echo "</table>";

7 个答案:

答案 0 :(得分:3)

如果几行共享“点”的相同值,这有点棘手:

points | title | submissionid
------ + ----- + ------------
    ...
   50  | 'foo' | ABCD01234     <-- If (50, 'foo') is the "midpoint" record,
   50  | 'bar' | EF7654321     <-- does (50, 'bar') come before or after?
    ...

在这种情况下,我们需要下订单。为方便起见,我们将按“点数”降序排列,然后按“标题”降序排序。

假设您的“中点记录”的分值为“@points”且标题为“@title”,我们会说“中间点”之前的记录是那些(点,标题)&gt;的记录。 (@points,@ title)。类似地,那些“在”中点之后的记录具有它们(点,标题)&lt; (@points,@ title)。

把它放在一起,我们有:

-- First, initialize user variables with the points and
-- title of our midpoint (including midpoint)
--
SELECT @title := title,
       @points := points
  FROM submission
 WHERE title = ?          -- bind your $submission variable bere
 LIMIT 1;

-- Now, select six records greater than or equal to our
-- midpoint.
--
  SELECT title, points, submissionid
        FROM (  SELECT title, points, submissionid
                  FROM submission
                 WHERE (points, title) >= (@points, @title)
              ORDER BY points ASC, title ASC
                 LIMIT 6) gte
-- and UNION those records with five records less than
-- our midpoint
--
   UNION
  SELECT title, points, submissionid
        FROM (  SELECT title, points, submissionid
                  FROM submission
                 WHERE (points, title) < (@points, @title)
              ORDER BY points DESC, title DESC
                 LIMIT 5) lt
-- Finally sort the result set
--
ORDER BY points DESC, title DESC

答案 1 :(得分:2)

您需要使用UNION

(
    SELECT title, points, submissionid
    FROM submission
    WHERE points < (SELECT points FROM submission WHERE title = <row_title> LIMIT 1)
    ORDER BY points DESC LIMIT 5
)UNION(
    SELECT title, points, submissionid
    FROM submission
    WHERE points > (SELECT points FROM submission WHERE title = <row_title> LIMIT 1)
    ORDER BY points ASC LIMIT 5
) ORDER BY points DESC

我还没有测试过,但这是它的要点。您将获得10条记录(或更少),您必须在PHP中弄清楚哪些记录应该高于您的$submission以及下面的记录,因为如果您获得9条记录,您将不知道4条记录是否更高,或者5条记录

或者你可以用2个查询当然&gt; _&gt;

来做

答案 2 :(得分:2)

SELECT 
    title, points, submissionid
FROM
    submission 
WHERE 
    ROWID >= (SELECT ROWID FROM submission WHERE ..... ORDER BY points DESC) - 5
AND
    ROWID <= (SELECT ROWID FROM submission WHERE ..... ORDER BY points DESC) + 5
ORDER BY 
    points DESC

答案 3 :(得分:1)

我认为对数据库进行2次查询最简单,但你可以用PHP来完成:

$entryFound = false;
$counter = 0;
$priorEntries = array();

while ($row = mysql_fetch_array($result)) { 
    $rowHtml = '<tr >';
    $rowHtml .=  '<td>'.$count++.'.</td>';
    $rowHtml .= '<td class="sitename1">'.$row["title"].'</td>';
    $rowHtml .= '<td class="sitename2"><div class="pointlink2">'.number_format($row["points"]).'</div></td>';
    $rowHtml .= '</tr>';

    if ($entryFound) {
            if ($counter < 5) {
                $counter++;
                echo $rowHtml;
            }
    } else {            
            array_unshift($priorEntries, $rowHtml); 
            if (strcmp($row["title"], $submission) == 0) {
                echo implode(array_reverse($priorEntries));
                $entryFound = true;
            }   
    array_splice($priorEntries, 5);
    }
}

答案 4 :(得分:1)

我认为以下可能会做你所要求的。它将运行查询并开始从结果集中读取记录,直到找到标题等于您拥有的$submission变量的记录的第一个出现(如果我理解它)正确地,您的表的主键是submitid,标题是一个简单的字段 - 即您没有唯一的键,因此可能有多个具有相同标题的记录。)

在找到第一条记录后,它会再读取5条记录并停止。然后它将为您提供您希望打印的记录数组的一部分,最后它将打印出来。

$sqlStr = "SELECT title, points, submissionid FROM submission ORDER BY points DESC"; 

$result = mysql_query($sqlStr);

$count = 1;

$found = false;

$continue = true;

$records = array();

$row = mysql_fetch_array($result);
while ($row !== false && $continue === true)
{ 
    if($found == false && $row['title'] == $submission)
    {
        $found = true;
    }
    elseif($found == true && $count < 6) 
    {
        $count++;
    }
    elseif($found == true && $count >= 6)
    {
        $continue = false;
    }

    $records[] = $row;
    $row = mysql_fetch_array($result);
}

if($found === true)
{
    if(array_count($records) > 11)
        $records = array_splice($records, -11);
}
else
{
    $records = array();
}

echo "<table class=\"samplesrec\">";

for($i = 1; $i <= count($records); $i++)
{
    echo '<tr >';

    echo '<td>'.$i.'.</td>';

    echo '<td class="sitename1">'.$records[$i]["title"].'</td>';

    echo '<td class="sitename2"><div class="pointlink2">'.number_format($records[$i]["points"]).'</div></td>';

    echo '</tr>';
}

echo "</table>";

答案 5 :(得分:1)

如果没有重复的"title"值,并且没有重复"points"

SELECT title, points, submissionid
FROM submission 
WHERE points <=
      ( SELECT points
        FROM submission
        WHERE points >= 
              ( SELECT points
                FROM submission
                WHERE title = @row_title
              )
        ORDER BY points ASC
        LIMIT 1 OFFSET 5
      )
ORDER BY points DESC
LIMIT 11 OFFSET 0

答案 6 :(得分:0)

如果打算使用单个查询并且表格很小且性能不重要则使用

SELECT b.title, b.points FROM (
  SELECT @rank1 := @rank1 + 1 as slno, temp1.* FROM (
    SELECT s1.title, s1.points, COUNT(s2.title) rank
    FROM submission s1
    JOIN submission s2 ON s1.points <= s2.points
    GROUP BY s1.title, s1.points
    ORDER BY rank, s1.title ASC
  ) as temp1 
  JOIN( SELECT @rank1 := 0 ) AS init
  ORDER BY slno 
) a
LEFT JOIN (
  SELECT @rank2 := @rank2 + 1 as slno, temp1.* FROM (
    SELECT s1.title, s1.points, COUNT(s2.title) rank
    FROM submission s1
    JOIN submission s2 ON s1.points <= s2.points
    GROUP BY s1.title, s1.points
    ORDER BY rank, s1.title ASC
  ) as temp1 
  JOIN( SELECT @rank2 := 0 ) AS init
  ORDER BY slno 
) b ON a.slno BETWEEN b.slno - 5 AND b.slno + 5
WHERE a.title = <row_title>;

这将返回上面选择的行和最多5行(如果存在)和最多5行(如果存在)。

但是,强烈建议使用临时表来存储排名并使用它来显示排名。

有关上述查询的更多详情@ http://www.artfulsoftware.com/infotree/queries.php?&bw=1280#460

有关优化方法的更多细节@ http://onlamp.com/pub/a/mysql/2007/03/01/optimize-mysql-rank-data.html