将 aukciono_id 从一张表转移到另一张表

Question

我正在尝试将“aukciono_id”从一张桌子转移到另一张桌子。我试图使用单个while循环来做到这一点，也许它会起作用，但主要问题是当用户悬停在元素类“placiau”上时，另一个表显示不是悬停 strong> 元素的“aukciono_id”。我只是不能使用 while 循环，因为 while 循环会一一遍历查询。我必须以某种方式识别每个<td class="placiau"></td>“aukciono_id”。此外，表格应该出现在 hovered 元素旁边，但我会自己修复它。

---

<table>
<?php
$aukciono_laimetojai_ir_aukciono_istorija_while = mysql_query("SELECT taw.id AS taw_id,
    tai.aukciono_id AS tai_aukciono_id, taw.user_id AS taw_user_id, taw.win AS taw_win,
    tai.date AS tai_date, taw.date AS taw_date, COALESCE(SUM(tai.bid), 0) AS tai_bid
FROM tb_auction_winners taw JOIN tb_aukciono_istorija tai
ON taw.id = tai.aukciono_id
WHERE tai.user_id = 206 GROUP BY aukciono_id");

$aukciono_istorija1_while = mysql_query("SELECT tab.id, tu.level, tab.user_id, tab.user,
    tab.bid, tab.date, tab.aukciono_id FROM tb_aukciono_istorija tab JOIN tb_users tu
ON tab.user_id = tu.id
WHERE tab.aukciono_id = 47 ORDER BY tab.id DESC");

while ($r1 = mysql_fetch_assoc($aukciono_laimetojai_ir_aukciono_istorija_while)) {
    if ($r1['taw_user_id'] == 206) { ?>
        <tr>
            <td><?php echo $r1['taw_date']; ?></td>
            <td><?php echo $r1['taw_win'] - $r1['tai_bid']; ?> Eur</td>
            <td>0 Eur</td>
            <td class="placiau">Plačiau <?php echo $r1['tai_aukciono_id']; ?></td>
        </tr>
<?php
    } else { ?>
            <tr>
                <td><?php echo $r1['tai_date']; ?></td>
                <td>0 Eur</td>
                <td><?php echo $r1['tai_bid']; ?> Eur</td>
                <td class="placiau">Plačiau <?php echo $r1['tai_aukciono_id']; ?></td>
            </tr>
<?php
    }
}
?>
</table>
<table id="aukciono_apzvalga" style="
width: 69.7%;
left: 632px;
position: absolute;
top: 590px;
z-index: 1;
opacity: 0;
-webkit-transition: opacity .5s ease-out;
transition: opacity .5s ease-out;
display: block;
overflow-y: auto;
max-height: 480px;">
<?php
    while ($r1 = mysql_fetch_assoc($aukciono_istorija1_while)) { ?>
        <tr>
            <td><?php echo $r1['aukciono_id']; ?></td>
        </tr>
<?php } ?>
</table>

<script>
    $('.placiau').hover(function() {
        $('#aukciono_apzvalga').css('opacity', 1);
    }, function() {
        $('#aukciono_apzvalga').css('opacity', 0);
    });
</script>

画出它的样子：

注意：47 只是一个随机数，例如它的样子

例如，当用户悬停“42”时，它应该在另一个表中显示42，而不是在“$aukciono_istorija1_while”查询中显示47。有可能吗？

PS：我知道，我应该使用mysqli_* 或PDO 而不是mysql_*

Answer 1

你只有一个循环生成的表，你应该在循环内构建所有可能的表，并给每个表一个与aukciono_id相关的不同id，然后在悬停函数中，获取目标@987654323 @ 并显示与此 id 相关的表。 试试这个：

<?php

$aukciono_laimetojai_ir_aukciono_istorija_while = mysql_query("SELECT taw.id AS taw_id,
    tai.aukciono_id AS tai_aukciono_id, taw.user_id AS taw_user_id, taw.win AS taw_win,
    tai.date AS tai_date, taw.date AS taw_date, COALESCE(SUM(tai.bid), 0) AS tai_bid
    FROM tb_auction_winners taw JOIN tb_aukciono_istorija tai
    ON taw.id = tai.aukciono_id
    WHERE tai.user_id = 206 GROUP BY aukciono_id");

$aukciono_istorija1_while = mysql_query("SELECT tab.id, tu.level, tab.user_id, tab.user,
    tab.bid, tab.date, tab.aukciono_id FROM tb_aukciono_istorija tab JOIN tb_users tu
    ON tab.user_id = tu.id
    WHERE tab.aukciono_id = 47 ORDER BY tab.id DESC");

?>
<table>
<?php
while ($r1 = mysql_fetch_assoc($aukciono_laimetojai_ir_aukciono_istorija_while)) {
    if ($r1['taw_user_id'] == 206) { ?>
    <tr>
        <td><?php echo $r1['taw_date']; ?></td>
        <td><?php echo $r1['taw_win'] - $r1['tai_bid']; ?> Eur</td>
        <td>0 Eur</td>
        <td class="placiau" data-aukciono-id="<?php echo $r1['tai_aukciono_id']; ?>">Plačiau <?php echo $r1['tai_aukciono_id']; ?></td>
    </tr>
<?php
    } else { ?>
    <tr>
        <td><?php echo $r1['tai_date']; ?></td>
        <td>0 Eur</td>
        <td><?php echo $r1['tai_bid']; ?> Eur</td>
        <td class="placiau" data-aukciono-id="<?php echo $r1['tai_aukciono_id']; ?>">Plačiau <?php echo $r1['tai_aukciono_id']; ?></td>
    </tr>
<?php
    }
}
?>
</table>
<?php
    while ($r1 = mysql_fetch_assoc($aukciono_istorija1_while)) { ?>
<table class="aukciono_apzvalga" data-aukciono-id="<?php echo $r1['aukciono_id']; ?>" style="
    width: 69.7%;
    left: 632px;
    position: absolute;
    top: 590px;
    z-index: 1;
    opacity: 0;
    -webkit-transition: opacity .5s ease-out;
    transition: opacity .5s ease-out;
    display: block;
    overflow-y: auto;
    max-height: 480px;">
    <tr>
        <td><?php echo $r1['aukciono_id']; ?></td>
    </tr>
</table>
<?php } ?>
<script>
    $('.placiau').hover(function() {
        var aukciono_id = $(this).attr('data-aukciono-id');
        $('.aukciono_apzvalga[data-aukciono-id="'+aukciono_id+'"]').css('opacity', 1);
    }, function() {
        $('.aukciono_apzvalga').css('opacity', 0);
    });
</script>

Answer 2

问题解决了！ :) 这是 while 循环中的错误。

</table>
<?php
while ($r1 = mysql_fetch_assoc($aukciono_istorija1_while)) { ?>
<table class="aukciono_apzvalga" data-aukciono-id="<?php echo $r1['aukciono_id']; ?>" style="
width: 69.7%;
left: 632px;
position: absolute;
top: 590px;
z-index: 1;
opacity: 0;
display: block;
overflow-y: auto;
max-height: 480px;">
    <tr>
        <th colspan="5">Aukciono apžvalga</th>
    </tr>
    <tr>
        <th>Nr.</th>
        <th>Žaidėjas</th>
        <th>Lygis</th>
        <th>Suma</th>
        <th>Laikas</th>
    </tr>
<?php
    $i = mysql_num_rows($aukciono_istorija1_while);
    $apa = $r1['aukciono_id'];
    $aukciono_istorija2_while = mysql_query("SELECT tab.id, tu.level, tab.user_id, tab.user, tab.bid, tab.date, tab.aukciono_id FROM tb_aukciono_istorija tab JOIN tb_users tu ON tab.user_id = tu.id
    WHERE aukciono_id = $apa ORDER BY tab.id DESC LIMIT 10");
    while ($r2 = mysql_fetch_assoc($aukciono_istorija2_while)) { ?>
        <tr>
            <td><?php echo $r2['aukciono_id']; ?></td>
            <td><?php echo $r2['user']; ?></td>
            <td><?php echo $r2['level']; ?> lvl.</td>
            <td><?php echo $r2['bid']; ?> Eur</td>
            <td><?php echo date('H:i:s', strtotime($r2['date'])); ?></td>
        </tr>
<?php
        $i--;
    }
}
?>
</table>