PHP/MYSQL：将数据从 MySQL 显示到 PHP 表
我有一个关于我当前项目的问题要问。从上面的问题，我想显示数据到我的 PHP 表是这样的：
------- ------ -------
1 2 3
------- ------ -------
4 5 6
------- ------ -------
7 8
------- ------ -------
目前，我只有 8 个数据。问题是，<td>尽管数据不存在，但仍然出现 9 号。
下面是我目前的代码。
$sql = "SELECT * FROM factory ORDER BY Fac_ID ASC";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
echo "<table class='table table-bordered' bgcolor='#ffffff' width='50%'>";
$td=0;
while($row = mysqli_fetch_array($result)){
if ($td==0){
echo "<tr>";
}
echo "<td>" . $row['Fac_ID'] . "</td>";
$td++;
if ($td==3){
echo "</tr>";
$td=0;
}
}
echo "</tbody>";
echo "</table>";
// Free result set
mysqli_free_result($result);
} else{
echo "<p class='lead'><em>No records were found.</em></p>";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
如何确保<td>当有现有数据时会出现？
函数式编程2回答
-
隔江千里
关于您的代码风格的两个方面的评论：1) 尽量避免使用echo静态 HTML 内容 - 禁用 PHP。2) 尝试将您的数据库逻辑与 HTML 分开 - 代码会更干净。$rows = [];if ($result = mysqli_query($link, "SELECT * FROM factory ORDER BY Fac_ID ASC")){ while ($row = mysqli_fetch_array($result)) { $rows[] = $row; }}$rows_chunked = array_chunk($rows, 3);?><table class="table table-bordered" bgcolor="#ffffff" width="50%"><?phpforeach ($rows_chunked as $rows){ ?><tr><?php foreach ($rows as $row) { ?><td><?php echo $row['Fac_ID'] ?></td><?php } echo str_repeat('<td></td>', 3 - count($rows)); ?></tr><?php}?></table> -
吃鸡游戏
如果您的数据不会自然地填满表格的所有单元格，您应该质疑表格是否是显示它的正确方式。CSS 可以稍微调整一下就可以显示出来，而不必担心将结果拆分成行；运行纯 html 表示的示例：.lined { border-top-style: dashed; border-bottom-style: dashed; margin-top: -3px; }<link href="https://stackpath.bootstrapcdn.com/bootstrap/4.3.1/css/bootstrap.min.css" rel="stylesheet"/> <div class="row"> <div class="col-sm-3">Page Column 1</div> <div class="col-sm-3">Page Column 2</div> <div class="col-sm-3">Page Column 3</div> <div class="col-sm-3">Page Column 4</div></div><div class="row"> <div class="col-sm-3"></div> <div class="col-sm-3"> <div class="row"> <div class="col-sm-4 lined">1</div> <div class="col-sm-4 lined">2</div> <div class="col-sm-4 lined">3</div> <div class="col-sm-4 lined">4</div> <div class="col-sm-4 lined">5</div> <div class="col-sm-4 lined">6</div> <div class="col-sm-4 lined">7</div> <div class="col-sm-4 lined">8</div> </div> </div> <div class="col-sm-3"></div> <div class="col-sm-3"></div></div>您需要做的就是将您的数据合并到 html 中：<?php$sql = "SELECT * FROM factory ORDER BY Fac_ID ASC";if(!$result = mysqli_query($link, $sql)){ echo "ERROR: Could not able to execute $sql. " . mysqli_error($link); }$rows = array();while($row = mysqli_fetch_array($result)){ $rows[] = $row;}mysqli_free_result($result);?><div class="row"> <div class="col-sm-3">Page Column 1</div> <div class="col-sm-3">Page Column 2</div> <div class="col-sm-3">Page Column 3</div> <div class="col-sm-3">Page Column 4</div></div><div class="row"> <div class="col-sm-3"></div> <div class="col-sm-3"> <div class="row"> <?php foreach($rows as $r): ?> <div class="col-sm-4 lined"><?= $r['Fac_ID'] ?></div> <?php endforeach; ?> <?php if(sizeof($rows) == 0): ?> <p class='lead'><em>No records were found.</em></p> <?php endif; ?> </div> </div> <div class="col-sm-3"></div> <div class="col-sm-3"></div></div>
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