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PHP脚本返回MySQL数据到表中,需要回车

Dan • 5 年前 • 1533 次点击  

我有一个PHP脚本,它从MySQL数据库表中选择信息,并使用while循环将其流到一个表中,然后对每个单元格使用for each循环创建信息列(日期、图像、句子)。我的问题是这些列是并排的,在我的页面上跑来跑去( http://dankirchoff.com/drawingweek.php ). 我需要一个“回车”的某种形式,以保持他们在网页上。我试着用“索引”字段作为计数器,这不起作用。有什么建议吗?

<?php
	include ("inc_connect.php");
	if ($link !== FALSE) {
    	$db = "dkfineart_01";
    	mysqli_select_db($db);
    	$TableName = "drawingoftheweek";
    	$sql = "SELECT * FROM $TableName";
		if ($res = mysqli_query($link, $sql))	{
			if (mysqli_num_rows($res) > 0)	{
				while ($drawresults = mysqli_fetch_array($res))	{
					$drawing[] = $drawresults;
				}
				echo "<table style='margin-left:190px; width:500px;'><tr>";
				foreach ($drawing as $draw)	{
					echo "<td style='padding:20px 20px 0 0'>" . date('m/d/Y', strtotime($draw['date'])) . "</td>";
				}
				echo "</tr><tr>";
				foreach ($drawing as $draw)	{
					echo "<td style='padding:0 20px 0 0'>" . "<img src='images/" . $draw['image'] . ".jpg' width='400px'>" . "</td>";
				}
				echo "</tr><tr>";
				foreach ($drawing as $draw)	{
					echo "<td width='400px' style='padding:0 20px 0 0'>" . $draw['drawingtext'] . "</td>";
				}
				echo "</tr>";
				foreach ($drawing as $draw)	{
					$index = $draw['index'];
						if ($index % 2 === 0)	{
							echo "<br />";
						}
					}
				echo "</table>";
			}
				mysqli_free_result($res);
			}
			else	{
				echo "No matching records are found.";
			}
		}
		else	{
			echo "Unable to select from table. " . mysqli_error($link);
		}
	mysqli_close($link);
?>
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