並排兩個PHP表

Question

我是PHP新手。 請幫我。

我試圖將兩個PHP表與它們中的圖像並排對齊，但是它們在另一個下面顯示一個。 我想在一個標題下並排放置兩張桌子，在第二個標題下並排放置兩張桌子。 我看到了一些HTML解決方案，但我正在尋找PHP。 請在下面找到我的錯誤和代碼的屏幕截圖：

請隨時要求任何澄清。

$prodcatSQL="select prodcatid, prodcatname, prodcatimage from prodcat"; // create an $sql variable and store the sql statement

$exeprodcatSQL=mysql_query($prodcatSQL) or die (mysql_error());

while ($arrayprod=mysql_fetch_array($exeprodcatSQL))


{

    echo "<strong>Using display: inline-block; </strong><br>\n"; 
    echo "<table border=1 class=\"inlineTable\">\n"; 
    echo "<tr>\n"; 
    echo "<td><p><a href=products.php?u_prodcatid=".$arrayprod['prodcatid'].">";
    echo $arrayprod['prodcatname'];
    echo "<p><img src=images/".$arrayprod['prodcatimage']."></p>";
    echo "</a></p></td>\n";
    echo "</tr>\n";
    echo "</table>\n";
}


echo "<h3><center>".$subheading."</center></h3>"; 

$treatcatSQL="select treatcatid, treatcatname, treatcatimage from treatcat"; // create an $sql variable and store the sql statement

$exetreatcatSQL=mysql_query($treatcatSQL) or die (mysql_error());

while ($arrayprod=mysql_fetch_array($exetreatcatSQL))


{


    echo "<strong>Using display: inline-block; </strong><br>\n"; 
    echo "<table border=1 class=\"inlineTable\">\n"; 
    echo "<tr>\n"; 
    echo "<td><p><a href=treatmentpackages.php?u_treatcatid=".$arrayprod['treatcatid'].">";
    echo $arrayprod['treatcatname'];
    echo "<p><img src=images/".$arrayprod['treatcatimage']."></p>";
    echo "</a></p></td>\n";
    echo "</tr>\n";
    echo "</table>\n";
}

Answer 1

如果您不需要外部CSS，則需要將此添加到表中

分別替換

echo "<table border=1 class=\"inlineTable\">\n"; 

有了這個

echo "<table border=1 class=\"inlineTable\" style=\"width:50%;float:left;\">\n"; 

Answer 2

您嵌套了<p>標記，這些標記可能會引入不必要的新行。 去除

標簽，並用單獨的<td>元素替換它們，對齊方式應該很好。 $ prodcatSQL =“從prodcat中選擇prodcatid，prodcatname，prodcatimage”; //創建$ sql變量並存儲sql語句

$exeprodcatSQL=mysql_query($prodcatSQL) or die (mysql_error());

while ($arrayprod=mysql_fetch_array($exeprodcatSQL))


{
    echo "<strong>Using display: inline-block; </strong><br>\n"; 
    echo "<table border=1 class=\"inlineTable\">\n"; 
    echo "<tr>\n"; 
    echo "<td><a href=products.php?u_prodcatid=".$arrayprod['prodcatid'].">";
    echo $arrayprod['prodcatname'];
    echo "</a></td><td><a href=products.php?u_prodcatid=".$arrayprod['prodcatid']."><img src=images/".$arrayprod['prodcatimage'].">";
    echo "</td></a>\n";
    echo "</tr>\n";
    echo "</table>\n";
}
$treatcatSQL="select treatcatid, treatcatname, treatcatimage from treatcat"; // create an $sql variable and store the sql statement

$exetreatcatSQL=mysql_query($treatcatSQL) or die (mysql_error());

while ($arrayprod=mysql_fetch_array($exetreatcatSQL))


{


    echo "<strong>Using display: inline-block; </strong><br>\n"; 
    echo "<table border=1 class=\"inlineTable\">\n"; 
    echo "<tr>\n"; 
    echo "<td><a href=treatmentpackages.php?u_treatcatid=".$arrayprod['treatcatid'].">";
    echo $arrayprod['treatcatname'];
    echo "</a></td><td><a href=treatmentpackages.php?u_treatcatid=".$arrayprod['treatcatid'].">";
    echo "<img src=images/".$arrayprod['treatcatimage']."></a>";
    echo "</td>\n";
    echo "</tr>\n";
    echo "</table>\n";
}

如果這不起作用，請在問題中編輯HTML輸出。