我必须制作一个动态表n * n,用户首先给出数字n,程序制作一个带有复选框的5 * 5表这部分我已经做到了,第二部分是用户检查相同的复选框和点击提交,程序再次制作一个表5 * 5,但在复选框的位置,检查是彩色的。我上传了图像。 抱歉我的英语不好,谢谢你的时间。 enter image description here
<form name="form" action="" method="get">
<input type="text" name="subject" id="subject" value="Give value">
</form>
<?php
$rows = $cols = $name = "";
if(isset($_GET['subject']))
$rows = $cols = $_GET['subject'];
if(isset($_POST['check_list']))
$name = $_POST['check_list'];
if(isset($_GET['subject'])){
echo "<form action='my.php' method='post'>";
echo "<table border='1'>";
for($tr=1;$tr<=$rows;$tr++){
echo "<tr>";
for($td=1;$td<=$cols;$td++){
echo "<td><input type='checkbox' name='check_list[]' value='value ".$td."'></td>";
}
echo "<tr>";
}
echo "</table>";
echo "<input type='submit' />
</form>";
}
// this part of code is not make the third excecution the number 3 image
echo $cols;
echo "<table border='1'>";
for($tr=1;$tr<=$rows;$tr++){
echo "<tr>";
foreach($_POST['check_list'] as $value){
if($tr == $value[td])echo "<td bgcolor='#FF0000'></td>";
else
echo "<td> </td>";
}
echo "</tr>";
}
echo "</table>";
?>