我已经显示了数据库中的数据表,左侧有复选框。我想找到一种方法将复选框链接到问题编号 (ID)。当我点击提交时,我希望回显所选的 id。我非常希望有人能够选择他们想要的问题然后显示它们。
<?php
$con=mysqli_connect("####","####","#####","###");
$result = mysqli_query($con,"SELECT * FROM q_and_a ");
echo "<table border='1'>
<tr>
<th>Add</th>
<th>#</th>
<th>Question</th>
<th>A</th>
<th>B</th>
<th>C</th>
<th>D</th>
<th>Answer</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo '<td><input type="checkbox" name="questions[]" value="$id"></td>';
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['question'] . "</td>";
echo "<td>" . $row['choiceA'] . "</td>";
echo "<td>" . $row['choiceB'] . "</td>";
echo "<td>" . $row['choiceC'] . "</td>";
echo "<td>" . $row['choiceD'] . "</td>";
echo "<td>" . $row['answer'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
提交按钮
<form method="POST" action="makeTest.php">
<input type="submit" name="make" value="Make Test">
</form>
最佳答案
对您的代码进行一些编辑,然后它就会起作用。 1.通过添加非*字段来更改查询(以确保性能和显示顺序)
$result = mysqli_query($con,"SELECT id,question,choiceA,choiceB,choiceC,choiceD,answer FROM q_and_a ");
然后在 while block 打开表单标签之前(HTML)
<?php //above codes will be there as you show before echo '<form method="POST" action="makeTest.php">'; while($row = mysqli_fetch_array($result)){ { $id=$row['id']; // initialize your id here, so as to pass it in checkbox too // then continue with your code } ?> <input type="submit" name="make" value="Make Test"> </form>
在 maketest.php 中,你可以使用 foreach 处理 ckeckbox,见下文
foreach($_POST['questions'] as $questions){
//do your job
}
关于php - 如何获取显示的数据库QUICKFIX上所有复选框的ID?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26583815/