按下按钮时如何从mysql获取数据？

Question

单击按钮时，我尝试从mysql随机获取数据，这是我的代码。

<?php

    $con=mysqli_connect("localhost","root","","school");

// Check connection

    if (mysqli_connect_errno())

{

    echo "Failed to connect to MySQL: " . mysqli_connect_error();

}

$result = mysqli_query($con,"SELECT description From task ORDER BY RAND() LIMIT 2");

    echo "<p><strong>Question:</strong></p>";

    while($row = mysqli_fetch_array($result))

{

    echo "<p>". $row['description'] ."</p>";

}

    mysqli_close($con);

?>

     <input type="button" value="Get Assignment" onclick="myfunc()">

</body>

</html>

我希望在我单击“获取分配”按钮时输出，它将显示随机问题，直到我将显示空框。

Answer 1

您应该将您的php代码包装为isset函数，该函数将检查是否已发送数据，然后运行，否则不运行。isset函数检查是否设置了变量且不为null或者，您可以调用Ajax来添加点击数据。这里<?php&nbsp; &nbsp; $con=mysqli_connect("localhost","root","","school");// Check connection&nbsp; &nbsp; if (mysqli_connect_errno()){&nbsp; &nbsp; echo "Failed to connect to MySQL: " . mysqli_connect_error();}if(isset($_POST['submit'])){$result = mysqli_query($con,"SELECT description From task ORDER BY RAND() LIMIT 2");&nbsp; &nbsp; echo "<p><strong>Question:</strong></p>";&nbsp; &nbsp; while($row = mysqli_fetch_array($result)){&nbsp; &nbsp; echo "<p>". $row['description'] ."</p>";}&nbsp; &nbsp; mysqli_close($con);}?>&nbsp; &nbsp; &nbsp;<input type="button" name="submit" value="Get Assignment" onclick="myfunc()"></body></html>编辑 固定我的代码您需要将输入按钮包装到一个表单中，该表单会将表单提交到页面本身<form method="post" action=""><input type="submit" name="submit" value="Get Assignment"></form>