單擊按鈕時，我嘗試從mysql隨機獲取數據，這是我的代碼。<?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>我希望在我單擊“獲取分配”按鈕時輸出，它將顯示隨機問題，直到我將顯示空框。
1 回答
四季花海
TA貢獻1811條經驗 獲得超5個贊
您應該將您的php代碼包裝為isset函數，該函數將檢查是否已發送數據，然后運行，否則不運行。
isset函數檢查是否設置了變量且不為null
或者，您可以調用Ajax來添加點擊數據。這里
<?php
$con=mysqli_connect("localhost","root","","school");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_POST['submit'])){
$result = mysqli_query($con,"SELECT description From task ORDER BY RAND() LIMIT 2");
echo "<p><strong>Question:</strong></p>";
while($row = mysqli_fetch_array($result))
{
echo "<p>". $row['description'] ."</p>";
}
mysqli_close($con);
}
?>
<input type="button" name="submit" value="Get Assignment" onclick="myfunc()">
</body>
</html>
編輯 固定我的代碼
您需要將輸入按鈕包裝到一個表單中，該表單會將表單提交到頁面本身
<form method="post" action="">
<input type="submit" name="submit" value="Get Assignment">
</form>
- 1 回答
- 0 關注
- 181 瀏覽
添加回答
舉報
0/150
提交
取消