我是 PHP 新手，所以這可能是我錯過的明顯事情。我試圖制作一個按鈕來增加數據庫中的值：<?php $alist = mysqli_query($conn, "SELECT * FROM `posts` ORDER BY `posts`.`id` DESC"); $results = mysqli_num_rows($alist); if ($results > 0){ while($row = mysqli_fetch_array($alist)) { echo $row['uid']. " says: ".$row['postText']." <button onclick=".mysqli_query($conn, "UPDATE `posts` SET `postLikes` = postLikes+1 WHERE uid = ".$row['uid'])." name='likebtn'>??</button>".$row['postLikes']."<br>"; } } ?>制作按鈕的部分位于第 6 行，我只是想找到如何使用按鈕 mysqli_query單擊， 我已經嘗試過：“https://stackoverflow.com/questions/3862462/php-mysql-run -query-on-button-press-click”但沒有結果提前致謝
2 回答
ABOUTYOU
TA貢獻1812條經驗 獲得超5個贊
如果您當前的頁面名為first.php，請放入按鈕內
<button><a href="first.php?p=like"></a></button>
<?php
if ( isset($_GET['p']) && $_GET['p']=="like") {
do your query
}
?>
如果你不想重新加載那么你需要ajax，
慕妹3242003
TA貢獻1824條經驗 獲得超6個贊
你這里有語法錯誤 echo $row['uid']. " says: ".$row['postText']." <button onclick=".mysqli_query($conn, "UPDATE posts SET postLikes = postLikes+1 WHERE uid = ".$row['uid'])." name='likebtn'>??</button>".$row['postLikes']."<br>";
你可以這樣做
$q = mysqli_query($conn, "UPDATE posts SET postLikes = postLikes+1 WHERE uid = ".$row['uid']);
echo $row['uid']. " says: ".$row['postText']." <button onclick=".$q." name='likebtn'>??</button>".$row['postLikes']."<br>";
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