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這是一場真正的鬥爭。我一直試圖調試1小時的更好的一部分,但我找不到任何解決方案。我正在製作一個與博客帖子關聯的評論系統。當您發表評論時,您將返回到相同的帖子。帖子頁面根據postid動態填充。評論是使用相同的postid添加的,所以他們只填寫他們的相關帖子。我查詢評論是錯誤的,但我沒有得到如何。如何將評論與postid與該郵件的帖子綁定?
Warning
: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given
in C:\xampp\htdocs\prog1\comments.php on line 31
代碼:
<?php
require("connect.php");
$postid = $_GET['id'];
if(isset($_POST['postcomment'])){
$comment = $_POST['comment'];
$author = $_SESSION['username'];
$commentQuery = mysqli_query($conn, "INSERT INTO comments (com_content, com_timestamp, com_author, com_postid) VALUES ('$comment', now(), '$author', '$postid')");
if($commentQuery){
header('Location: postpage.php?id='.$postid);
}
}
?>
<div class="comments-separator">
</div>
<div class="flex-enable comments-wrapper flex-column">
<span class="comment-header">Comments</span>
<div class="comment-new-wrapper">
<form class="flex-enable flex-column" method="POST">
<textarea class="blog-input-text comment-entry" cols="40" rows="3" name="comment"></textarea>
<input class="comment-button small-white-subtitle" type="submit" name="postcomment" value="Post Comment">
</form>
</div>
</div>
<?php
$summonComm = "SELECT * FROM comments WHERE com_postid='".$postid."' ORDER BY id DESC";
$resultComm = mysqli_query($conn, $summonComm);
while ($row = mysqli_fetch_array($resultComm)){
$comment= $row['com_content'];
$timestamp= $row['timestamp'];
$author= $row['com_author'];
?>
<span class="blog-timestamp"><?php echo $timestamp; ?> • Written by <a href="<?php echo $author; ?>.php"><?php echo $author; ?></a></span>
<span class="blog-entry"><?php echo $comment; ?></span>
</div>
<?php
}
?>
那麼它在做什麼錯? – RiggsFolly
_Small point_你需要和'exit;'後面的'header()'作爲頭不停止執行當前腳本,它只是向瀏覽器發送一個頭文件 – RiggsFolly
新增了我的錯誤。 – mechanicarts