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我最近做了一個代碼,在我的博客上更新我的帖子。它在localhost上完美運行。但是當我上傳它時,它不再工作。奇怪的是它甚至不顯示錯誤,所以我不知道在哪裏看。有人可以幫幫我嗎 ?更新帖子不工作
require('config.php');
$query = "SELECT * FROM project ORDER BY idproject DESC";
$result = mysqli_query($verbinding, $query) or die (mysqli_error('kan geen verbinding maken met de database'));
if(isset($_POST['editBut'])){
$editTitle = $_POST['editName'];
$editThis = mysqli_query($verbinding, "SELECT * FROM project WHERE title = '".$editTitle."'");
$values = mysqli_fetch_assoc($editThis);
}
if(isset($_POST['update'])){
$editedTitle = $_POST['newTitle'];
$editedText = $_POST['newTekst'];
$oldTitle = $_POST['oldTitle'];
$date = $_POST['datum'];
$updater = mysqli_query($verbinding, "UPDATE Project SET title='".$editedTitle."', content='".$editedText."' WHERE title='".$oldTitle."' AND datum='".$date."'");
echo $updater;
header('location:editPost.php?id=1');
}
if(isset($_GET['id'])){
echo 'post has been succesfully updated';
}
<?php if(isset($_POST['editBut'])){ ?>
<form action="" method="post">
Title: <input type="text" name="newTitle" value="<?php echo $values['title'] ?>"><br>
Text: <textarea type="text" name="newTekst" id="newTekst"><?php echo $values['content'] ?></textarea><br>
<input type="hidden" value="<?php echo $values['title'] ?>" name="oldTitle">
<input type="hidden" value="<?php echo $values['datum'] ?>" name="datum">
<input type="submit" name="update" value="Edit post">
</form>
<?php } else { ?>
<p>Find the post you want to edit:</p>
<form action="" method="post">
<select name="editName">
<?php
while ($row = mysqli_fetch_assoc($result)) {
?> <option value="<?php echo $row['title'] ?>"><?php echo $row['title'] ?></option>
<?php } ?>
</select>
<input type="submit" name="editBut" value="Choose">
</form>
<?php } ?>
是隻更新不工作更換
Project
??? – Deep123由於安全原因,您的網頁未顯示錯誤。它很可能會被關閉。要啓用它,請在頁面頂部添加'error_reporting(E_ALL);'(JUST用於測試目的)。 **不要忘記在發現錯誤後將其刪除!! ** – Jules