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我目前陷入困境,因为我可以成功地更新我的查询,当我输入一个值然后提交,但最终的结果是输出“消息”记录更新成功“甚至在我点击提交按钮之前添加,我尝试添加一个计数器,只有在达到某个特定配额时才回复成功消息,但它无济于事,请帮助!非常感谢。如何在php中插入sql查询时返回正确的消息?
我想要做的只是输出“记录成功更新”我点击提交按钮后,没有出现直接的消息甚至在我点击提交按钮。
HTML代码
<div class = "form-group">
Book Title
<input type="text" class="form-control" id="bookTitle" name="bookTitle"
value= "<?php if (isset($_POST['bookTitle'])) echo($_POST['bookTitle']); ?>"
placeholder="" required>
PHP代码
<?php
$sql = "SELECT * FROM `allbooks` WHERE `allbooks`.`id` = $id ";
if ($result = mysqli_query($connection, $sql)) {
while ($row = mysqli_fetch_assoc($result)) {
if ($connection) {
if (isset($_POST['bookTitle'])) {
$bookTitle = mysql_real_escape_string($_POST['bookTitle']);
$_SESSION['bookTitle'] = $bookTitle;
}
}
$sql = "UPDATE `allbooks` SET `allbooks`.bookTitle = '$bookTitle', WHERE `allbooks`.id = $id";
if (mysqli_query($connection, $sql)) {
echo "Record updated successfully";
} else {
echo "Please enter the correct values for the boxes!";
}
}
}
?>