我必須創建一個表單,將答案發送到數據庫,但是當我填寫表單時數據庫沒有更新,並且我收到了自制錯誤:「出錯了」。任何人都可以看到任何錯謝謝。不插入數據庫。 PHP和mysql
形式:
<form id="contact-form" method="post" action="sentEnquiries.php" name="enquiries">
<div class="row">
<div class="col-md-6">
<div class="form-group">
<label for="name">
Name</label>
<div class="input-group">
<span class="input-group-addon"><span class="glyphicon glyphicon-user"></span>
</span>
<input type="text" class="form-control" id="name" name="name" placeholder="Enter name" required="required" /></div>
</div>
<div class="form-group">
<label for="email">
Email Address</label>
<div class="input-group">
<span class="input-group-addon"><span class="glyphicon glyphicon-envelope"></span>
</span>
<input type="email" class="form-control" id="email" name="email" placeholder="Enter email" required="required" /></div>
</div>
<div class="form-group">
<label for="phoneNumber">
Phone Number</label>
<div class="input-group">
<span class="input-group-addon"><span class="glyphicon glyphicon-earphone"></span>
</span>
<input type="tel" class="form-control" id="phoneNumber" name="phone" placeholder="Enter phone number" required="required" /></div>
</div>
<div class="form-group">
<label for="partySize">
Party Size</label>
<input type="number" min="1" max="6" class="form-control" id="partySize" name="partySize" required="required" />
</div>
</div>
<div class="col-md-6">
<div class="form-group">
<label for="arrivalDate">
Arrival Date</label>
<input type="date" class="form-control" id="arrivalDate" name="arrivalDate" />
</div>
<div class="form-group">
<label for="departureDate">
Departure Date</label>
<input type="date" class="form-control" id="departureDate" name="departureDate"/>
</div>
<div class="form-group">
<label for="name">
Message</label>
<textarea name="message" id="message" class="form-control" rows="9" cols="25" required="required"
placeholder="Message"></textarea>
</div>
</div>
<div class="col-md-12">
<button type="submit" class="btn btn-skin pull-right" id="btnContactUs">
Send enquiry</button>
</div>
</div>
</form>
答案:
<?php
include("conn.php");
$sentName = $_POST['name'];
$sentEmail = $_POST['email'];
$sentPhone = $_POST['phone'];
$sentPartySize = $_POST['partySize'];
$sentArrivalDate = $_POST['arrivalDate'];
$sentDepartureDate = $_POST['departureDate'];
$sentMessage = $_POST['message'];
$insertQuery = "INSERT INTO Enquiries(enquiryID, name, email, phone, partySize, arrivalDate, departureDate, message) VALUES(NULL, '$sentName', '$sentEmail', '$sentPhone, '$sentPartySize', $sentArrivalDate, '$sentDepartureDate', '$sentMessage')";
?>
再往下答案DOC:
<div class="descriptions">
<?php
if(mysqli_query($conn, $insertQuery)) {
echo "<p>Thank you for your enquiry.</p>";
mysqli_close($conn);
} else {
echo "<p>Something went wrong.</p>";
mysqli_close($conn);
}
?>
</div>
還沒有找錯誤運行查詢之前添加的條件,但是你的應用程序將不會被保存到SQL注入。使用準備好的語句使它們保存;) – Twinfriends
你不能在插入值上使用'name'將其改爲別的。並且該ID應該是自動增量,所以也刪除這些。 – Option
@選項哪個值?在帖子裏面? – monkey232