<?php
session_start();
$host = "localhost";
$user = "root";
$password = "";
$database = "tuition";
$conn = mysqli_connect($host, $user, $password, $database);
?>
<div role="tabpanel" class="tab-pane fade in" id="sign_up">
<form action="" method="post">
<input type="text" class="form-control" name="register_name" placeholder="Name" required />
<input type="text" class="form-control" name="register_ic" placeholder="IC" required />
<input type="telephone" class="form-control" name="register_phone" placeholder="Telephone" required />
<input type="email" class="form-control" name="register_email" placeholder="Email" required />
<input type="password" class="form-control" name="register_password" placeholder="Password" required />
<input type="submit" class="form-control" name="register_btn" value="Sign up" />
</form>
</div>
<?php
if(isset($_POST['register_btn'])){
$name = $_POST['register_name'];
$ic = $_POST['register_ic'];
$phone = $_POST['register_phone'];
$email = $_POST['register_email'];
$password = md5($_POST['register_password']);
$result = mysqli_query($conn,"select parent.parent_ic from parent where parent_ic ='$ic'");
if(mysqli_num_rows($result)!=1){
mysqli_query($conn,"insert into parent(parent_name,parent_ic,parent_email,parent_contact_num,parent_password) values ('$name','$ic','$email','$phone','$password')");
?>
<script type="text/javascript">
alert("success");
</script>
<?php
}
else{
?>
<script type="text/javascript">
alert("fail");
</script>
<?php
}
}
?>
它成功发出警报,但是没有将数据插入数据库。但是当我用硬代码将数据插入数据库时,它就可以工作了!为什么?
这是我单击“提交”按钮后的结果:
答案 0 :(得分:0)
插入数据查询后,您应该会看到打印错误
echo(“错误说明:”。mysqli_error($ conn));
我认为您的MySQL错误在那里。