我正在研究html和php表单,它将用户信息存储在mysql表中。 我无法将数据发送到mysql数据库。当我提交表单时,它不会返回任何错误,但同时它不反映对表的任何更改。 这是我的HTML代码。
<form method="post" action="datapost.php">
<div class="col-xl-12">
<div class="row">
<div class="col-sm-6 form-group">
<label>First Name</label>
<input type="text" placeholder="Enter First Name Here.." name="First_name" class="form-control">
</div>
<div class="col-sm-6 form-group">
<label>Last Name</label>
<input type="text" placeholder="Enter Last Name Here.." name="Last_Name" class="form-control">
</div>
</div>
<div class="form-group">
<label>Address</label>
<textarea placeholder="Enter Address Here.." rows="3" name="Address" class="form-control"></textarea>
</div>
<div class="row">
<div class="col-sm-4 form-group">
<label>City</label>
<input type="text" placeholder="Enter City Name Here.." name="City" class="form-control">
</div>
<div class="col-sm-4 form-group">
<label>State</label>
<input type="text" placeholder="Enter State Name Here.." class="form-control">
</div>
<div class="col-sm-4 form-group">
<label>Zip</label>
<input type="text" placeholder="Enter Zip Code Here.." name="Zip" class="form-control">
</div>
</div>
<div class="form-group">
<label>Phone Number</label>
<input type="text" placeholder="Enter Phone Number Here.." name="Phone_number"class="form-control">
</div>
<div class="form-group">
<label>Email Address</label>
<input type="text" placeholder="Enter Email Address Here.." name="Email" class="form-control">
</div>
<div class="panel panel-default credit-card-box">
<div class="panel-heading display-table" >
<div class="row display-tr" >
<h3 class="panel-title display-td" >Payment Details</h3>
<div class="display-td" >
<img class="img-responsive pull-right" src="http://i76.imgup.net/accepted_c22e0.png">
</div>
</div>
</div>
<div class="panel-body">
<form role="form" id="payment-form" method="POST" action="javascript:void(0);">
<div class="row">
<div class="col-xs-12">
<div class="form-group">
<label for="cardNumber">CARD NUMBER</label>
<div class="input-group">
<input
type="tel"
class="form-control"
name="cardNumber"
placeholder="Valid Card Number"
autocomplete="cc-number"
required autofocus
/>
<span class="input-group-addon"><i class="fa fa-credit-card"></i></span>
</div>
</div>
</div>
</div>
<div class="row">
<div class="col-xs-7 col-md-7">
<div class="form-group">
<label for="cardExpiry"><span class="hidden-xs">EXPIRATION</span><span class="visible-xs-inline">EXP</span> DATE</label>
<input
type="tel"
class="form-control"
name="cardExpiry"
placeholder="MM / YY"
autocomplete="cc-exp"
required
/>
</div>
</div>
<div class="col-xs-5 col-md-5 pull-right">
<div class="form-group">
<label for="cardCVC">CV CODE</label>
<input
type="tel"
class="form-control"
name="cardCVC"
placeholder="CVC"
autocomplete="cc-csc"
required
/>
</div>
</div>
</div>
<!-- <div class="row">
<div class="col-xs-12">
<button class="subscribe btn btn-success btn-lg btn-block" type="button">Start Subscription</button>
</div>
</div>-->
<button type="submit" class="btn btn-lg btn-info">Submit</button>
<div class="col-xs-12 col-md-4">
<div class="row" style="display:none;">
<div class="col-xs-12">
<p class="payment-errors"></p>
</div>
</div>
</form>
</div>
</div>
</div>
我的PHP代码在这里
<?php
include 'databaseconn.php';
$First_name=$_POST['First_name'];
$Last_Name=$_POST['Last_Name'];
$Address=$_POST['Address'];
$City=$_POST['City'];
$Zip=$_POST['Zip'];
$Phone_number=$_POST['Phone_number'];
$Email=$_POST['Email'];
$cardNumber=$_POST['cardNumber'];
$cardExpiry=$_POST['cardExpiry'];
$cardCVC=$_POST['cardCVC'];
mysqli_query($connect, "INSERT INTO user_data (`First_name`,`Last_Name`,`Address`,`City`,`Zip`,`Phone_number`,`Email`,`cardNumber`,`cardExpiry`,`cardCVC`) values ('$First_name','$Last_Name','$Address','$City','$Zip','$Phone_number','$Email','$cardNumber','$cardExpiry','$cardCVC') ");
if(mysqli_affected_rows($connect)>0)
{
echo'<p> User successfully registered </p>';
echo'<a href="index.php"> Go BAck </a>';
}
else
{
echo 'Registration not successfull';
echo 'mysqli_error($connect)';
}
?>
我的数据库文件在这里。
<?php
$servername = "localhost";
$username = "root";
$password = "";
// Create connection
$connect = mysqli_connect($servername, $username, $password);
// Check connection
if (!$connect) {
die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully";
?>
please someone help me out.
答案 0 :(得分:-1)
您需要添加数据库名称
$servername = "localhost";
$username = "root";
$password = "";
$database = 'databasename';
// Create connection
$connect = mysqli_connect($servername, $username, $password, $database);