无法通过PHP将新数据插入数据库
[英]Unable to insert new data into database via PHP
问题描述
I am unable to store new data into the database after the user submits the form. 
用户提交表单后，我无法将新数据存储到数据库中。 I check the database table via phpMyAdmin, but I do not see any new data. 我通过phpMyAdmin检查数据库表，但是没有看到任何新数据。 Can anyone show me the mistake in my code. 谁能告诉我我代码中的错误。
HTML code (listing.html): HTML代码 （listing.html）：
<html>
<form action="confirm_listing.php" method="POST">
<input name="Firstname" placeholder="Firstname">
<input name="Lastname" placeholder="Lastname">
...
<input type="submit" value="Submit">
</form>
</html>
php code (confirm_listing.php): php代码 （confirm_listing.php）：
<?php
$firstname = $_POST['Firstname'];
$lastname = $_POST['Lastname'];
$address = $_POST['Address'];
$suburb = $_POST['Suburb'];
$city = $_POST['City'];
$type = $_POST['Type'];
$room = $_POST['Room'];
$floorArea = $_POST['Floor'];
$landArea = $_POST['Land'];
$price = $_POST['Price'];
$contact = $_POST['Contact'];
$active = "1";
$server = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "summitrealestatedb";
$connection = mysqli_connect($server, $dbusername, $dbpassword, $dbname);
if (!$connection) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "INSERT INTO property (firstname, lastname, address, suburb, city, type, room, floor, land, price, contact, active) VALUES ('$firstname', '$lastname', '$address', '$suburb', '$city', '$type', '$room', '$floorArea', '$landArea', '$price', '$contact', '$active');";
mysqli_query($connection, $sql);
mysqli_close($connection);
?>
Here is a picture of what it currently looks like in the database table: 这是数据库表中当前外观的图片 ：
2 个解决方案
解决方案1
1 2015-04-27 10:57:18
instead of 代替
<?php
$firstname = $_POST['Firstname'];
$lastname = $_POST['Lastname'];
?>
use 采用
<?php
$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
?>
use the name that you have assigned to the name not to the placeholder 使用您分配给名称的名称而不是占位符
解决方案2
1 已采纳 2015-04-27 10:57:40
尝试打印插入错误
mysqli_query($connection, $sql) or die(mysqli_error($connection));
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