2016-11-23 69 views
0

我有一個表單,用戶可以將其聯繫信息(名字,姓氏,電子郵件,電話,網站,評論,託管)提交到數據庫。表單工作正常,直到我嘗試將HTML中的'comment'名稱更改爲'description'。現在,當我提交表單時,php迴應出了響應(因此導致我認爲它正確提交),但是當我在phpMYAdmin中查看錶時,它是空的。SQL查詢正在運行,但隨機停止

數據庫結構是astyle_lefteyebrow>聯繫

表結構:

CustomerID (Primary) int(11)    
FirstName text    
LastName text    
Email  varchar (128)   
Phone  varchar (20) NULL   
Website  varchar (500) NULL   
Description varchar (2000)   
Hosting  tinyint (1) 

PHP文件:

<?php 

$servername = "localhost"; 
$username = "astyle_quiggly"; 
$password = "**********"; 
$dbname = "astyle_lefteyebrow"; 

// Create connection 
$conn = new mysqli($servername, $username, $password, $dbname); 

// Check connection 
if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 

// Check that posts have values 
echo $_POST['first_name']; 
echo $_POST['last_name']; 
echo $_POST['email']; 
echo $_POST['phone']; 
echo $_POST['website']; 
echo $_POST['comment']; 
echo $_POST['hosting']; 

// prepare and bind 
$stmt = $conn->prepare("INSERT INTO Contact (FirstName, LastName, Email, Phone, Website, Description, Hosting) VALUES (?, ?, ?, ?, ?, ?, ?)"); 
$stmt->bind_param("ssssssi", $firstname, $lastname, $email, $phone, $website, $description, $hosting); 


// set parameters and execute 
$firstname = $_POST['first_name']; 
$lastname = $_POST['last_name']; 
$email = $_POST['email']; 
$phone = $_POST['phone']; 
$website = $_POST['website']; 
$description = $_Post['comment']; 
if ($_POST['hosting'] == 'yes') { 
$hosting = 1;} 
else { 
$hosting = 0; 
} 
$stmt->execute(); 


$stmt->close(); 
$conn->close(); 
?> 

我不知道爲什麼它會被工作,然後所有的突然除非它必須與編碼錯誤有關。有任何想法嗎?

+0

爲什麼你假設你的查詢成功執行('$ stmt-> execute();')?它返回一個布爾值的原因。 – Xorifelse

回答

1

它應該是:

$description = $_POST['comment']; 

而不是:

$description = $_Post['comment']; 

功能不區分大小寫,但變量不是。

+0

謝謝我甚至沒有看到! – strasbal