即使成功提交，值也不會從php表單插入MySQL表單

Question

我是PHP和MySQL的新手。 我創建了一個php表單來將一些數據保存到databse表中。 提交表格后，

“添加了1條記錄”

如php插入字符串中所指定。 但是在phpmyadmin中打開表后，它顯示添加了新行，但其中沒有數據。

php形式為：

<h2>Register Yourself</h2>
            <form method="post" action="get-member.php">
                <div>
                    <span><p>First Name:</p></span>
                    <span><input type="text" class="form-control" id="FirstName"></span>
                </div>
                <div>
                    <span>Last Name:</span>
                    <span><input type="text" class="form-control" id="LastName"></span>
                </div>
                <div>
                    <span>Father's Name: </span>
                    <span><input type="text" class="form-control" id="FatherName"></span>
                </div>
                <div>       
                    <span>Mother's Name: </span>
                    <span><input type="text" class="form-control" id="MotherName"></span>
                </div>
                <div>           
                    <span>Date of Birth: </span>
                    <span><input type="date" class="form-control" id="DOB"></span>
                </div>
                <div>
                    <span>Address: </span>
                    <span><input type="text" class="form-control" id="Address"></span>
                </div>
                <div>       
                    <span>City: </span>
                    <span><input type="text" class="form-control" id="City"></span>
                </div>
                <div>   
                    <span>District: </span>
                    <span><input type="text" class="form-control" id="District"></span>
                </div>
                <div>       
                    <span>Postal Code: </span>
                    <span><input type="text" class="form-control" id="PostalCode"></span>
                </div>
                <div>
                    <span>Personal Mobile #:</span>
                    <span><input type="number" class="form-control" id="Pmobile"></span>
                </div>
                <div>   
                    <span>Father's Contact #:</span>
                    <span><input type="number" class="form-control" id="Fmobile"></span>
                </div>
                <div>
                    <span>Mother's Contct #:</span>
                    <span><input type="number" class="form-control" id="Mmobile"></span>
                </div>
                <div>
                    <span>Home contact #:</span>
                    <span><input type="number" class="form-control" id="Hmobile"></span>
                </div>
                <div>
                    <span><input type="submit" value="Register"></span>
                </div>
            </form>

而get-member.php文件是：

<?php


//stablising connection
include("../database/connection.php"); 

//escape variables for security
$FirstName = mysqli_real_escape_string($con, $_POST['FirstName']);
$LastName = mysqli_real_escape_string($con, $_POST['LastName']);
$FatherName = mysqli_real_escape_string($con, $_POST['FatherName']);
$MotherName = mysqli_real_escape_string($con, $_POST['MotherName']);
$DOB = mysqli_real_escape_string($con, $_POST['DOB']);
$Address = mysqli_real_escape_string($con, $_POST['Address']);
$City = mysqli_real_escape_string($con, $_POST['City']);
$District = mysqli_real_escape_string($con, $_POST['District']);
$PostalCode = mysqli_real_escape_string($con, $_POST['PostalCode']);
$Pmobile = mysqli_real_escape_string($con, $_POST['Pmobile']);
$Fmobile = mysqli_real_escape_string($con, $_POST['Fmobile']);
$Mmobile = mysqli_real_escape_string($con, $_POST['Mmobile']);
$Hmobile = mysqli_real_escape_string($con, $_POST['Hmobile']);




$sql="INSERT INTO RUPmembers (FirstName, LastName, FatherName, MotherName, DOB, Address, City, District, PostalCode, Pmobile, Fmobile, Mmobile, Hmobile)

VALUES ('$FirstName','$LastName','$FatherName','$MotherName','$DOB','$Address','$City','$District','$PostalCode','$Pmobile','$Fmobile','$Mmobile','$Hmobile')";

if (!mysqli_query($con,$sql))
{
    die ('Error:' . mysqli_error($con));
}
echo "1 Record addedd";




//closing connection
include("../database/close-connection.php");

?>

插入3個數據后的phpmyadmin表：

[ 對不起，它要求獲得10個回購才能添加圖像，但這是截圖的鏈接http://oi60.tinypic.com/n6rtyu.jpg ]

Answer 1

您正在使用id =“ FirstName”

您需要為每個字段設置屬性name 。 例如：

name="FirstName"

要調試它，您應該使用var_dump($_POST)查看$_POST變量，以顯示正在從表單傳遞的內容。

注意：應該保護查詢免受從用戶收到的任何數據的影響，以防止sql注入。 請參閱這篇文章，以了解如何避免這種情況。

Answer 2

更改

輸入type =“ text” class =“ form-control” id =“ LastName”>

至

輸入type =“ text” class =“ form-control” name =“ LastName”>

這應該可以解決您的問題，但是請不要將用戶未處理的信息插入查詢中。

Answer 3

要通過php從html表單捕獲數據，應在$ _post ['']方法內指定名稱，而不是id。 使用name="FirstName" name="LastName" name="DOB"等。

它將解決問題。