Question

我是PHP和MySQL的新手。我创建了一个php表单来保存一些数据到数据库表。提交表格后，它会给出

“1记录添加”

在php插入字符串中指定。但是在phpmyadmin中打开表后，它显示添加了一行但没有数据..

php表单是：

<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>

并且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个数据后的phpmyadmin表：

[抱歉它要求获得10个回复才能添加图片，但这里是截屏链接http://oi60.tinypic.com/n6rtyu.jpg ]

Answer 1

您正在使用id =“FirstName”

您需要为每个字段设置属性name。例如：

name="FirstName"

要对此进行调试，您应该使用$_POST查看var_dump($_POST)变量，以显示表单中传递的内容。

注意：应保护查询不受从用户收到的任何数据的影响，以防止sql注入。请参阅此post以了解如何避免这种情况。

Answer 2

更改

输入类型=＆＃34;文字＆＃34;类=＆＃34;形状控制＆＃34; ID =＆＃34;名字＆＃34;＆GT;

到

输入类型=＆＃34;文字＆＃34;类=＆＃34;形状控制＆＃34;名称=＆＃34;名字＆＃34;＆GT;

这可以解决您的问题，但不要将用户未处理的信息插入您的查询。

Answer 3

要通过php从html表单中捕获数据，应该在$ _post [＆＃39;＆＃39;]方法中给出名称而不是id。使用name="FirstName" name="LastName" name="DOB" 等。

它将解决问题..

即使在成功提交之后，值也不会从php表单中以MySQL形式插入

3 个答案: