该问题已得到简要解决，现在数据库为信息添加了一个新行，我唯一的问题是数据库没有从表单中获取任何信息。
<?php
$con = mysql_connect("localhost","username","password");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("databasename", $con);
$sql="INSERT INTO breakdowns (username, breakdowndate, policyinception, customername, customersurname, covertype, vehiclemake, vehiclemodel, vehiclereg, vehicleage, excess, mileage, paid, HSRS, fault, garage, telephone)
VALUES
('$_POST[username]','$_POST[breakdowndate]','$_POST[policyinception]','$_POST[customername]','$_POST[customersurname]','$_POST[covertype]','$_POST[vehiclemake]','$_POST[vehiclemodel]','$_POST[vehiclereg]','$_POST[vehicleage]','$_POST[excess]','$_POST[mileage]','$_POST[paid]','$_POST[HSRS]','$_POST[fault]','$_POST[garage]','$_POST[telephone]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "1 breakdown added";
mysql_close($con)
?>
当在服务器上执行脚本时，我收到回显消息，数据库中出现一个空白行，但不存在表单中的信息。
现在正在运作：
<?php
// This sends the breakdown form to the database //
$con = mysql_connect("localhost","dgtlss20_orca1","qG!P}ubT.tuw");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("dgtlss20_orca", $con);
$sql="INSERT INTO breakdowns (username, breakdowndate, policyinception, customername, customersurname, covertype, vehiclemake, vehiclemodel, vehiclereg, vehicleage, excess, mileage, paid, HSRS, fault, garage, telephone)
VALUES ('$_POST[username]','$_POST[breakdowndate]','$_POST[policyinception]','$_POST[customername]','$_POST[customersurname]','$_POST[covertype]','$_POST[vehiclemake]','$_POST[vehiclemodel]','$_POST[vehiclereg]','$_POST[vehicleage]','$_POST[excess]','$_POST[mileage]','$_POST[paid]','$_POST[HSRS]','$_POST[fault]','$_POST[garage]','$_POST[telephone]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
header('Location: ../breakdown-added.php');
mysql_close($con)
?>
答案 0 :(得分：0)
您的mysqli_query函数未正确写入：
您有mysqli_query(connection, "query"后跟一些应该在查询中的代码，并且无法识别它应该是，
而不是mysqli_query(connection, "query");
mysqli_query($connect,"INSERT INTO posts (user, breakdowndate, policyinception, customername, customersurname, covertype, vehiclemake, vehiclemodel, vehiclereg, vehicleage, excess, mileage, paid, HSRS, fault, garage, telephone) VALUES ('".$_POST[post_user]."', '".$_POST[post_breakdowndate]."', '".$_POST[post_policyinception]."', '".$_POST[post_customername]."', '".$_POST[post_customersurname]."', '".$_POST[post_covertype]."', '".$_POST[post_vehiclemake]."', '".$_POST[post_vehiclemodel]."', '".$_POST[post_vehiclereg]."', '".$_POST[post_vehicleage]."', '".$_POST[post_excess]."', '".$_POST[post_mileage]."', '".$_POST[post_paid]."', '".$_POST[post_HSRS]."', '".$_POST[post_fault]."', '".$_POST[post_garage]."', '".$_POST[post_telephone]."')");
答案 1 :(得分：0)
您在查询以值结束之前终止了查询。试试 -
mysqli_query($connect,"INSERT INTO posts (user, breakdowndate, policyinception, customername, customersurname, covertype, vehiclemake, vehiclemodel, vehiclereg, vehicleage, excess, mileage, paid, HSRS, fault, garage, telephone) VALUES ('$_POST[post_user]', '$_POST[post_breakdowndate]', '$_POST[post_policyinception]', '$_POST[post_customername]', '$_POST[post_customersurname]', '$_POST[post_covertype]', '$_POST[post_vehiclemake]', '$_POST[post_vehiclemodel]', '$_POST[post_vehiclereg]', '$_POST[post_vehicleage]', '$_POST[post_excess]', '$_POST[post_mileage]', '$_POST[post_paid]', '$_POST[post_HSRS]', '$_POST[post_fault]', '$_POST[post_garage]', '$_POST[post_telephone]')");
答案 2 :(得分：0)
如果您的列数和值计数在查询中相同，则以下代码将起作用：
<?php
//Connecting to sql db.
$connect = mysqli_connect("localhost","username","password","databasename");
//Sending form data to sql db.
mysqli_query($connect,"INSERT INTO posts (user, breakdowndate, policyinception, customername, customersurname, covertype, vehiclemake, vehiclemodel, vehiclereg, vehicleage, excess, mileage, paid, HSRS, fault, garage, telephone)".
"VALUES ('$_POST[post_user]', '$_POST[post_breakdowndate]', '$_POST[post_policyinception]', '$_POST[post_customername]', '$_POST[post_customersurname]', '$_POST[post_covertype]', '$_POST[post_vehiclemake]', '$_POST[post_vehiclemodel]', '$_POST[post_vehiclereg]', '$_POST[post_vehicleage]', '$_POST[post_excess]', '$_POST[post_mileage]', '$_POST[post_paid]', '$_POST[post_HSRS]', '$_POST[post_fault]', '$_POST[post_garage]', '$_POST[post_telephone]')");
?>
答案 3 :(得分：0)
您的代码容易受到SQL注入攻击
您需要将参数传递为$_POST['post_breakdowndate']而不是$_POST[post_breakdowndate]
您需要将查询更新为
mysqli_query($connect,"INSERT INTO posts (user, breakdowndate, policyinception, customername, customersurname, covertype, vehiclemake, vehiclemodel, vehiclereg, vehicleage, excess, mileage, paid, HSRS, fault, garage, telephone) VALUES ('".$_POST['post_user']."', '".$_POST['post_breakdowndate']."', '".$_POST['post_policyinception']."', '".$_POST['post_customername']."', '".$_POST['post_customersurname']."', '".$_POST['post_covertype']."', '".$_POST['post_vehiclemake']."', '".$_POST['post_vehiclemodel']."', '".$_POST['post_vehiclereg']."', '".$_POST['post_vehicleage']."', '".$_POST['post_excess']."', '".$_POST['post_mileage']."', '".$_POST['post_paid']."', '".$_POST['post_HSRS']."', '".$_POST['post_fault']."', '".$_POST['post_garage']."', '".$_POST['post_telephone']."')");
您正在终止函数
之外的值