解析错误：语法错误，C：\xampp\htdocs\form1 中意外的"$fname"(T_VARIABLE.php第 20 行

<html>
<head>
    <title>Purchaser Info</title></head>
<body>
<?php include 'connect.php';
if (isset($_POST['submit'])) {
    $fname   = $_POST['fname'];
    $mid     = $_POST['mname'];
    $last    = $_POST['lname'];
    $add     = $_POST['add'];
    $mail    = $_POST['email'];
    $contact = $_POST['cont'];
    if (mysqli_connect_errno()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    mysqli_query($con, ("INSERT INTO tblpurchaser (fname, mi, lname, contact, email, homeadd)
            VALUES ( "$fname", "$mid", "$last", "$add", "$mail", "$contact")"));
    echo "Your record was added";
    mysqli_close($con); }?> </body>
</html>

注意：我不是一个好的程序员。无论如何。我前段时间一直在研究这个问题。搜索了很多，但我仍然找不到有什么问题。如果有人愿意/可以纠正我，我将不胜感激。嘿嘿嘿提前谢谢！

更改此设置

mysqli_query($con, ("INSERT INTO tblpurchaser (fname, mi, lname, contact, email, homeadd)
            VALUES ( "$fname", "$mid", "$last", "$add", "$mail", "$contact")"));

自

mysqli_query($con, ("INSERT INTO tblpurchaser (fname, mi, lname, contact, email, homeadd)
            VALUES (  "'.$fname.'",  "'.$mid.'", "'.$last.'", "'.$add.'", "'.$mail.'", "'.$contact.'")"));

你可以试试这个

   $con = mysqli_connect("localhost","USERNAME","PASSWORD","DB") or die("Error " . mysqli_error($con));

if (isset($_POST['submit'])) {
    $fname = mysqli_escape_string($con, $_POST['fname']);
    $mid = mysqli_escape_string($con, $_POST['mname']);
    $last = mysqli_escape_string($con, $_POST['lname']);
    $add = mysqli_escape_string($con, $_POST['add']);
    $mail = mysqli_escape_string($con, $_POST['email']);
    $contact = mysqli_escape_string($con, $_POST['cont']);
    $query = mysqli_query($con, ("INSERT INTO tblpurchaser (fname, mi, lname, contact, email, homeadd)
    VALUES ( '$fname', '$mid', '$last', '$add', '$mail', '$contact')"));
    if($query){
        echo "Your record was added";
    }
    else{
        echo  'Query Error: '.mysqli_error($con);
    }
    mysqli_close($con);
}

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