如何通过更新代码来防止SQL注入

Question

This code is vulnerable to SQL injection. How can I improve it to prevent injections ? And How can I test this code with injections ? This is my code: if (isset ($_POST['name'])) { $name = $_POST['name']; }

if (isset ($_POST['name'])) { 
$name = $_POST['name']; 
} 
if (isset ($_POST['remarcs'])) { 
$remarcs = $_POST['remarcs']; 
} 
if (isset ($_POST['test_res'])) { 
$test_res = $_POST['test_res']; 
} 
if (isset ($_POST['address'])) { 
$address = $_POST['address']; 
} 

if (isset ($_POST['date'])) { 
$date = $_POST['date']; 
} 

if (isset ($_POST['phone_num'])) { 
$phone = $_POST['phone_num']; 
}

if (isset ($_POST['illness'])) { 
$illness = $_POST['illness']; 
} 
if (isset ($_POST['echo'])) { 
$echo = $_POST['echo']; 
} 
if (isset ($_POST['pe'])) { 
$pe = $_POST['pe']; 
} 
if (isset ($_POST['pmhx'])) { 
$pmhx = $_POST['pmhx']; 
} 
if (isset ($_POST['pshx'])) { 
$pshx = $_POST['pshx']; 
} 
if (isset ($_POST['habbits'])) { 
$habbits = $_POST['habbits']; 
} 
if (isset ($_POST['occup'])) { 
$occup = $_POST['occup']; 
} 
if (isset ($_POST['allergy'])) { 
$allergy = $_POST['allergy']; 
} 

//Check file is uploaded or not 
//if (isset($_FILES['file']['name']) && $_FILES['file']['name']!='' && $_FILES['file']['error']=='') {
//$path2 = ... ; 
//move_uploaded_file(...);
if(is_uploaded_file($_FILES["file"]["tmp_name"]))
{
    $path = "../uploads/".$_FILES['file']['name'];
    move_uploaded_file($_FILES["file"]["tmp_name"], $path);
        $new_path = "./uploads/".$path;

}
else{
    $new_path = $_POST['org_path'];
//$path2 = "../uploads/".$_FILES['echo_photo']['name']; 
}
//move_uploaded_file($_FILES["file"]["tmp_name"], $path);
//$new_path = $path; 
$sql="UPDATE $tbl_name SET  
name = '$name', 
echo_files = '$new_path', 
remarcs = '$remarcs',
test_res = '$test_res', 
date = '$date', 
address = '$address', 
phone_num = '$phone',
illness = '$illness',
echo = '$echo', 
pmhx = '$pmhx', 
pshx = '$pshx', 
habbits = '$habbits', 
occup = '$occup', 
allergy = '$allergy',
pe = '$pe'
WHERE id = ".$id; 

$result=mysqli_query($con,$sql) or die('Unable to execute query. '. mysqli_error($con));

Answer 1

1) You can use Prepared Statements

A prepared statement or a parameterised statement is used to execute statement securely with high efficiency. Eg:PDO

2) For advanced hardening techniques you may refer OWASP SQL Injection Prevention