php - MySQL所有变量不传递数据

Question

我在向 mysql 发送变量数据时遇到问题，我的一个变量正在传递，但另一个没有。

<?php 
session_start();
if(!isset($_SESSION["PAT_ID"])){
ob_start();
header("location: login.php");
ob_end_flush(); 
}

$patient_id ="";
$complainID = "";

$patient_id = $_SESSION["PAT_ID"];

if(isset($_GET["ComplainID"])){

$complainID = $_GET["ComplainID"];
}

include "config/connect_to_mysql.php";
?>

<?php 
//Diesease photo\

echo "Hello Patient Your Complain ID: " . $complainID;

$imageerr = "";
$typeerr = "";
$sizeerr = "";

if ($_SERVER["REQUEST_METHOD"] == "POST") {

 if( (isset($_FILES['galleryField_1']) && $_FILES['galleryField_1']['error'] == 0)
    || (isset($_FILES['galleryField_2'])) 
        || (isset($_FILES['galleryField_3']))
            || (isset($_FILES['galleryField_4'])) 
   ){

        $allowedExts = array("JPEG", "jpeg", "jpg", "JPG");
        $temp = explode(".", $_FILES["galleryField_1"]["name"]);
        $extension = end($temp);

        if ((
           ($_FILES["galleryField_1"]["type"] == "image/JPEG")
        || ($_FILES["galleryField_1"]["type"] == "image/jpeg")
        || ($_FILES["galleryField_1"]["type"] == "image/jpg")
        || ($_FILES["galleryField_1"]["type"] == "image/JPG"))
        && in_array($extension, $allowedExts)){



            if($_FILES['galleryField_1']['size'] > 1048576) { //1 MB (size is also in bytes)

                $sizeerr = "Photo must be within 1 MB";

            } else{ 

              // Add this image into the database now

              $sql = mysql_query("INSERT INTO `shasthojito`.`sdispic` (`sdis_pic_id`, `pat_id`,        `comp_id`) VALUES (NULL, '$patient_id', '$complainID')") 
                   or die (mysql_error());

              $gallery_id = mysql_insert_id();
              // Place image in the folder 
              $newgallery = "$gallery_id.jpg";

              move_uploaded_file( $_FILES['galleryField_1']['tmp_name'], "dpic/$newgallery");

                }
        }else{
            $typeerr = "You have to put JPEG Image file";
        }
    }else{
            $imageerr = "No Image Selected";
    }

此处变量 $patientID 工作正常并将数据传递给它，但是 $complainID 不适用于 sql 查询，但它在 echo 中显示值 ...

Answer 1

您将 GET 与 POST 混合使用，因为您使用的是这一行:

if ($_SERVER["REQUEST_METHOD"] == "POST") 

你需要改变这个:

if(isset($_GET["ComplainID"])){

$complainID = $_GET["ComplainID"];
}

收件人:

if(isset($_POST["ComplainID"])){

$complainID = $_POST["ComplainID"];
}

或者也许你只需要改变这个:

if ($_SERVER["REQUEST_METHOD"] == "POST") 

收件人:

if ($_SERVER["REQUEST_METHOD"] == "GET") 

请确定您将日期传输到实际文件所使用的方法。

编辑 1:

根据您在上面评论中的回答，请更改此设置:

$sql = mysql_query("INSERT INTO `shasthojito`.`sdispic` (`sdis_pic_id`, `pat_id`,        `comp_id`) VALUES (NULL, '$patient_id', '$complainID')") 
                   or die (mysql_error());

收件人:

$sql = mysql_query("INSERT INTO `shasthojito`.`sdispic` (`sdis_pic_id`, `pat_id`,        `comp_id`) VALUES (NULL, '$patient_id', '".$complainID."')") 
                   or die (mysql_error());

编辑 2:

在插入变量之前，确保它与表的 comp_id 列具有相同的类型:

if (isset($_GET['ComplainID']) && ctype_digit($_GET['ComplainID']))
{
  $complainID = $_GET["ComplainID"];
}