如何正确地将一个PHP变量传递给另一个PHP页面？

Question

I have an android app where it uploads the file along with other variables using PHP to a server. I want to insert the uploaded file name(from the PHP file) to the PHP file which handles the inserting of variables to the database.

I want to get file_path to the another php file. UploadToServer.php

<?php
session_start();

   $file_path = "uploads/";

   $file_path = $file_path . basename( $_FILES['uploaded_file']['name']);
   if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $file_path)) {
   $_SESSION['image'] = "$file_path";
       echo "success";
   } else{
       echo "fail";
   }

?>

create_product.php

<?php
session_start();
/*
 * Following code will create a new report row
 * All report details are read from HTTP Post Request
 */

// array for JSON response
$response = array();
$image = $_SESSION['image'];

// check for required fields
if (isset($_POST['name']) && isset($_POST['location']) && isset($_POST['description']) && isset($_POST['type'])) {

    $type = $_POST['type'];
    $name = $_POST['name'];
    $location = $_POST['location'];
    $description = $_POST['description'];

    // include db connect class
    require_once __DIR__ . '/db_connect.php';

    // connecting to db
    $db = new DB_CONNECT();

    // mysql inserting a new row
    $result = mysql_query("INSERT INTO reports(type, name, location, description, image) VALUES('$type', '$name', '$location', '$description', '$file_path')");

    // check if row inserted or not
    if ($result) {
        // successfully inserted into database
        $response["success"] = 1;
        $response["message"] = "report successfully created.";

        // echoing JSON response
        echo json_encode($response);
    } else {
        // failed to insert row
        $response["success"] = 0;
        $response["message"] = "Oops! An error occurred.";

        // echoing JSON response
        echo json_encode($response);
    }
} else {
    // required field is missing
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";

    // echoing JSON response
    echo json_encode($response);
}
?>

Please tell me what's wrong.

Answer 1

There should be no session involved at all, you need to combine the 2 scripts:

<?php
// array for JSON response
$response = array();

// check for required fields
if (isset($_POST['name']) && isset($_POST['location']) && isset($_POST['description']) && isset($_POST['type'])) {

    $file_path = "uploads/";

    $file_path = $file_path . basename( $_FILES['uploaded_file']['name']);
    if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $file_path)) {
        echo "success save to db";
        $type = $_POST['type'];
        $name = $_POST['name'];
        $location = $_POST['location'];
        $description = $_POST['description'];

        // include db connect class
        require_once __DIR__ . '/db_connect.php';

        // connecting to db
        $db = new DB_CONNECT();

        // mysql inserting a new row
        $result = mysql_query("INSERT INTO reports(type, name, location, description, image) 
                               VALUES('$type', '$name', '$location', '$description', '$file_path')");

        // check if row inserted or not
        if ($result) {
            // successfully inserted into database
            $response["success"] = 1;
            $response["message"] = "report successfully created.";

            // echoing JSON response
            echo json_encode($response);
        } else {
            // failed to insert row
            $response["success"] = 0;
            $response["message"] = "Oops! An error occurred.";

            // echoing JSON response
            echo json_encode($response);
        }
    } else{
       echo "fail to upload will not save to DB";
    }
} else {
    // required field is missing
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";

    // echoing JSON response
    echo json_encode($response);
}
?>

Please note:

• This code will insert the record in the database only, it it successfully uploaded the file.
• mysql is deprecated
• use mysqli or PDO, and prepare query to avoid SQL injection