var_dump array returning my string, double?












0















After researching for hours, no proper fix has been found by myself so far, thus here I am!



I'm currently working on a school project that uses a login.php validator file to check if the user filled in all inputs, etc. etc., and also to create sessions.



require "config.inc.php";

$query = ("SELECT username, password, active FROM *DATABASE* WHERE username='".$username."' AND password='".$password."' AND active='1'") or die(mysql_error());
$result = mysqli_query($mysqli, $query);
$match = mysqli_num_rows($result);

if($match > 0){
//Directing to account page
header("Location:account.php");

//User info search
$queryID = ("SELECT id FROM *DATABASE* WHERE username='".$username."' AND password='".$password."' AND active='1'") or die(mysql_error());
$resultID = mysqli_query($mysqli, $queryID);
$id = mysqli_fetch_assoc($resultID);

$queryNAME = ("SELECT name FROM *DATABASE* WHERE username='".$username."' AND password='".$password."' AND active='1'") or die(mysql_error());
$resultNAME = mysqli_query($mysqli, $queryNAME);
$naam = mysqli_fetch_assoc($resultNAME);

//Session user info
$_SESSION['id'] = $id;
$_SESSION['naam'] = $naam;


This is just a minor piece of the validator, but as you can see it redirects to the account page (This works fine), and should create sessions for the name and ID found in the database.



The account page then again receives and translates the sessions to variables with the following code;



  //SESSIONS
require 'session.php'; //CHECKS IF SESSION IS VALID, IF NOT: RETURN TO HOME

//VARIABLES
if (isset($_SESSION['naam']) && isset($_SESSION['id']) && isset($_SESSION['username'])) {
print_r($_SESSION['naam']); //Debug purposes
print_r($_SESSION['id']);
print_r($_SESSION['username']);
$id = implode($_SESSION['id']); //Convert to String
$naam = implode($_SESSION['naam']);
$username = ($_SESSION['username']); //Already a string, this was a variable used in the POST form
var_dump($naam);
} else {
echo "No sessions"; //Debug purposes
}


When I echo the username on my account page, it displays the username perfectly fine, and I assume that's because username has always been a string to begin with.



Yet when I echo the naam variable, it displays "AdministratorAdministrator" even though the naam in the database is just Administrator.



This is my print_r result:



Array ( [0] => Administrator [name] => Administrator )
Array ( [0] => 23 [id] => 23 )
admin
string(26) "AdministratorAdministrator"


Any clues on how I can ensure that whenever I echo the naam variable, it shows just Administrator ONCE, and not twice? thanks!










share|improve this question

























  • //header("Location:account.php");// You are redirecting before store values in session!!!, redirect should be at last line in your code. Also use exit; after redirect

    – user1844933
    Nov 23 '18 at 18:28













  • whoa! How awkward is that, thanks so much :)

    – SessionPHP
    Nov 23 '18 at 18:33











  • You also can't mix different mysql apis.

    – Funk Forty Niner
    Nov 23 '18 at 18:35











  • This is open to an SQL injection, use a prepared statement. Also, do not store plain text passwords, use password_hash().

    – Funk Forty Niner
    Nov 23 '18 at 18:37






  • 1





    Your or die(mysql_error() code is on setting a string value, usually they should be on the mysqli_query() call (if indeed at all) and as @FunkFortyNiner pointed out it should be mysqli_error($mysqli).

    – Nigel Ren
    Nov 23 '18 at 18:39


















0















After researching for hours, no proper fix has been found by myself so far, thus here I am!



I'm currently working on a school project that uses a login.php validator file to check if the user filled in all inputs, etc. etc., and also to create sessions.



require "config.inc.php";

$query = ("SELECT username, password, active FROM *DATABASE* WHERE username='".$username."' AND password='".$password."' AND active='1'") or die(mysql_error());
$result = mysqli_query($mysqli, $query);
$match = mysqli_num_rows($result);

if($match > 0){
//Directing to account page
header("Location:account.php");

//User info search
$queryID = ("SELECT id FROM *DATABASE* WHERE username='".$username."' AND password='".$password."' AND active='1'") or die(mysql_error());
$resultID = mysqli_query($mysqli, $queryID);
$id = mysqli_fetch_assoc($resultID);

$queryNAME = ("SELECT name FROM *DATABASE* WHERE username='".$username."' AND password='".$password."' AND active='1'") or die(mysql_error());
$resultNAME = mysqli_query($mysqli, $queryNAME);
$naam = mysqli_fetch_assoc($resultNAME);

//Session user info
$_SESSION['id'] = $id;
$_SESSION['naam'] = $naam;


This is just a minor piece of the validator, but as you can see it redirects to the account page (This works fine), and should create sessions for the name and ID found in the database.



The account page then again receives and translates the sessions to variables with the following code;



  //SESSIONS
require 'session.php'; //CHECKS IF SESSION IS VALID, IF NOT: RETURN TO HOME

//VARIABLES
if (isset($_SESSION['naam']) && isset($_SESSION['id']) && isset($_SESSION['username'])) {
print_r($_SESSION['naam']); //Debug purposes
print_r($_SESSION['id']);
print_r($_SESSION['username']);
$id = implode($_SESSION['id']); //Convert to String
$naam = implode($_SESSION['naam']);
$username = ($_SESSION['username']); //Already a string, this was a variable used in the POST form
var_dump($naam);
} else {
echo "No sessions"; //Debug purposes
}


When I echo the username on my account page, it displays the username perfectly fine, and I assume that's because username has always been a string to begin with.



Yet when I echo the naam variable, it displays "AdministratorAdministrator" even though the naam in the database is just Administrator.



This is my print_r result:



Array ( [0] => Administrator [name] => Administrator )
Array ( [0] => 23 [id] => 23 )
admin
string(26) "AdministratorAdministrator"


Any clues on how I can ensure that whenever I echo the naam variable, it shows just Administrator ONCE, and not twice? thanks!










share|improve this question

























  • //header("Location:account.php");// You are redirecting before store values in session!!!, redirect should be at last line in your code. Also use exit; after redirect

    – user1844933
    Nov 23 '18 at 18:28













  • whoa! How awkward is that, thanks so much :)

    – SessionPHP
    Nov 23 '18 at 18:33











  • You also can't mix different mysql apis.

    – Funk Forty Niner
    Nov 23 '18 at 18:35











  • This is open to an SQL injection, use a prepared statement. Also, do not store plain text passwords, use password_hash().

    – Funk Forty Niner
    Nov 23 '18 at 18:37






  • 1





    Your or die(mysql_error() code is on setting a string value, usually they should be on the mysqli_query() call (if indeed at all) and as @FunkFortyNiner pointed out it should be mysqli_error($mysqli).

    – Nigel Ren
    Nov 23 '18 at 18:39
















0












0








0








After researching for hours, no proper fix has been found by myself so far, thus here I am!



I'm currently working on a school project that uses a login.php validator file to check if the user filled in all inputs, etc. etc., and also to create sessions.



require "config.inc.php";

$query = ("SELECT username, password, active FROM *DATABASE* WHERE username='".$username."' AND password='".$password."' AND active='1'") or die(mysql_error());
$result = mysqli_query($mysqli, $query);
$match = mysqli_num_rows($result);

if($match > 0){
//Directing to account page
header("Location:account.php");

//User info search
$queryID = ("SELECT id FROM *DATABASE* WHERE username='".$username."' AND password='".$password."' AND active='1'") or die(mysql_error());
$resultID = mysqli_query($mysqli, $queryID);
$id = mysqli_fetch_assoc($resultID);

$queryNAME = ("SELECT name FROM *DATABASE* WHERE username='".$username."' AND password='".$password."' AND active='1'") or die(mysql_error());
$resultNAME = mysqli_query($mysqli, $queryNAME);
$naam = mysqli_fetch_assoc($resultNAME);

//Session user info
$_SESSION['id'] = $id;
$_SESSION['naam'] = $naam;


This is just a minor piece of the validator, but as you can see it redirects to the account page (This works fine), and should create sessions for the name and ID found in the database.



The account page then again receives and translates the sessions to variables with the following code;



  //SESSIONS
require 'session.php'; //CHECKS IF SESSION IS VALID, IF NOT: RETURN TO HOME

//VARIABLES
if (isset($_SESSION['naam']) && isset($_SESSION['id']) && isset($_SESSION['username'])) {
print_r($_SESSION['naam']); //Debug purposes
print_r($_SESSION['id']);
print_r($_SESSION['username']);
$id = implode($_SESSION['id']); //Convert to String
$naam = implode($_SESSION['naam']);
$username = ($_SESSION['username']); //Already a string, this was a variable used in the POST form
var_dump($naam);
} else {
echo "No sessions"; //Debug purposes
}


When I echo the username on my account page, it displays the username perfectly fine, and I assume that's because username has always been a string to begin with.



Yet when I echo the naam variable, it displays "AdministratorAdministrator" even though the naam in the database is just Administrator.



This is my print_r result:



Array ( [0] => Administrator [name] => Administrator )
Array ( [0] => 23 [id] => 23 )
admin
string(26) "AdministratorAdministrator"


Any clues on how I can ensure that whenever I echo the naam variable, it shows just Administrator ONCE, and not twice? thanks!










share|improve this question
















After researching for hours, no proper fix has been found by myself so far, thus here I am!



I'm currently working on a school project that uses a login.php validator file to check if the user filled in all inputs, etc. etc., and also to create sessions.



require "config.inc.php";

$query = ("SELECT username, password, active FROM *DATABASE* WHERE username='".$username."' AND password='".$password."' AND active='1'") or die(mysql_error());
$result = mysqli_query($mysqli, $query);
$match = mysqli_num_rows($result);

if($match > 0){
//Directing to account page
header("Location:account.php");

//User info search
$queryID = ("SELECT id FROM *DATABASE* WHERE username='".$username."' AND password='".$password."' AND active='1'") or die(mysql_error());
$resultID = mysqli_query($mysqli, $queryID);
$id = mysqli_fetch_assoc($resultID);

$queryNAME = ("SELECT name FROM *DATABASE* WHERE username='".$username."' AND password='".$password."' AND active='1'") or die(mysql_error());
$resultNAME = mysqli_query($mysqli, $queryNAME);
$naam = mysqli_fetch_assoc($resultNAME);

//Session user info
$_SESSION['id'] = $id;
$_SESSION['naam'] = $naam;


This is just a minor piece of the validator, but as you can see it redirects to the account page (This works fine), and should create sessions for the name and ID found in the database.



The account page then again receives and translates the sessions to variables with the following code;



  //SESSIONS
require 'session.php'; //CHECKS IF SESSION IS VALID, IF NOT: RETURN TO HOME

//VARIABLES
if (isset($_SESSION['naam']) && isset($_SESSION['id']) && isset($_SESSION['username'])) {
print_r($_SESSION['naam']); //Debug purposes
print_r($_SESSION['id']);
print_r($_SESSION['username']);
$id = implode($_SESSION['id']); //Convert to String
$naam = implode($_SESSION['naam']);
$username = ($_SESSION['username']); //Already a string, this was a variable used in the POST form
var_dump($naam);
} else {
echo "No sessions"; //Debug purposes
}


When I echo the username on my account page, it displays the username perfectly fine, and I assume that's because username has always been a string to begin with.



Yet when I echo the naam variable, it displays "AdministratorAdministrator" even though the naam in the database is just Administrator.



This is my print_r result:



Array ( [0] => Administrator [name] => Administrator )
Array ( [0] => 23 [id] => 23 )
admin
string(26) "AdministratorAdministrator"


Any clues on how I can ensure that whenever I echo the naam variable, it shows just Administrator ONCE, and not twice? thanks!







php mysql arrays session login






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 23 '18 at 18:36









Funk Forty Niner

1




1










asked Nov 23 '18 at 18:23









SessionPHPSessionPHP

11




11













  • //header("Location:account.php");// You are redirecting before store values in session!!!, redirect should be at last line in your code. Also use exit; after redirect

    – user1844933
    Nov 23 '18 at 18:28













  • whoa! How awkward is that, thanks so much :)

    – SessionPHP
    Nov 23 '18 at 18:33











  • You also can't mix different mysql apis.

    – Funk Forty Niner
    Nov 23 '18 at 18:35











  • This is open to an SQL injection, use a prepared statement. Also, do not store plain text passwords, use password_hash().

    – Funk Forty Niner
    Nov 23 '18 at 18:37






  • 1





    Your or die(mysql_error() code is on setting a string value, usually they should be on the mysqli_query() call (if indeed at all) and as @FunkFortyNiner pointed out it should be mysqli_error($mysqli).

    – Nigel Ren
    Nov 23 '18 at 18:39





















  • //header("Location:account.php");// You are redirecting before store values in session!!!, redirect should be at last line in your code. Also use exit; after redirect

    – user1844933
    Nov 23 '18 at 18:28













  • whoa! How awkward is that, thanks so much :)

    – SessionPHP
    Nov 23 '18 at 18:33











  • You also can't mix different mysql apis.

    – Funk Forty Niner
    Nov 23 '18 at 18:35











  • This is open to an SQL injection, use a prepared statement. Also, do not store plain text passwords, use password_hash().

    – Funk Forty Niner
    Nov 23 '18 at 18:37






  • 1





    Your or die(mysql_error() code is on setting a string value, usually they should be on the mysqli_query() call (if indeed at all) and as @FunkFortyNiner pointed out it should be mysqli_error($mysqli).

    – Nigel Ren
    Nov 23 '18 at 18:39



















//header("Location:account.php");// You are redirecting before store values in session!!!, redirect should be at last line in your code. Also use exit; after redirect

– user1844933
Nov 23 '18 at 18:28







//header("Location:account.php");// You are redirecting before store values in session!!!, redirect should be at last line in your code. Also use exit; after redirect

– user1844933
Nov 23 '18 at 18:28















whoa! How awkward is that, thanks so much :)

– SessionPHP
Nov 23 '18 at 18:33





whoa! How awkward is that, thanks so much :)

– SessionPHP
Nov 23 '18 at 18:33













You also can't mix different mysql apis.

– Funk Forty Niner
Nov 23 '18 at 18:35





You also can't mix different mysql apis.

– Funk Forty Niner
Nov 23 '18 at 18:35













This is open to an SQL injection, use a prepared statement. Also, do not store plain text passwords, use password_hash().

– Funk Forty Niner
Nov 23 '18 at 18:37





This is open to an SQL injection, use a prepared statement. Also, do not store plain text passwords, use password_hash().

– Funk Forty Niner
Nov 23 '18 at 18:37




1




1





Your or die(mysql_error() code is on setting a string value, usually they should be on the mysqli_query() call (if indeed at all) and as @FunkFortyNiner pointed out it should be mysqli_error($mysqli).

– Nigel Ren
Nov 23 '18 at 18:39







Your or die(mysql_error() code is on setting a string value, usually they should be on the mysqli_query() call (if indeed at all) and as @FunkFortyNiner pointed out it should be mysqli_error($mysqli).

– Nigel Ren
Nov 23 '18 at 18:39














1 Answer
1






active

oldest

votes


















0














You need to save onlye the data you need in the Session



$_SESSION['id'] = $id['id'];
$_SESSION['naam'] = $naam['name'];


In your case you are saving the entire result from the function mysqli_fetch_assoc, which is an array, as you can see from the docs:



http://php.net/manual/en/mysqli-result.fetch-assoc.php






share|improve this answer























    Your Answer






    StackExchange.ifUsing("editor", function () {
    StackExchange.using("externalEditor", function () {
    StackExchange.using("snippets", function () {
    StackExchange.snippets.init();
    });
    });
    }, "code-snippets");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "1"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53451425%2fvar-dump-array-returning-my-string-double%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    You need to save onlye the data you need in the Session



    $_SESSION['id'] = $id['id'];
    $_SESSION['naam'] = $naam['name'];


    In your case you are saving the entire result from the function mysqli_fetch_assoc, which is an array, as you can see from the docs:



    http://php.net/manual/en/mysqli-result.fetch-assoc.php






    share|improve this answer




























      0














      You need to save onlye the data you need in the Session



      $_SESSION['id'] = $id['id'];
      $_SESSION['naam'] = $naam['name'];


      In your case you are saving the entire result from the function mysqli_fetch_assoc, which is an array, as you can see from the docs:



      http://php.net/manual/en/mysqli-result.fetch-assoc.php






      share|improve this answer


























        0












        0








        0







        You need to save onlye the data you need in the Session



        $_SESSION['id'] = $id['id'];
        $_SESSION['naam'] = $naam['name'];


        In your case you are saving the entire result from the function mysqli_fetch_assoc, which is an array, as you can see from the docs:



        http://php.net/manual/en/mysqli-result.fetch-assoc.php






        share|improve this answer













        You need to save onlye the data you need in the Session



        $_SESSION['id'] = $id['id'];
        $_SESSION['naam'] = $naam['name'];


        In your case you are saving the entire result from the function mysqli_fetch_assoc, which is an array, as you can see from the docs:



        http://php.net/manual/en/mysqli-result.fetch-assoc.php







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 23 '18 at 18:30









        Felippe DuarteFelippe Duarte

        10.5k21524




        10.5k21524






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Stack Overflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53451425%2fvar-dump-array-returning-my-string-double%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Berounka

            Sphinx de Gizeh

            Different font size/position of beamer's navigation symbols template's content depending on regular/plain...