php sees, for example, 11 as 1












-2














function zwrot() {
var zwrot = document.getElementsByClassName("zwroc");
var i;
for (i = 0; i < zwrot.length; i++) {
if(zwrot[i].checked){
parseInt(zwrot[i])
var ajax = $.ajax({
url: 'php/request.php',
type: 'POST',
data: {zwrot: zwrot[i].value},
success: function(data)
{
$.growl.notice({
title: "INFO",
message: "Oddano książkę" });
}
})
}
}
setTimeout(function() {
location.reload()
}, 2000
);
}


php code:



if(isset($_POST['zwrot'])) { 
$zwrot=$_POST['zwrot'];
$n = count($zwrot);
for ($i=0;$i<$n; $i++){
$data=date("d-m-Y");
$zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]");
$zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");
}
}


everything is ok, until id is a one-digit number. the problem occurs with two-digit numbers, php sees, for example, 11 as 1










share|improve this question




















  • 2




    until id is a one-digit number What is id? There is no id in your code
    – CertainPerformance
    Nov 12 at 0:23










  • "php sees" - but we don't see any php!?
    – Jeff
    Nov 12 at 0:24










  • sry, i mean var zwrot = document.getElementsByClassName("zwroc"); this is getting value from my table, my fault not id
    – lysekk
    Nov 12 at 0:24










  • if(isset($_POST['zwrot'])) { $zwrot=$_POST['zwrot']; $n = count($zwrot); for ($i=0;$i<$n; $i++){ $data=date("d-m-Y"); $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]"); $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");}}
    – lysekk
    Nov 12 at 0:30






  • 1




    did you check what $zrot in php actually is? Hint: It's not an array. You are doing a seperate ajax call for each element with classname 'szroc'.
    – Jeff
    Nov 12 at 0:30


















-2














function zwrot() {
var zwrot = document.getElementsByClassName("zwroc");
var i;
for (i = 0; i < zwrot.length; i++) {
if(zwrot[i].checked){
parseInt(zwrot[i])
var ajax = $.ajax({
url: 'php/request.php',
type: 'POST',
data: {zwrot: zwrot[i].value},
success: function(data)
{
$.growl.notice({
title: "INFO",
message: "Oddano książkę" });
}
})
}
}
setTimeout(function() {
location.reload()
}, 2000
);
}


php code:



if(isset($_POST['zwrot'])) { 
$zwrot=$_POST['zwrot'];
$n = count($zwrot);
for ($i=0;$i<$n; $i++){
$data=date("d-m-Y");
$zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]");
$zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");
}
}


everything is ok, until id is a one-digit number. the problem occurs with two-digit numbers, php sees, for example, 11 as 1










share|improve this question




















  • 2




    until id is a one-digit number What is id? There is no id in your code
    – CertainPerformance
    Nov 12 at 0:23










  • "php sees" - but we don't see any php!?
    – Jeff
    Nov 12 at 0:24










  • sry, i mean var zwrot = document.getElementsByClassName("zwroc"); this is getting value from my table, my fault not id
    – lysekk
    Nov 12 at 0:24










  • if(isset($_POST['zwrot'])) { $zwrot=$_POST['zwrot']; $n = count($zwrot); for ($i=0;$i<$n; $i++){ $data=date("d-m-Y"); $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]"); $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");}}
    – lysekk
    Nov 12 at 0:30






  • 1




    did you check what $zrot in php actually is? Hint: It's not an array. You are doing a seperate ajax call for each element with classname 'szroc'.
    – Jeff
    Nov 12 at 0:30
















-2












-2








-2







function zwrot() {
var zwrot = document.getElementsByClassName("zwroc");
var i;
for (i = 0; i < zwrot.length; i++) {
if(zwrot[i].checked){
parseInt(zwrot[i])
var ajax = $.ajax({
url: 'php/request.php',
type: 'POST',
data: {zwrot: zwrot[i].value},
success: function(data)
{
$.growl.notice({
title: "INFO",
message: "Oddano książkę" });
}
})
}
}
setTimeout(function() {
location.reload()
}, 2000
);
}


php code:



if(isset($_POST['zwrot'])) { 
$zwrot=$_POST['zwrot'];
$n = count($zwrot);
for ($i=0;$i<$n; $i++){
$data=date("d-m-Y");
$zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]");
$zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");
}
}


everything is ok, until id is a one-digit number. the problem occurs with two-digit numbers, php sees, for example, 11 as 1










share|improve this question















function zwrot() {
var zwrot = document.getElementsByClassName("zwroc");
var i;
for (i = 0; i < zwrot.length; i++) {
if(zwrot[i].checked){
parseInt(zwrot[i])
var ajax = $.ajax({
url: 'php/request.php',
type: 'POST',
data: {zwrot: zwrot[i].value},
success: function(data)
{
$.growl.notice({
title: "INFO",
message: "Oddano książkę" });
}
})
}
}
setTimeout(function() {
location.reload()
}, 2000
);
}


php code:



if(isset($_POST['zwrot'])) { 
$zwrot=$_POST['zwrot'];
$n = count($zwrot);
for ($i=0;$i<$n; $i++){
$data=date("d-m-Y");
$zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]");
$zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");
}
}


everything is ok, until id is a one-digit number. the problem occurs with two-digit numbers, php sees, for example, 11 as 1







javascript php mysql arrays






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 at 0:26









Jeff

6,18911024




6,18911024










asked Nov 12 at 0:18









lysekk

1




1








  • 2




    until id is a one-digit number What is id? There is no id in your code
    – CertainPerformance
    Nov 12 at 0:23










  • "php sees" - but we don't see any php!?
    – Jeff
    Nov 12 at 0:24










  • sry, i mean var zwrot = document.getElementsByClassName("zwroc"); this is getting value from my table, my fault not id
    – lysekk
    Nov 12 at 0:24










  • if(isset($_POST['zwrot'])) { $zwrot=$_POST['zwrot']; $n = count($zwrot); for ($i=0;$i<$n; $i++){ $data=date("d-m-Y"); $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]"); $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");}}
    – lysekk
    Nov 12 at 0:30






  • 1




    did you check what $zrot in php actually is? Hint: It's not an array. You are doing a seperate ajax call for each element with classname 'szroc'.
    – Jeff
    Nov 12 at 0:30
















  • 2




    until id is a one-digit number What is id? There is no id in your code
    – CertainPerformance
    Nov 12 at 0:23










  • "php sees" - but we don't see any php!?
    – Jeff
    Nov 12 at 0:24










  • sry, i mean var zwrot = document.getElementsByClassName("zwroc"); this is getting value from my table, my fault not id
    – lysekk
    Nov 12 at 0:24










  • if(isset($_POST['zwrot'])) { $zwrot=$_POST['zwrot']; $n = count($zwrot); for ($i=0;$i<$n; $i++){ $data=date("d-m-Y"); $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]"); $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");}}
    – lysekk
    Nov 12 at 0:30






  • 1




    did you check what $zrot in php actually is? Hint: It's not an array. You are doing a seperate ajax call for each element with classname 'szroc'.
    – Jeff
    Nov 12 at 0:30










2




2




until id is a one-digit number What is id? There is no id in your code
– CertainPerformance
Nov 12 at 0:23




until id is a one-digit number What is id? There is no id in your code
– CertainPerformance
Nov 12 at 0:23












"php sees" - but we don't see any php!?
– Jeff
Nov 12 at 0:24




"php sees" - but we don't see any php!?
– Jeff
Nov 12 at 0:24












sry, i mean var zwrot = document.getElementsByClassName("zwroc"); this is getting value from my table, my fault not id
– lysekk
Nov 12 at 0:24




sry, i mean var zwrot = document.getElementsByClassName("zwroc"); this is getting value from my table, my fault not id
– lysekk
Nov 12 at 0:24












if(isset($_POST['zwrot'])) { $zwrot=$_POST['zwrot']; $n = count($zwrot); for ($i=0;$i<$n; $i++){ $data=date("d-m-Y"); $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]"); $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");}}
– lysekk
Nov 12 at 0:30




if(isset($_POST['zwrot'])) { $zwrot=$_POST['zwrot']; $n = count($zwrot); for ($i=0;$i<$n; $i++){ $data=date("d-m-Y"); $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot[$i]"); $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot[$i];");}}
– lysekk
Nov 12 at 0:30




1




1




did you check what $zrot in php actually is? Hint: It's not an array. You are doing a seperate ajax call for each element with classname 'szroc'.
– Jeff
Nov 12 at 0:30






did you check what $zrot in php actually is? Hint: It's not an array. You are doing a seperate ajax call for each element with classname 'szroc'.
– Jeff
Nov 12 at 0:30














1 Answer
1






active

oldest

votes


















0














I dont think you're passing any array to your php. and that's not how you pass and receive an array through ajax if ever, you need json for that.



Anyway, try this:



if(isset($_POST['zwrot'])) { 

$zwrot=$_POST['zwrot'];

$data=date("d-m-Y");
$zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
$zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");

}


Or if you want to pass by array:



function zwrot() {
var zwrot = document.getElementsByClassName("zwroc");
var i;

var zwrot_array = ;

for (i = 0; i < zwrot.length; i++) {
if(zwrot[i].checked){
zwrot_array[i] = zwrot[i].value;
}
}

var ajax = $.ajax({
url: 'php/request.php',
type: 'POST',
data: {zwrot: JSON.stringify(zwrot_array)},
success: function(data)
{
$.growl.notice({
title: "INFO",
message: "Oddano książkę"
});

setTimeout(function() {
location.reload()
}, 2000
);
}
});
}


PHP:



if(isset($_POST['zwrot'])) { 

$zwrots = json_decode($_POST['zwrot']);
$data=date("d-m-Y");

foreach ($zwrots as $zwrot) {
$zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
$zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");
}
}





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%2f53254561%2fphp-sees-for-example-11-as-1%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














    I dont think you're passing any array to your php. and that's not how you pass and receive an array through ajax if ever, you need json for that.



    Anyway, try this:



    if(isset($_POST['zwrot'])) { 

    $zwrot=$_POST['zwrot'];

    $data=date("d-m-Y");
    $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
    $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");

    }


    Or if you want to pass by array:



    function zwrot() {
    var zwrot = document.getElementsByClassName("zwroc");
    var i;

    var zwrot_array = ;

    for (i = 0; i < zwrot.length; i++) {
    if(zwrot[i].checked){
    zwrot_array[i] = zwrot[i].value;
    }
    }

    var ajax = $.ajax({
    url: 'php/request.php',
    type: 'POST',
    data: {zwrot: JSON.stringify(zwrot_array)},
    success: function(data)
    {
    $.growl.notice({
    title: "INFO",
    message: "Oddano książkę"
    });

    setTimeout(function() {
    location.reload()
    }, 2000
    );
    }
    });
    }


    PHP:



    if(isset($_POST['zwrot'])) { 

    $zwrots = json_decode($_POST['zwrot']);
    $data=date("d-m-Y");

    foreach ($zwrots as $zwrot) {
    $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
    $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");
    }
    }





    share|improve this answer




























      0














      I dont think you're passing any array to your php. and that's not how you pass and receive an array through ajax if ever, you need json for that.



      Anyway, try this:



      if(isset($_POST['zwrot'])) { 

      $zwrot=$_POST['zwrot'];

      $data=date("d-m-Y");
      $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
      $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");

      }


      Or if you want to pass by array:



      function zwrot() {
      var zwrot = document.getElementsByClassName("zwroc");
      var i;

      var zwrot_array = ;

      for (i = 0; i < zwrot.length; i++) {
      if(zwrot[i].checked){
      zwrot_array[i] = zwrot[i].value;
      }
      }

      var ajax = $.ajax({
      url: 'php/request.php',
      type: 'POST',
      data: {zwrot: JSON.stringify(zwrot_array)},
      success: function(data)
      {
      $.growl.notice({
      title: "INFO",
      message: "Oddano książkę"
      });

      setTimeout(function() {
      location.reload()
      }, 2000
      );
      }
      });
      }


      PHP:



      if(isset($_POST['zwrot'])) { 

      $zwrots = json_decode($_POST['zwrot']);
      $data=date("d-m-Y");

      foreach ($zwrots as $zwrot) {
      $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
      $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");
      }
      }





      share|improve this answer


























        0












        0








        0






        I dont think you're passing any array to your php. and that's not how you pass and receive an array through ajax if ever, you need json for that.



        Anyway, try this:



        if(isset($_POST['zwrot'])) { 

        $zwrot=$_POST['zwrot'];

        $data=date("d-m-Y");
        $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
        $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");

        }


        Or if you want to pass by array:



        function zwrot() {
        var zwrot = document.getElementsByClassName("zwroc");
        var i;

        var zwrot_array = ;

        for (i = 0; i < zwrot.length; i++) {
        if(zwrot[i].checked){
        zwrot_array[i] = zwrot[i].value;
        }
        }

        var ajax = $.ajax({
        url: 'php/request.php',
        type: 'POST',
        data: {zwrot: JSON.stringify(zwrot_array)},
        success: function(data)
        {
        $.growl.notice({
        title: "INFO",
        message: "Oddano książkę"
        });

        setTimeout(function() {
        location.reload()
        }, 2000
        );
        }
        });
        }


        PHP:



        if(isset($_POST['zwrot'])) { 

        $zwrots = json_decode($_POST['zwrot']);
        $data=date("d-m-Y");

        foreach ($zwrots as $zwrot) {
        $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
        $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");
        }
        }





        share|improve this answer














        I dont think you're passing any array to your php. and that's not how you pass and receive an array through ajax if ever, you need json for that.



        Anyway, try this:



        if(isset($_POST['zwrot'])) { 

        $zwrot=$_POST['zwrot'];

        $data=date("d-m-Y");
        $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
        $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");

        }


        Or if you want to pass by array:



        function zwrot() {
        var zwrot = document.getElementsByClassName("zwroc");
        var i;

        var zwrot_array = ;

        for (i = 0; i < zwrot.length; i++) {
        if(zwrot[i].checked){
        zwrot_array[i] = zwrot[i].value;
        }
        }

        var ajax = $.ajax({
        url: 'php/request.php',
        type: 'POST',
        data: {zwrot: JSON.stringify(zwrot_array)},
        success: function(data)
        {
        $.growl.notice({
        title: "INFO",
        message: "Oddano książkę"
        });

        setTimeout(function() {
        location.reload()
        }, 2000
        );
        }
        });
        }


        PHP:



        if(isset($_POST['zwrot'])) { 

        $zwrots = json_decode($_POST['zwrot']);
        $data=date("d-m-Y");

        foreach ($zwrots as $zwrot) {
        $zapytanie5 = mysqli_query ($link, "UPDATE zamowienie SET data_zwrotu='$data' WHERE id_zamowienie=$zwrot");
        $zapytanie6 = mysqli_query ($link, "UPDATE ksiazka INNER JOIN zamowienie ON ksiazka.id_ksiazka=zamowienie.id_ksiazka SET ilosc=ilosc+1 WHERE id_zamowienie=$zwrot;");
        }
        }






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 12 at 1:22

























        answered Nov 12 at 1:13









        ACD

        854111




        854111






























            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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%2f53254561%2fphp-sees-for-example-11-as-1%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

            Florida Star v. B. J. F.

            Danny Elfman

            Lugert, Oklahoma