Array compare

[gmc1103]

Hi

 

I have an  for examplo {1,2,3,4,5,6,7,8,9,10}and another one from a fprm  {1,4,7,9,10} , i would like to compafe both and create a new one with null where there are differences..

 

any help?

 

Thanks

[blacknight]

http://php.net/manual/en/function.array-diff.php

[Barand]

try

$a = [1,2,3,4,5,6,7,8,9,10];
$b = [1,4,7,9,10];
$c = array_fill_keys(array_keys(array_diff($a,$b)),null) + $a;
ksort($c);

result:

$c = Array
(
    [0] => 1
    [1] => 
    [2] => 
    [3] => 4
    [4] => 
    [5] => 
    [6] => 7
    [7] => 
    [8] => 9
    [9] => 10
)

[gmc1103]

Hi

 

Thank you for your answer, i'm foloowing this example and it seems to be what i need but i'm having an issue

$a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17];
$c = array_fill_keys(array_keys(array_diff($a,$hora)),null) + $a;
ksort($c);
foreach ($c as $key => $val) {
$stmt = $mysqli->prepare("INSERT INTO `ebspma`.`sala_ocupacao` (id_dia, id_sala, id_tempo) VALUES (?, ?, ?);")or die(mysql_error($mysqli));
$horas = $val;
echo $mysqli->error, PHP_EOL;
$stmt->bind_param('ssi', $dia, $sala, $horas);
if(!$stmt->execute()) echo $stmt->error;
$response = "Registo gravado";
echo json_encode($response);
$stmt->close();

But instead of writing to the bd the values of teh array $c....the loop is executed more than 1000 times..

 

This is what a i get from my debuger

 

http://postimg.org/image/739yp0qfr/

 

What's wrong after the ksort?

[Barand]

What does $hora (or $_POST['hora']) contain?

[gmc1103]

$hora (or $_POST['hora']) contain in this example

Array[4]

[0] String  = "1"

[1] String  = "2"

[2] String  = "5"

[3] String  = "6"

 

 

 

I have tested in php online with this code

$a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17];
$hora = [1,4,7,9,10];
$c = array_fill_keys(array_keys(array_diff($a,$hora)),null) + $a;
ksort($c);
foreach ($c as $key => $val) {
    echo "$val\n";
} 

And i receive the correct values

So the problem should be here inside the loop

 

$horas = $val; because i just get one value

 

Thanks

[Barand]

I can't see a reason for 1000+ writes in the code posted. Is that code inside another loop?

[gmc1103]

Hi Barand

 

Thanks again for your help

I have fixed now, the code is now like this

<?php
header('Content-type: application/json');
error_reporting(E_ALL | E_NOTICE);
ini_set('display_errors', '1');
$mysqli = new mysqli('localhost', 'root', '', 'ebspma');
 
// Works as of PHP 5.2.9 and 5.3.0.
if ($mysqli->connect_error) {
    die('Connect Error: ' . $mysqli->connect_error);
}
$num= $_POST['num'];
$dia = $_POST['dia'];
$sala = $_POST['sala'];
$hora = explode(",", $_POST['hora']);
$a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17];
$c = array_fill_keys(array_keys(array_diff($a,$hora)),null) + $a;
ksort($c);
for ($i=0; $i<17; $i++){
$stmt = $mysqli->prepare("INSERT INTO `ebspma`.`sala_ocupacao` (id_dia, id_sala, id_tempo) VALUES (?, ?, ?);")or die(mysql_error($mysqli));
$horas = $c[$i];
echo $mysqli->error, PHP_EOL;
$stmt->bind_param('ssi', $dia, $sala, $horas);
if(!$stmt->execute()) echo $stmt->error;
$response = "Registo gravado";
echo json_encode($response);
$stmt->close();
}

I'm sure the code can be improved 

 

I just have another small issue 

I have no json_encode response in my javascript, can you see why?

<script>
             function postData(){
                var dia = document.getElementById('dia').value;
                var sala = document.getElementById('sala').value;
                var tempos = [];
                var s = document.getElementById('hora');
                for (var i = 0; i < s.options.length; i++) {
                    if (s.options[i].selected == true) {
                        var valores = s.options[i].value;
                        tempos.push(valores);
                    }
                }
                console.log(tempos);
                var num = document.getElementById('num').value;
                var data = 'dia='+ dia + '&sala='+ sala + '&hora='+ tempos + '&num='+ num;
                    $.ajax({ type: "POST",
                            url: "registerBd.php",
                            data: data, 
                            cache: false,
                            success: function(data) {
                            y = data.y;
                            console.log(y);
                            document.getElementById('ajaxDivOk').style.display = "block";
                            document.getElementById('ajaxDivOk').innerHTML = "Registo(s) gravado(s) com sucesso";
                            alert("Form submitted successfully.\nReturned json: " + y );
                          }
                        });
                }
        </script>

Thanks again for your help,

[Barand]

 

$response = "Registo gravado";

echo json_encode($response);

Why json_encode plain text?

 

As you are already using jquery, why use getElementById() and not

$('#ajaxDivOk').css("display", "block");
$('#ajaxDivOk').html("Registo(s) gravado(s) com sucesso");

[gmc1103]

Hi Barnad

 

With this code

<?php
header('Content-type: application/json');
error_reporting(E_ALL | E_NOTICE);
ini_set('display_errors', '1');
$mysqli = new mysqli('localhost', 'root', '', 'ebspma');
 
// Works as of PHP 5.2.9 and 5.3.0.
if ($mysqli->connect_error) {
    die('Connect Error: ' . $mysqli->connect_error);
}
$num= $_POST['num'];
$dia = $_POST['dia'];
$sala = $_POST['sala'];
$hora = explode(",", $_POST['hora']);
$a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17];
$c = array_fill_keys(array_keys(array_diff($a,$hora)),null) + $a;
ksort($c);
for ($i=0; $i<17; $i++){
$stmt = $mysqli->prepare("INSERT INTO `ebspma`.`sala_ocupacao` (id_dia, id_sala, id_tempo) VALUES (?, ?, ?);")or die(mysql_error($mysqli));
$horas = $c[$i];
echo $mysqli->error, PHP_EOL;
$stmt->bind_param('ssi', $dia, $sala, $horas);
if(!$stmt->execute()) echo $stmt->error;
$stmt->close();
}

and this function

<script>
             function postData(){
                var dia = document.getElementById('dia').value;
                var sala = document.getElementById('sala').value;
                var tempos = [];
                var s = document.getElementById('hora');
                for (var i = 0; i < s.options.length; i++) {
                    if (s.options[i].selected == true) {
                        var valores = s.options[i].value;
                        tempos.push(valores);
                    }
                }
                console.log(tempos);
                var num = document.getElementById('num').value;
                var data = 'dia='+ dia + '&sala='+ sala + '&hora='+ tempos + '&num='+ num;
                    $.ajax({ type: "POST",
                            url: "registerBd.php",
                            data: data, 
                            cache: false,
                            success: function(data) {
                            $('#ajaxDivOk').css("display", "block");
                            $('#ajaxDivOk').html("Registo(s) gravado(s) com sucesso");
                         }
                       });
                       return false;
                }
        </script>

Nothing is show...in my div...what i missing??

 

Regards and many thanks

[Barand]

Your PHP doesn't echo anything to go into the response (unless you have a query error).

[gmc1103]

No the query is working because the data is being inserted into the db.

 

using

echo "Gravado";

My function is not getting the echo

$.ajax({ type: "POST",
                            url: "registerBd.php",
                            data: data, 
                            cache: false,
                            dataType: 'html',
                            success: function(data) {
                            var dataR = data;
                            console.log(data);
                            console.log(dataR);
                            $('#ajaxDivOk').css("display", "block");
                            $('#ajaxDivOk').html("Registo(s) gravado(s) com sucesso");
                         }
                       });

the data variable doesn't receive anything....the console log is empty...i'm confused

[Barand]

You removed the echo. It used to be

 

echo json_encode($response);

 

So it should still

 

echo $response;

[gmc1103]

Hi Barand

 

Sorry...not even with this, i have changed  the php...to 

<?php
header('Content-type: application/json');
error_reporting(E_ALL | E_NOTICE);
ini_set('display_errors', '1');
$mysqli = new mysqli('localhost', 'root', '', 'ebspma');
 
// Works as of PHP 5.2.9 and 5.3.0.
if ($mysqli->connect_error) {
    die('Connect Error: ' . $mysqli->connect_error);
}
$num= $_POST['num'];
$dia = $_POST['dia'];
$sala = $_POST['sala'];
$hora = explode(",", $_POST['hora']);
$a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17];
$c = array_fill_keys(array_keys(array_diff($a,$hora)),null) + $a;
ksort($c);
for ($i=0; $i<17; $i++){
$stmt = $mysqli->prepare("INSERT INTO `ebspma`.`sala_ocupacao` (id_dia, id_sala, id_tempo) VALUES (?, ?, ?);")or die(mysql_error($mysqli));
$horas = $c[$i];
echo $mysqli->error, PHP_EOL;
$stmt->bind_param('ssi', $dia, $sala, $horas);
if($stmt->execute()){
   $response = "Registo gravado";
   echo ($response); 
}
else{
  echo $stmt->error;  
}
$stmt->close();
}

and the javascript is

$.ajax({ type: "POST",
                            url: "registerBd.php",
                            data: data, 
                            cache: false,
                            dataType:'html',
                            success: function(data) {
                            var dataR = data;
                            console.log(dataR);
                            console.log(data);
                            $('#ajaxDivOk').css("display", "block");
                            $('#ajaxDivOk').html("Registo(s) gravado(s) com sucesso");
                         }
                       });

Nothing is show...not even in console 

[gmc1103]

Hi Barand

 

Sorry...not even with this, i have changed  the php...to 

<?php
header('Content-type: application/json');
error_reporting(E_ALL | E_NOTICE);
ini_set('display_errors', '1');
$mysqli = new mysqli('localhost', 'root', '', 'ebspma');
 
// Works as of PHP 5.2.9 and 5.3.0.
if ($mysqli->connect_error) {
    die('Connect Error: ' . $mysqli->connect_error);
}
$num= $_POST['num'];
$dia = $_POST['dia'];
$sala = $_POST['sala'];
$hora = explode(",", $_POST['hora']);
$a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17];
$c = array_fill_keys(array_keys(array_diff($a,$hora)),null) + $a;
ksort($c);
for ($i=0; $i<17; $i++){
$stmt = $mysqli->prepare("INSERT INTO `ebspma`.`sala_ocupacao` (id_dia, id_sala, id_tempo) VALUES (?, ?, ?);")or die(mysql_error($mysqli));
$horas = $c[$i];
echo $mysqli->error, PHP_EOL;
$stmt->bind_param('ssi', $dia, $sala, $horas);
if($stmt->execute()){
   $response = "Registo gravado";
   echo ($response); 
}
else{
  echo $stmt->error;  
}
$stmt->close();
}

and the javascript is

$.ajax({ type: "POST",
                            url: "registerBd.php",
                            data: data, 
                            cache: false,
                            dataType:'html',
                            success: function(data) {
                            var dataR = data;
                            console.log(dataR);
                            console.log(data);
                            $('#ajaxDivOk').css("display", "block");
                            $('#ajaxDivOk').html("Registo(s) gravado(s) com sucesso");
                         }
                       });

Nothing is show...not even in console 

[gmc1103]

Sorry.....duplicate

 

And i have

echo json_encode($response);

[Barand]

Is the div visible?

[gmc1103]

<tr>
                      <td colspan="2" align="center" valign="middle" class="tg-6997">
                        <div class="col-sm-6">     
                            <div id="ajaxDivOk" style="display:none" class="alert alert-success"></div>	
                        </div>
                      </td>
                </tr>  

Well...now or i'm stupid or i don't see anything

 

In netbeans if i run normally....nothing is showed...

 

In netbeans if i run in debug mode...the div comes to light...

 

check the image in debug mode

 

http://postimg.org/image/qf6huodjb/

 

What's is going wrong??, does the main page do a quick refresh or something else??

[gmc1103]

Well after some experience i getting nut...

 

i receive  the error alert in my function but the data is inserted into database

 <script>
             function postData(){
                var dia = document.getElementById('dia').value;
                var sala = document.getElementById('sala').value;
                var tempos = [];
                var s = document.getElementById('hora');
                for (var i = 0; i < s.options.length; i++) {
                    if (s.options[i].selected == true) {
                        var valores = s.options[i].value;
                        tempos.push(valores);
                    }
                }
                console.log(tempos);
                var num = document.getElementById('num').value;
                var data = 'dia='+ dia + '&sala='+ sala + '&hora='+ tempos + '&num='+ num;
                    $.ajax({ 
                            type: "POST",
                            dataType: "json",
                            url: "registerBd.php",
                            data: data, 
                            cache: false,
                            success: function () {
                            console.log("Aqui");
                            $('#ajaxDivOk').css("display", "block");
                            $('#ajaxDivOk').html("Registo(s) gravado(s) com sucesso");
                         },
                            error:function(){
                            console.log("Aqui 2");
                            alert("something went wrong");
                            }
                       });
                       return false;
                }
        </script> 

 So...is not receiving the json_encode from my php....but why?

<?php
header('Access-Control-Allow-Origin: *');
header('Content-type: application/json');
ini_set('max_execution_time', 300);
error_reporting(E_ALL | E_NOTICE);
ini_set('display_errors', '1');
$mysqli = new mysqli('localhost', 'root', '', 'ebspma');
 
// Works as of PHP 5.2.9 and 5.3.0.
if ($mysqli->connect_error) {
    die('Connect Error: ' . $mysqli->connect_error);
}
$num= $_POST['num'];
$dia = $_POST['dia'];
$sala = $_POST['sala'];
$hora = explode(",", $_POST['hora']);
$a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17];
$c = array_fill_keys(array_keys(array_diff($a,$hora)),null) + $a;
ksort($c);
for ($i=0; $i<17; $i++){
$horas = $c[$i];
$stmt = $mysqli->prepare("INSERT INTO `ebspma`.`sala_ocupacao` (id_dia, id_sala, id_tempo) VALUES (?, ?, ?);")or die(mysql_error($mysqli));
$stmt->bind_param('ssi', $dia, $sala, $horas);
if(!$stmt->execute()) echo $stmt->error;
$stmt->close();
}
$response = "Registo gravado";
echo json_encode($response);
?>

Any help??

 

Thanks