Array compare

gmc1103 · April 27, 2015

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 · April 27, 2015

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

Barand · April 27, 2015

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 · April 27, 2015

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 · April 27, 2015

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

gmc1103 · April 27, 2015

$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

Edited April 27, 2015 by gmc1103

Barand · April 27, 2015

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

gmc1103 · April 27, 2015

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 · April 27, 2015

$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 · April 27, 2015

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 · April 27, 2015

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

gmc1103 · April 27, 2015

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 · April 27, 2015

You removed the echo. It used to be

echo json_encode($response);

So it should still

echo $response;

gmc1103 · April 27, 2015

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 · April 27, 2015

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 · April 27, 2015

Sorry.....duplicate

And i have

echo json_encode($response);

Barand · April 27, 2015

Is the div visible?

gmc1103 · April 27, 2015

<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 · April 27, 2015

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

Sign In

Array compare

Recommended Posts

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Join the conversation

Important Information