Posts by Eduardo Worrel • 126 points
9 posts
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1
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answers41
viewsA: Do not display variable value in the Laravel view
to run a php command without displaying it use @php @foreach ($respostas as $resposta) @if($reclamacao->id == $resposta->reclamacao_id) <td>{{$resposta->resposta}}</td> @php…
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answers57
viewsA: Is it possible to use jQuery effects with the <option> tag of a <select>?
Hi, it is possible yes. Try something like: <script> $(document).ready(function() { $(".cat").change(function() { if($(this).val() == 0){ $(".div").siblings().hide(); } if($(this).val() == 1){…
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answer41
viewsA: Fatal error: Call to Undefined Function query
all right? buddy very likely the php file that has "my functions" is not being included in the page. To confirm use in your main code include "MenuzinCandi.php"; include "conexao.php"; if…
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answer70
viewsA: Change image according to specific date
Good morning dear, I understand that your doubt is punctual so I will give a more objective answer, with some tips. // -> Two bars means a comment in javascript code, I will use it to explain…
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answer27
viewsA: Notice: Undefined index: pg in C: xampp htdocs pirataflixS Assets Movies.php on line 36
Good afternoon dear, I believe the following code solves the problem: $pg = 1; //default if(isset($_GET['pg'])){ $pg = $pg; }
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answer69
viewsA: change a foreach view according to the data it receives
Dear follows a suggested solution, I hope you understand, any doubt I reply again. First if you must confirm that the two values are being sent, I recommend using comma to concatenate //concatena os…
phpanswered Eduardo Worrel 126 -
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votes1
answer679
viewsA: Remove "[ ]" brackets from json return
Dear a solution suggestion. First a json must have key and value "a":"b" soon: return json_encode(array('Método inexistente!')); } } else { return json_encode(array('Classe inexistente!')); will not…
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answers263
viewsA: Assign a database value to a php variable
Good afternoon dear I have some suggestions. First the instance of the connection is called as follows <?php $link = mysqli_connect("127.0.0.1", "my_user", "my_password", "my_db"); if (!$link) {…
phpanswered Eduardo Worrel 126 -
0
votes2
answers60
viewsA: Cookie is not being created
Good afternoon dear, I have a suggestion. first create the following php file //arquivo setCookie.php <?php setcookie("Changed","Ok", time() + 172800, $path = "/"); echo 1; now with jquery relate…