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Hey, everything cool with you guys? I recently started a form where some information is sent to the database, but when I edit some information it changes everything except the image, before it only erased and left blank, now it does nothing... if someone can give a force I thank from now.
consultation page
<?php
//inicia a conexao com o banco
include_once("conexao.php");
//filtro para pesquisa por palavra chave
$filtro = isset($_GET['filtro'])?$_GET['filtro']:"";
//consulta slq
$sql = "SELECT * FROM eqp";
$consulta = mysqli_query($conexao,$sql);
$registros = mysqli_num_rows($consulta);
?>
<html lang ="pt-br">
<head>
<meta charset="utf-8">
<title>Filtro de Busca </title>
<link rel="stylesheet" href="../agoravai/edicao/form_consultate.css">
</head>
<body>
<div class="container">
<nav>
<ul class="menu">
<li></li>
<a href="inicio.php"><li>Inicio</li></a>
<a href="filtros.php"><li>Filtros:</li></a>
<a href="telalog_ed_cons.php"><li>Editar Dados:</li></a>
</ul>
</nav>
<section>
<h1>Consulta por Formularios Enviados Realizada:</h1>
<br>
<hr>
<?php
//verifica os itens do banco e os exibe
print" <h1>$registros Registros encontrados:</h1>";
while($exebirregistros = mysqli_fetch_array($consulta)){
$codigo = $exebirregistros[0];
$nome = $exebirregistros[1];
$setor =$exebirregistros[2];
$equipamento =$exebirregistros[3];
$numeroserie =$exebirregistros[4];
$numeropatrimonio =$exebirregistros[13];
$locais =$exebirregistros[5];
$motivo =$exebirregistros[6];
$enviando = $exebirregistros[7];
$recebendo =$exebirregistros[8];
$defeito =$exebirregistros[9];
$transporte =$exebirregistros[10];
$dataid =$exebirregistros[11];
$dataret =$exebirregistros[12];
$numeropatrimonio =$exebirregistros[13];
$arquivo =$exebirregistros[14];
$ordemdeservico =$exebirregistros[15];
//alerme para data de envio e entrega de equipamento
$data_atual = new DateTime(date('Y-m-d'));
$data_expiracao = new DateTime($dataret);
$intervalo_em_dias = $data_atual->diff($data_expiracao);
echo $intervalo_em_dias->format('<h4>%R%a DIAS PARA VENCER O RETORNO</h4>');
echo (" <table border='2px' cellpacing='15px'>
<tr>
<td>Codigo: </td>
<td>Nome:</td>
<td>Setor:</td>
<td>Equipamento:</td>
<td>Anexo N° ".$codigo.":</td>
<td>Numero de Serie:</td>
<td>Numero de Patrimonio:</td>
<td>Local de Envio:</td>
<td>Motivo Envio:</td>
<td>Quem esta Enviando:</td>
<td>Quem esta Recebendo:</td>
<td>Defeito:</td>
<td>Meio de Transporte:</td>
<td>Data ida:</td>
<td>Data Retorno:</td>
</tr> ");
echo ("
<tr>
<td>" . $codigo . " </td><br>
<td>" . $nome . " </td>
<td>" . $setor . " </td>
<td>" . $equipamento . " </td>
<td class= ok >" . "<img src='upload/".$arquivo."'width=50px height=50px /></td>
<td>" . $numeroserie . " </td>
<td>" . $numeropatrimonio . " </td>
<td>" . $locais . " </td>
<td>" . $motivo . " </td>
<td>" . $enviando . " </td>
<td>" . $recebendo . " </td>
<td>" . $defeito . " </td>
<td>" . $transporte . " </td>
<td>" . $dataid . " </td>
<td>" . $dataret . " </td>
</tr>
<td> <a href='ed_cons_.php?codigo=" . $codigo . "'>editar</td>
")
;
}
mysqli_close($conexao);
?>
<br>
</section>
</div>
</body>
</html>
ed_cons_.php
<?php
//inicia a conexao com o banco
include ("conexao.php");
?>
<html lang ="pt-br">
<html>
<head>
<meta charset="utf-8">
<title>Formulario</title>
<link rel="stylesheet" href="../agoravai/css/estilo.css">
</head>
<body>
<nav>
<ul class="menu">
<a href="../agoravai/inicio.php"><li>Inicio</li></a>
<a href="../aff/consultaeqp.php"><li>Consulta</a>
</ul>
</nav>
</body>
<?php
//recebe o codigo da pag de consulta
$iden = isset($_GET['iden'])?$_GET['iden']:"";
$iden = $_GET ['codigo'];
// consulta no banco de dados
$sql = "select * from eqp where codigo = '$iden'";
$consulta = mysqli_query($conexao,$sql);
$registros = mysqli_num_rows($consulta);
while($linhas = mysqli_fetch_array($consulta)){
//recebe os dados
$codigo = $linhas ['codigo'];
$arquivo = $linhas['arquivo'];
echo ("
<tr>
<form method='post' action='salva_ed.php''>
<td>Codigo:</td><td> <input type='text' name='codigo' value='" . $codigo . "'> </td>
<td>Arquivo em anexo: <input type='file' class= 'anexo' name='arquivo'> </td>
<br><br>
</tr>
<br><br>
<input type='submit' class='bnt salvar' value='Salvar'>
");
}
?>
</html>
salva_ed
<?php
//inicia conexão com o banco
include ("conexao.php");
// recebe o codigo do registro
$iden = isset($_POST['iden'])?$_POST['iden']:"";
// codigo registro
$iden = $_POST ['codigo'];
// aqui seria onde ele veifica se existe algo no campo arquivo e o substitui por um valor em branco
if($arquivo == ""){
$query =("update eqp set arquivo = '' WHERE codigo='$iden'");
}else{
// aqui seria onde fazia o update da imagem, dando um novo nome e movendo para a pasta de upload
if(isset($_FILES['imagem']))
{
$sql = mysql_query("SELECT * FROM eqp WHERE codigo = '$id' LIMIT 1");
$resultado = mysql_fetch_assoc($result);
date_default_timezone_set("Brazil/East"); //Definindo timezone padrão
$ext = strtolower(substr($_FILES['imagem']['name'],-4)); //Pegando extensão do arquivo
$new_name = $resultado['foto']; //Definindo um novo nome para o arquivo
$dir = 'upload'; //Diretório para uploads
move_uploaded_file($_FILES['imagem']['tmp_name'], $dir.$new_name); //Fazer upload do arquivo
}
$imagem = $new_name;
$query = mysql_query("UPDATE eqp set arquivo = '$imagem' WHERE codigo='$id'");
$result1 = mysqli_query($conexao,$sql,$query);
// Verifica se o comando foi executado com sucesso
if(!$result1)
echo "Registro NÃO alterado.";
else
echo "Registro Alterado com sucesso.";
}
?>
I think that problem is on this page where saved, because I saw that I would have to delete the image of the bank first to then upload but so far without success (on the editing page more variables are received however as I am only having trouble editing the image I left only the code and image to be displayed to facilitate understanding )
Hello, welcome! I believe the title of your question does not match what you are asking. because in fact you are not uploading an image in the database(blob) you just save a file name and are having problems with the algorithm. Vc can use the code
var_dump($imagem,$_FILES); exit;before the last update to check what is coming out of these variables, it is likely that it is returning an invalid value.– LeonanCarvalho
hello leonancarvalho, thank you so much for your attention!! Dude, I didn’t quite understand the code you sent me, what would be the function of it? I’m sorry about the haha snafu, but in time we learn.. i added it like you said, but still the same problem is giving this error "Notice: Undefined variable: file in C: xammp htdocs.
– noobphp
Quiet @noobphp! var_dump serves for you to debug the output of variables through a dump(screen print) more details here: https://answall.com/questions/172775/dif%C3%A7a-entre-var-dump-e-print-r It is a very useful function for debugging during development. As I imagined the reason you’re not saving is because your variable
$arquivois empty, review the treatment logic because it is incorrect.– LeonanCarvalho