Error updating image "Array to string conversion"

Question

Error updating image "Array to string conversion"

Navigation

#1 by (0 votes)

-1

I have a form where information is sent to the database, all are changed correctly except the image that is not changed.

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' enctype='multipart/form-data' 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

<?php
//inicia conexão com o banco 
include ("conexao.php");
$new_name = "";

 // recebe o codigo do registro 

$iden = isset($_POST['iden'])?$_POST['iden']:""; 

$iden = $_POST ['codigo']; 

$arquivo = isset($_FILES['arquivo'])?$_FILES['arquivo']:""; 
$arquivo = $_FILES ['arquivo']; 

// 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']))
   {
      $sql =  mysqli_query("SELECT * FROM eqp WHERE codigo = '$iden' 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

   }
$new_name = $_FILES['arquivo'];
$imagem = $new_name;

var_dump($imagem);
echo '<pre>';
print_r($imagem);
echo '</pre>';
$query=("UPDATE eqp SET arquivo= '$imagem' WHERE codigo='$iden'");

  $result = mysqli_query($conexao,$query);

  // Verifica se o comando foi executado com sucesso
  if(!$result)
    echo "Registro NÃO alterado.";
  else
    echo "Registro Alterado com sucesso.";
}
?>

Error occurred:

array(5) { ["name"]=> string(10) "images.jpg" ["type"]=> string(10) "image/jpeg" ["tmp_name"]=> string(24) "C:\xammp\tmp\phpE054.tmp" ["error"]=> int(0) ["size"]=> int(7225) }

Array
(
    [name] => images.jpg
    [type] => image/jpeg
    [tmp_name] => C:\xammp\tmp\phpE054.tmp
    [error] => 0
    [size] => 7225
)


Notice: Array to string conversion in C:\xammp\htdocs\agoravai\salva_ed.php on line 41
Registro Alterado com sucesso.

I have the var_dump($imagem); function that shows that it is receiving the image, but does not change in the database, what could be causing it?

php sql-update

asked by anonymous 14.09.2018 / 15:20

1 answer

"Submit" button, check the number of digits in the password and if it is the same as another password in html / javascript Display a string in a text field that is locked for editing

score 0 · Accepted Answer

When you try to concatenate the $ image variable within the string of your UPDATE command it is not possible because $ image is an array and not a string.

I think you're trying to get the name of the image. In this case, in salva_ed.php, instead of $new_name = $_FILES['arquivo']; , because $_FILES['arquivo'] is an array, just get the name of the file that is inside that array. So:

$new_name = $_FILES['arquivo']["name"];

Now, $ new_name is a string that can be passed to $ image and should no longer give concatenation problem in your UPDATE command.