Save name and image extension and upload to a specific folder in php and then show in a table

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Save name and image extension and upload to a specific folder in php and then show in a table

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#2 by (0 votes)

1

When displaying a query in a table I create these three fields for the user to update to the line that shows the query data:

<?php
$tabela1 .= '<td> <input type="file" name= "Imagem['.$rows_cursos['Id'].']" value="'.$rows_cursos['Imagem'].'"></td>';

$tabela1 .= '<td> <input type="text" name= "Tratamento['.$rows_cursos['Id'].']" value="'.$rows_cursos['Tratamento'].'"></td>';

$tabela1 .= '<td> <input type="radio" name= "Id['.$rows_cursos['Id'].']" value="Pendente"> Pendente  <input type="radio" name= "Id['.$rows_cursos['Id'].']" value="Concluido">Concluido</td>';
?>

In this next step, I keep the path of the image in this way, but it does not save the complete path, it only saves the image name and the format "DSCF2712.JPG":

    <?php  
    if(isset($_POST['registar']))
    {
    $servername = "xxx.xxx.x.xx";
    $username = "xxxxx";
    $password = "xxxxxxx";
    $dbname = "xxxxxxxx";

    $conn = new mysqli($servername, $username, $password, $dbname);
    $conn->set_charset('utf8');

    $registro = $_POST['Id'];
    $imagem = $_POST['Imagem'];
    $tratamento = $_POST['Tratamento'];


    foreach($registro as $Id => $estado) { 


        $conn->query("UPDATE RegistoManutencao SET Estado='$registro[$Id]', Imagem = '$imagem[$Id]', Tratamento = '$tratamento[$Id]' WHERE Id='".$Id."'"); 
    } 


    }
    ?>

I've been searching and from what I realized the ideal is just save the name of the image and the extension, as I have, but now how do I show it? I'm doing it this way:

<?php
$result_cursos = "SELECT centrodb.RegistoManutencao.Imagem FROM centrodb.RegistoManutencao";

$resultado_cursos = mysqli_query($conn, $result_cursos);

$tabela1 .= '<div style="float: center" table align="center">';

$tabela1 .= '<table border="5">';

$tabela1 .= '<tr>';

$tabela1 .='<thead>';

$tabela1 .= '<tr>';

$tabela1 .= '<th>Imagem</th>';

$tabela1 .= '</tr>';

$tabela1 .='</thead>'; 

$tabela1 .='<tbody>';

    while($rows_cursos = mysqli_fetch_array($resultado_cursos)) {   

$tabela1 .= '<tr>';

$tabela1 .= '<td><img src="' .$rows_cursos['Imagem']. '" /></td>';

$tabela1 .= '</tr>'; 
}
$tabela1 .= '</tr>';

$tabela1 .='</tbody>'; 

$tabela1 .= '</table>';

$tabela1 .= '</div>';

echo $tabela1;

?>

And the result is this, does not show the image, but if you do inspect it shows the name and the extension that is in the database:

I think the problem is not being able to find the full path of the image

php mysql html

asked by anonymous 20.02.2018 / 10:05

2 answers

0

This is missing the path of the image,

$tabela1 .= '<td><img src="' .$rows_cursos['Imagem']. '" /></td>';

This would only work if your image is in the same folder as the file if you use an image folder structure that is next to the file you should do:

$tabela1 .= '<td><img src="/imagens/'.$rows_cursos['Imagem']. '" /></td>';

20.02.2018 / 12:47

Color background of row (row) in django admin Cros Origin + Spring + Angular 4/5

score 0 · Accepted Answer

In PHP

$registro = $_POST['Id'];

$tratamento = $_POST['Tratamento'];

//Pasta onde as imagens serão salvas
$pasta = "img";
//se não existir será criada
if (!file_exists($pasta)){
    mkdir("$pasta", 0777);
}

$diretorio = "img/";

    foreach ($registro as $Id => $estado) {

        $url = $diretorio .  $_FILES['Imagem']['name'][$Id];

        $nome_arquivo = $_FILES['Imagem']['name'][$Id];

        // salva as imagens na pasta
        move_uploaded_file($_FILES['Imagem']['tmp_name'][$Id], $url);

        // insere o nome com extensão no banco de dados
        $conn->query("UPDATE RegistoDiario SET Estado='$registro[$Id]', Imagem = '$nome_arquivo', Tratamento = '$tratamento[$Id]' WHERE Id='".$Id."'");

    }

No HTML

In the form tag you need to define the enctype as "multipart / form-data" which indicates that the form we are using will work with sending files

echo "<form method='POST' action='' enctype='multipart/form-data'>";

$tabela1 .= '<td><img src="img/' .$rows_cursos['Imagem']. '" /></td>';