put a column in the table to mark as view

Asked

Viewed 40 times

0

I have this query and show in a table:

<form method="POST" action="">
<fieldset>
 <table cellspacing="10">
 <tr>
   <td>
	<strong>Insira data de inicio:</strong> <input type="Date" name="inicio" placeholder="PESQUISAR">
	</td>
</tr>
 </table>
</fieldset>

<fieldset>
 <table cellspacing="10">
 <tr>
   <td>
	<strong>Insira data de fim:</strong> <input type="Date" name="fim" placeholder="PESQUISAR">
	</td>
</tr>
 </table>
</fieldset>
	<input type="submit" name="pesquisa" value="ENVIAR">
</form>

<?php  

$servername = "xxx.xxx.x.xx";
$username = "xxxxx";
$password = "xxxxxxxx";
$dbname = "xxxxxxx";

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

if(isset($_POST['inicio']) && isset($_POST['fim']))
{

    $inicio = $_POST['inicio'];
    $fim = $_POST['fim'];
    $result_cursos = "SELECT DataRegisto,
       Dia,
       TipoRefeicao,
       Refeicao,
       Hora,
       Motivo,
       Sugestao,
       Colaborador

FROM centrodb.RegistoDiario

WHERE DataRegisto >= '$inicio' AND DataRegisto <= '$fim'";
    $resultado_cursos = mysqli_query($conn, $result_cursos);

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

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

$tabela1 .= '<tr>';

$tabela1 .='<thead>';

$tabela1 .= '<tr>';

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

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

$tabela1 .= '<th>Tipo de Refeição</th>';

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

$tabela1 .= '<th>Hora da Refeição</th>';

$tabela1 .= '<th>Motivo do incumprimento</th>';

$tabela1 .= '<th>Sugestões/Observações</th>';

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

$tabela1 .= '</tr>';

$tabela1 .='</thead>'; 

$tabela1 .='<tbody>';

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

$tabela1 .= '<tr>';

$tabela1 .= '<td>'.$rows_cursos['DataRegisto'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['Dia'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['TipoRefeicao'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['Refeicao'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['Hora'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['Motivo'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['Sugestao'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['Colaborador'].'</td>';

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

$tabela1 .='</tbody>'; 

$tabela1 .= '</table>';

$tabela1 .= '</div>';

echo $tabela1;

?>

I want to add a column to this table to record in the database as a view.

Like the example I put this image, have a column to mark as view:

inserir a descrição da imagem aqui

php html mysql

  • how so 'how view'? and in which column in the first or second? if in the second table, in which query field that returns this information?

  • Do you have an image of how you would like it to look? Your question is confused...

  • Already put, an image as an example. is how to mark as completed when consulting the information to know that we have already seen that information

1 answer

0


<?php  

$servername = "xxx.xxx.x.xx";
$username = "xxxxx";
$password = "xxxxxxxx";
$dbname = "xxxxxxx";

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

if(isset($_POST['inicio']) && isset($_POST['fim']))
{

    $inicio = $_POST['inicio'];
    $fim = $_POST['fim'];
    $result_cursos = "SELECT DataRegisto,
       Dia,
       TipoRefeicao,
       Refeicao,
       Hora,
       Motivo,
       Sugestao,
       Colaborador

FROM centrodb.RegistoDiario

WHERE DataRegisto >= '$inicio' AND DataRegisto <= '$fim'";
    $resultado_cursos = mysqli_query($conn, $result_cursos);

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

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

$tabela1 .= '<tr>';

$tabela1 .='<thead>';

$tabela1 .= '<tr>';

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

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

$tabela1 .= '<th>Tipo de Refeição</th>';

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

$tabela1 .= '<th>Hora da Refeição</th>';

$tabela1 .= '<th>Motivo do incumprimento</th>';

$tabela1 .= '<th>Sugestões/Observações</th>';

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

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

$tabela1 .= '</tr>';

$tabela1 .='</thead>'; 

$tabela1 .='<tbody>';

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

$tabela1 .= '<tr>';

$tabela1 .= '<td>'.$rows_cursos['DataRegisto'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['Dia'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['TipoRefeicao'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['Refeicao'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['Hora'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['Motivo'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['Sugestao'].'</td>';

$tabela1 .= '<td>'.$rows_cursos['Colaborador'].'</td>';

$tabela1 .= '<td> <input type='checkbox' name= 'Vista'> </td>';

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

$tabela1 .='</tbody>'; 

$tabela1 .= '</table>';

$tabela1 .= '</div>';

echo $tabela1;

  • Thanks, now I just need you to do the update in the database table

Browser other questions tagged php html mysql

You are not signed in. Login or sign up in order to post.