Export Excel PHP Workbench Variable data

Question

I’m having doubts regarding generating an excel with the query.

<?php
include("conectar.php");
$id = $_GET['id'];
$arquivo = 'Não autorizado .xls';

First question is this: In the $file part I can put a variable to rename the file?

// -- Cabeçalho do arquivo -------------------------------------
$html .= '<tr><td align="center"><b>Instalador</b></td>';
$html .= '<td align="center"><b>Morada</b></td>';
$html .= '<td align="center"><b>Email</b></td>';
$html .= '<td align="center"><b>Validade</b></td>';

</tr>';

for($i=1; $i<=$aux; $i ++){
$sql = mysql_query("SELECT tb_trabalhador.Nome,Morada,Email,Validade from             
          tb_detalhe_trabalhador inner join tb_trabalhador on 
          tb_detalhe_trabalhador.id = tb_trabalhador.id inner join 
          tb_equipamentos on tb_detalhe_trabalhador.id = tb_equipamentos.id 
          inner join tb_funcoes on tb_detalhe_trabalhador.id = tb_funcoes.id
          WHERE tb_trabalhador.id = $id = ".$i) or die(mysql_error());
          $row=mysql_fetch_array($sql);
          $html .= '<tr><td>'.$row[0].'</td><td>'.$row[1].'</td>
          <td>'.$row[2].'</td><td>'.$row[3].'</td></tr>';
  }
  $html .= '</table>';

The second question is: I will export data to Excel with many dates. It is possible for me to export the data to Excel and where the dates have been passed in relation to today’s day become a red rectangle?

Answer 1

In the Code

<?php
include("conectar.php");
$id = $_GET['id'];
$arquivo = $_GET['arquivo'];

you can use another $_GET['arquivo'] and assign the variable $arquivo so that you can put the name of your preference. NOTE: I would filter_input for security reasons.

Example with filter_input

<?php
    include("conectar.php");
    $id      = filter_input(INPUT_GET, 'id', FILTER_SANITIZE_NUMBER_INT)
    $arquivo = filter_input(INPUT_GET, 'arquivo', FILTER_SANITIZE_STRING)

And in the code, apparently it’s the Validity field so it’s the row[3] I compared the current date with the date coming from the bank (date('Y-m-d')), there’s no way I know if the code has an error, I just did the actual processing.

<?php

// -- Cabeçalho do arquivo -------------------------------------
$html .= '<tr><td align="center"><b>Instalador</b></td>';
$html .= '<td align="center"><b>Morada</b></td>';
$html .= '<td align="center"><b>Email</b></td>';
$html .= '<td align="center"><b>Validade</b></td>';

$html .= '</tr>';

for($i=1; $i<=$aux; $i ++){
        $sql = mysql_query("SELECT tb_trabalhador.Nome,Morada,Email,Validade from             
              tb_detalhe_trabalhador inner join tb_trabalhador on 
              tb_detalhe_trabalhador.id = tb_trabalhador.id inner join 
              tb_equipamentos on tb_detalhe_trabalhador.id = tb_equipamentos.id 
              inner join tb_funcoes on tb_detalhe_trabalhador.id = tb_funcoes.id
              WHERE tb_trabalhador.id = $id = ".$i) or die(mysql_error());

          $row=mysql_fetch_array($sql);       

          $html .= '<tr><td>'.$row[0].'</td><td>'.$row[1].'</td><td>'.$row[2].'</td>';
          //SE FOR DATA IGUAL A DE HOJE PINTE AS BORDAS DE VERMELHO
          if (date('Y-m-d') == $row[3]){
            $html .= '<td bordercolor="#FF0000" style="border:1px solid #FF0000; border-color:#FF0000">'.$row[3].'</td>';
          } else {
            $html .= '<td>'.$row[3].'</td>';
          }
          $html .= '</tr>';

  }
$html .= '</table>';