How can I display the results of a query in a one-page html table?
Below is what I was able to do. However, the code only shows the first record.
include("conectar.php");
$sql = mysql_query("Select * From tb_trabalhador and tb_detalhe_trabalhador");
$exibe = mysql_fetch_assoc($sql);
echo "<table>";
echo "<tr><td>Nome:</td>";
echo "<td>".$exibe["Nome"]."</td></tr>"; e os outros campos
When more than one result is returned by your query, use while to get all of them.
$sql = mysql_query("Select * From tb_trabalhador and tb_detalhe_trabalhador");
while($exibe = mysql_fetch_assoc($sql)){
echo $exibe['nome'] .'<br>';
}
I recommend using the mysql_error () to debug your code and get a bank error if this is a legacy project.
mysql_query($sql) or die(mysql_error());
It's highly recommended to use the PDO or mysqli to connect to the database, since mysql_* functions have already been deprecated and will soon be removed.
All mysql_fetch_ * functions return a single line and advance the internal cursor to the next record.
To get all the records, you need to use them within some repeat structure.
Example:
while ($exibe = mysql_fetch_assoc($sql) ) { // Obtém os dados da linha atual e avança para o próximo registro
echo $exibe["nome"];
}
As mentioned, consider using a PDO extension to manage communication between application
To get all the records you need a loopback:
include("conectar.php");
$sql = mysql_query("Select * From tb_trabalhador and tb_detalhe_trabalhador");
echo "<table>";
while($exibe = mysql_fetch_assoc($sql)){
echo "<tr><td>Nome:</td>";
echo "<td>".$exibe["Nome"]."</td></tr>";
}
echo "</table>";
It is important to note that queries should not be executed in a view. The correct thing is that you apply an MVC (Model View Controller) pattern and separate things. A good example, using the Framework Code Igniter, would be:
model.php
function carregar_trabalhador(){
$banco = new mysqli("servidor","usuário","senha");
$sql="Select * From tb_trabalhador and tb_detalhe_trabalhador";
$resultadoConsulta=$banco->query($sql);
$trabalhadores = array();
$i=0;
while (!empty($resultadoConsulta[$i])) {
$row = $resultadoConsulta[$i]=
$trabalhadores[] = $row;
$i++;
}
return $resultadoConsulta
}
controller.php
function exibir_trabalhador(){
$model= new Model();
$trabalhadores = $model->carregar_trabalhador();
$this->load->view("trabalhador.html", $trabalhadores);
}
trabalhador.html
<?php
foreach ($trabalhadores as $trabalhador) {
?>
<table>
<tr><td>Nome:</td>
<td><?= $trabalhador["Nome"] ?></td></tr>; e os outros campos
<?php }
?>
$connection = new mysqli("localhost", "user", "pass", "database");
$sql = $connection->query("Select * From tb_trabalhador and tb_detalhe_trabalhador");
if($sql){ // If $sql is True
while($exibe = $sql->fetch_assoc()){
foreach($exibe as $key => $value){
echo "<br />" . $value;
}
}
}
This form exempts you from using these html tags in your code. The foreach displays all information, without specifying what should be displayed, simplifying your code, eg. $ displays ["Name"].