how to associate two tables in query? [duplicate] - php

Question

This question already has answers here:
Object of class mysqli_result could not be converted to string
(5 answers)
Closed 1 year ago.
I saved a value as a cookie and then checked if there exists in the perfiles_vinculados table to get all the data that has the same id in the perfil table.
Then I create an array of the $vinculado result and show it in a HTML table as a row.
The problem is that the console returns:
Catchable fatal error:
Object of class mysqli_result could not be converted to string in
C:\xampp\htdocs\miramonteapp\api\modal.php
The cookie:
document.cookie = "vinculaciones=" + $("#mod_id_perfil").val();
PHP:
//querys
<?php
include 'api/conexion.php';
$ides = $_COOKIE['vinculaciones'];
$juridicos = "SELECT perfil_juridica FROM perfiles_vinculados where perfil_fisica = '$ides'";
$con = mysqli_query($conexion, $juridicos);
$vinculado = mysqli_query($conexion, "SELECT * FROM perfil where id = '$con'");
?>
//table
<?php
while($reg = mysqli_fetch_array($vinculado)) {
$id = $reg['id'];
?>
<tr id="<?php echo " tr_ ".$reg['id']; ?>">
<td class="" data-id="<?php echo $reg['usuario'] ?>">
<?php echo $reg['nombre']; ?>
</td>
<td class="" data-id="<?php echo $reg['usuario'] ?>">
<?php echo $reg['cuit']; ?>
</td>
<td class="td-actions text-right">
<button type="button" rel="tooltip" class="btn btn-danger">
<i class="material-icons">close</i>
</button>
</td>
<?php } ?>

Answer 1

You have to learn about join sql statement.
As for you current approach, first you need to fetch perfil_juridica value from a result of $juridicos execution and then pass this value to your second query:
// first query
$juridicos = "SELECT perfil_juridica FROM perfiles_vinculados where perfil_fisica = '$ides'";
$result = mysqli_query($conexion, $juridicos);
$row = mysqli_fetch_array($result);
$perfil_juridica = $row['perfil_juridica'];
// second query
$vinculado = mysqli_query($conexion, "SELECT * FROM perfil where id = '$perfil_juridica'");
What you should do next is move to prepared statements instead of putting unsafe values into query texts. This question will help you.