sql query error using php

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sql query error using php

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<tr id="campo-select">
        <td>    
            <strong>Personas:</strong>
        </td>
        <td>
        <?php $consulta_listar_usuario = $usuario->listUserpordepartamento($conexion); ?>
            <span id="error-persona">
                <select class="browser-default" name="persona0" id="persona0" required>
                    <option value="0">-</option>
                    <?php while($resultado_listar_usuario = $consulta_listar_usuario->fetch_array(MYSQLI_ASSOC)): ?>
                                  <option value="<?php echo $resultado_listar_usuario['idUsuario'];?>"><?php echo $resultado_listar_usuario['personaNombre'];?> <?php echo $resultado_listar_usuario['apellido'];?>   <?php echo $resultado_listar_usuario['email'];?> <?php echo $resultado_listar_usuario['nombreDepartamento'];?></option>

                    <?php endwhile; ?>
                 </select>
            </span>
            <td><button type="button" class="btn" id="suma_persona" ><i class="fa fa-plus" aria-hidden="true"></i></button></td>
            <td><button type="button" class="btn" id="borra_persona" ><i class="fa fa-minus" aria-hidden="true"></i></button></td>
            <input type="hidden" name="cantidad_persona" id="cantidad_persona">
        </td>

    </tr>

I have an element in a table which brings me with an sql query the values of registered users.

I'm trying to make a filter that only brings me the fields that contain a specific id, like the one used by the user entered

function listUserpordepartamento($conexion){

        $consulta = (mysqli_query($conexion, "SELECT *, p.nombre as personaNombre, p.id as idPersona, d.nombre as nombreDepartamento, u.id as idUsuario, u.nombre as nombreUsuario
                                              FROM persona as p
                                              JOIN usuario as u
                                              on p.id = u.id_persona
                                              JOIN departamento as d
                                              on u.id_departamento = d.id
                                              ORDER BY p.nombre ASC")) or die("Error listando Usuarios: ".mysqli_error($conexion));

        return $consulta;
    }

I have errors in bringing the fields, I declare the variable privately

private $departamento;

and the department

function setDepartamento($departamento){

        $this->departamento = $departamento;
    }

I hope you understand, I already get involved. thanks for the help

sql php mysqli

asked by Juan Ortiz 06.04.2018 в 21:48

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2 answers

1

Seeing how you have your function and your tables is as follows:

function listUserForDepartament($conexion, $departamento){
    $consulta = (
        mysqli_query($conexion, 
        "SELECT p.nombre AS personaNombre, p.id AS idPersona, d.nombre AS nombreDepartamento, u.nombre AS nombreUsuario
            FROM persona AS p
            JOIN usuario AS u ON p.id = u.id_persona
            JOIN departamento AS d ON u.id_departamento = d.id
            WHERE d.id = ".$departamento.""))
            or die("Error listando usuarios: ".mysqli_error($conexion));
    return $consulta;
}

If you are bringing the id of departamento well, it should not cause you any inconvenience.

answered by 06.04.2018 в 22:04

Problem when printing a factorial series [closed] Spring MVC Error running builder

score -1 · Accepted Answer

$id_departamento = $resultado_usuario['id_departamento'] ;

had not declared at any time the ID_department of the person who entered the system

then

<tr id="campo-select">
        <td>    
            <strong>Personas:</strong>
        </td>
        <td>
        <?php $consulta_listar_usuario = $usuario->listUserpordepartamentodos($conexion, $id_departamento ); ?>
            <span id="error-persona">
                <select class="browser-default" name="persona0" id="persona0" required>
                    <option value="0">SELECCIONE A QUIEN NOTIFICAR</option>
                    <?php while($resultado_listar_usuario = $consulta_listar_usuario->fetch_array(MYSQLI_ASSOC)): ?>
                     <option value="<?php echo $resultado_listar_usuario['idUsuario'];?>"><?php echo $resultado_listar_usuario['personaNombre'];?> <?php echo $resultado_listar_usuario['apellido'];?>   <?php echo $resultado_listar_usuario['email'];?> <?php echo $resultado_listar_usuario['nombreDepartamento'];?></option>





                    <?php endwhile; ?>
                 </select>
            </span>

What I did was assign the names and place them in my method

to the end modify my fuction a bit as you recommend me @Eduardo Javier Maldonado thanks (and)

function listUserpordepartamentodos($conexion, $id_departamento){

        $consulta = (mysqli_query($conexion, "SELECT *, p.nombre as personaNombre, p.id as idPersona, d.nombre as nombreDepartamento, u.id as idUsuario, u.nombre as nombreUsuario
                                              FROM persona as p
                                              JOIN usuario as u
                                              on p.id = u.id_persona
                                              JOIN departamento as d
                                              on u.id_departamento = d.id
                                               WHERE  u.id_departamento = ".$id_departamento)) or die("Error listando Usuarios: ".mysqli_error($conexion));

        return $consulta;
    }