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Juan Mejía Karma: 15 Colombia
 
 Date: 06-06-2019 06:59:03 (In Spanish)

Error cuando voy a realizar un inner join: Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result[Resolved]

Buenas, tengo el siguiente código, cuando lo ejecuto.... me sale el siguiente error y no lo he podido arreglar... quien me ayuda

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given in 


<?php

$conexion=mysqli_connect('localhost','root','','pms24_db');

?>

<!DOCTYPE html>
    <html>
        <head>
        <meta charset="UTF-8">
        <meta name="viewport" content="width=device-width, user-scalable=no, initial-scale=1.0, maximum-scale=1.0, minimum-scale=1.0">
            <link rel="stylesheet" href="css/estilos.css">
                <title>INVENTARIO</title>
            <body background="img/fondook.png">
                <script scr="js/main1.js"></script>
                <script type="text/javascript"></script>
              
         </head>
         
    <header class="header2">
        <img src="img/logo_marcazeta.png">
            <center>
                
                <h1 class="titulo1">INVENTARIO MARCAZETA LINEA</h1>
                    <table class="table" id="mitabla">
                        <thead class="titulo">
                              <tr>
                                <th>CLIENTE</th>
                                <th>REFERENCIA</th>
                                <th>DESCRIPCION</td>
                                <th>MODELO</th>
                                <th>CANTIDAD</th>
                                <th>PRECIO</th>
                             </tr>
                        </thead>
     </header>
     
<?php 
//$sql="SELECT * FROM mzt_reference";
//$result=mysqli_query($conexion,$sql);

$query="SELECT r.reference, r.description, r.modelo , r.modelo, r.precio, c.client_id FROM mzt_reference r INNER JOIN mzt_client c  ON  r.client_id=c.client_id";
                //$consulta=$conexion mysqli_query($conexion,$sql);
                $consulta=$conexion->query($query);
                while ($mostrar=mysqli_fetch_array($query))
                    {
          echo'
            <tr>
                <td style="font-size: 19px">'.$mostrar['reference'].'</td>
                <td style="font-size: 19px">'.$mostrar['description'].'</td>
                <td style="font-size: 19px">'.$mostrar['modelo'].'</td>
                <td style="font-size: 19px">'.$mostrar['modelo'].'</td>
                <td style="font-size: 19px">'.$mostrar['precio'].'</td>
                <td style="font-size: 19px">'.$mostrar['name'].'></td>
            </tr>
           ';
                    }


        ?>
      </table>

</center>
</body>
</html>.
Tags: Error - MySQL - PHP - PHP MySQLi - Warning - Web Votes: 0 - Answers: 2 - Views: 12 Share on: Google Facebook Twitter LinkedIn Link
 

Answers:

  •  
      0  
     
     
    Fernando Jimenez Karma: 599 Mexico
     
    Date: 06-06-2019 07:20:18 Verifica bien esas dos lineas...

    $consulta=$conexion->query($query);
while ($mostrar=mysqli_fetch_array($query))

    Saludos....


    Stryfe™       Votes: 0 - Link answer
     
  •  
      2  
     
     
    Juan Vasquez Karma: 952 Guatemala
     
    Date: 09-06-2019 14:23:55 La línea que dice
    while ($mostrar=mysqli_fetch_array($query))


    debe ser

    while ($mostrar=mysqli_fetch_array($consulta))
       Votes: 2 - Link answer
     
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