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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




<!DOCTYPE html>
        <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">
            <body background="img/fondook.png">
                <script scr="js/main1.js"></script>
                <script type="text/javascript"></script>
    <header class="header2">
        <img src="img/logo_marcazeta.png">
                <h1 class="titulo1">INVENTARIO MARCAZETA LINEA</h1>
                    <table class="table" id="mitabla">
                        <thead class="titulo">
//$sql="SELECT * FROM mzt_reference";

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


Tags: Error - MySQL - PHP - PHP MySQLi - Warning - Web Votes: 0 - Answers: 2 - Views: 12 Share on: Google Facebook Twitter LinkedIn Link


  • Date: 06-06-2019 07:20:18 Verifica bien esas dos lineas...

    while ($mostrar=mysqli_fetch_array($query))


      Votes: 0 - Link answer
  • 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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