Relationship between Join tables

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I'm trying to do a JOIN (it's the first time) and I have this code but it shows me the following errors.

  

Warning: mysqli_query () expects at least 2 parameters, 1 given in C: \ xampp \ htdocs \ inventory \ int.php on line 5

     

Warning: mysqli_fetch_array () expects parameter 1 to be mysqli_result, null given in C: \ xampp \ htdocs \ inventory \ int.php on line 7 

<?php
require("connect_db.php");

$query = "SELECT DISTINCT id_usuarios, nombre, ap, am, fecha, localidad, departamento FROM usuarios";
$res =mysqli_query($query);
$option = '';
while ($row = mysqli_fetch_array($res)){

    $option.="option value=\"$row[id_usuarios]\">$row[nombre] </option>";   
    $option.="option value=\"$row[id_usuarios]\">$row[ap] </option>";
    $option.="option value=\"$row[id_usuarios]\">$row[am] </option>";
    $option.="option value=\"$row[id_usuarios]\">$row[fecha] </option>";
    $option.="option value=\"$row[id_usuarios]\">$row[localidad] </option>";
    $option.="option value=\"$row[id_usuarios]\">$row[departamento] </option>";
}
?>

<select>
<option value="-">Selecciona el usuario</option>
<?php echo $option; ?>
</select>

File connect_db.php 

<?php
$mysqli = new MySQLi("", "","", "");
if ($mysqli -> connect_errno) {
  die( "Fallo la conexión a MySQL: (" . $mysqli -> mysqli_connect_errno() . ") " . $mysqli -> mysqli_connect_error());
} else //echo "Conexión exitossa!";

//  $link =mysqli_connect("localhost","","");
//  if($link){
//  mysqli_select_db($link,"academ");
// }
?>
    
asked by isac martinez 06.04.2017 в 21:46
source

1 answer

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Errors:

  • Viewing your file connect_db.php , we can see that you have decided to use the style Orientado a objeto of the library mysqli . But when you execute both the query and the fetch , you are using the style by procedimientos

  • In the while loop, you have not successfully opened the tags options

Solutions:

  • Decide on a style, in this case I will choose the orientado a objeto , that is why we are going to use:
    • $mysqli->query instead of mysqli_query .
    • $res->fetch_assoc() instead of mysqli_fetch_array .
  • Open option correctly, that is, <option...

Example: 

<?php
require("connect_db.php");

$query = "SELECT DISTINCT id_usuarios, nombre, ap, am, fecha, localidad, departamento FROM usuarios";
$res = $mysqli->query($query);
$option = '';
while ($row = $res->fetch_assoc()){

    $option.="<option value=\"$row[id_usuarios]\">$row[nombre] </option>";   
    $option.="<option value=\"$row[id_usuarios]\">$row[ap] </option>";
    $option.="<option value=\"$row[id_usuarios]\">$row[am] </option>";
    $option.="<option value=\"$row[id_usuarios]\">$row[fecha] </option>";
    $option.="<option value=\"$row[id_usuarios]\">$row[localidad] </option>";
    $option.="<option value=\"$row[id_usuarios]\">$row[departamento] </option>";
}
?>

<select>
<option value="-">Selecciona el usuario</option>
<?php echo $option; ?>
</select>
    
answered by 06.04.2017 / 22:17
source
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