Fatal error: Uncaught ArgumentCountError: 3 arguments are required, 2 given in /var/www/html/cocoa/index.php:13 Stack trace: #0 /var/www/html/cocoa/index.php(13): printf('%s %s\n', '<a href='cocoa_...') #1 {main} thrown in /var/www/html/cocoa/index.php o

[bertrc]

 

printf error line

<?php

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

$mysqli = mysqli_connect("localhost", "root", "P-11fl32fg14", "cocoa");

$query = "SELECT type_chocolate.type_chocolate,type_chocolate.type_chocolate FROM type_chocolate";

$result = mysqli_query($mysqli, $query);

/* numeric array */

while ($row = mysqli_fetch_array($result, MYSQLI_BOTH))

{

printf("%s %s\n","<a href='cacau_type_chocolate.php?type_chocolate=" . $row['type_chocolate'] . ">,". $row['type_chocolate'] . "</a>");?> <br>

<?php

}

?>

 

[Barand]

https://www.php.net/print_f

You don't need to select the same column twice.

The %s in the print_f format argument are placeholders for the string variables ($row['type_chocolate'])

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$mysqli = mysqli_connect("localhost", "root", "P-11fl32fg14", "cocoa");

$query = "SELECT type_chocolate FROM type_chocolate";

$result = mysqli_query($mysqli, $query);

while ($row = mysqli_fetch_assoc($result))

{
    printf("<a href='cacau_type_chocolate.php?type_chocolate=%s'>%s</a>", $row['type_chocolate'], $row['type_chocolate']);
}

 

[bertrc]

it works

printf("%s %s\n","<a href='cacau_tipus_xocolata.php?tipus_xocolata=" . $row["tipus_xocolata"] . "'>" , $row["tipus_xocolata"] . "</a>");?> <br>

 

[Barand]

Probably the wierdest (and worst) way of using printf that I have ever seen.

[Phi11W]

Or even, delving a little deeper into the Documentation:  

printf( '<a href="cacau_type_chocolate.php?type_chocolate=%1$s">%1$s</a>', $row[ 'type_chocolate' ] );

Regards, 
   Phill  W.