Mysqli While loop: print variables inside html

Question

My php While loop run the query, but the results must be print inside html. In this moment I unknow the way to make this:

My php while loop

<?php
include "connect.php";
$username=$_SESSION['Username'];

$result = mysqli_query($dbconn,"
 SELECT *
  FROM books
 WHERE username = '$Username'
");
while($rows = mysqli_fetch_array($result));
?>

After this code there is a Html code where I want print the variables:

<a href="editBook.php?bookid=<?php echo $book_id; ?>&book_name=<?php echo $book_name; ?>">Edit</a>

In this moment the variable is empty

How to fix this?

[Resolved] Update I have resolve my problem. This is the correct php script. Work fine:

<?php
include "connect.php";
$username=$_SESSION['Username'];

$result = mysqli_query($dbconn,"
 SELECT *
  FROM books
 WHERE username = '$Username'
");
global $book_id, $book_name
while($row = mysqli_fetch_array($result)) {
$book_id = row['book_id'];
$book_name = row['book_name'];
?>

Outside while loop. Print variable inside Html:

<?php echo $row['book_id']; ?> <br>
<?php echo $row['book_name']; ?>

Close while loop and connection:

<?php
}
mysqli_close($dbconn);
?>

1st of all, you should really think about prepared statements, then, add error_reporting(E_ALL); ini_set('display_errors', 1); on top of the page, and see what error shows up. Then, echo $rows['my_column_name_whatever_is_name_is'] and see what it returns — OldPadawan
– OldPadawan, Commented Apr 8, 2017 at 19:19
@OldPadawan he is not searching for an error, he is looking for a solution. — Aloso
– Aloso, Commented Apr 8, 2017 at 19:21

OldPadawan · Accepted Answer · 2017-04-08 19:47:37Z

1

with prepared statements :

<?php
session_start();
$username = $_SESSION['Username'];

error_reporting(E_ALL);
ini_set('display_errors', 1);

include"config.inc.php";

/* connect to DB */
$mysqli = mysqli_connect("$host", "$user", "$mdp", "$db");

if (mysqli_connect_errno()) { echo "Error connecting : " . mysqli_connect_error($mysqli); }

$query = " SELECT * FROM books WHERE username=? ";
$stmt = $mysqli->prepare($query);
$stmt->bind_param("s", $username);
$results = $stmt->execute();
$stmt->bind_result($book_id, $book_name);
$stmt->store_result();

if ($stmt->num_rows > 0) {
while($stmt->fetch()){
?>
<p><?php echo"$book_name"; ?> > <a href="editBook.php?bookid=<?php echo"$book_id"; ?>&book_name=<?php echo"$book_name"; ?>">Edit</a></p>
<?php
}
}
else
{ echo"[ no data ]"; }
?>

edited Apr 8, 2017 at 19:47

answered Apr 8, 2017 at 19:39

OldPadawan

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Comments

Rick James · Accepted Answer · 2017-04-09 15:41:06Z

1

(Rewriting)

The real issue is:

while ($rows = ...) ;

This loops until $rows is NULL. So there is nothing to display afterwards.

Since you are fetching the entire array in a single function call, there is no need for the loop!

$rows = ...;

But then you need to reference the first(?) row to get the desired data:

$row = $rows[0];

So, another approach is to just fetch one row, then close the fetch process.

edited Apr 9, 2017 at 15:41

answered Apr 9, 2017 at 4:21

Rick James

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Comments

Kuntal Gupta · Accepted Answer · 2017-04-08 19:31:48Z

0

In your html you have to do this

<a href="editBook.php?bookid=<?php echo $rows['book_id']; ?>&book_name=<?php echo $rows['book_name']; ?>">Edit</a>

answered Apr 8, 2017 at 19:31

Kuntal Gupta

116 bronze badges

Comments

Aloso · Accepted Answer · 2017-04-08 20:40:41Z

0

As there might be multiple books for each user, you have to print the link inside the while loop, or store it in a string:

<?php
include "connect.php";
$username = $_SESSION['Username'];

$result = mysqli_query($dbconn,"
    SELECT * FROM books
    WHERE username = '$Username'
");

$links = ""; // all links are stored in this string
while($rows = mysqli_fetch_array($result)) {
    // I assume that the columns are called `id` and `name`
    $links .= '<a href="editBook.php?bookid='. $rows["id"] .'&book_name='. $rows["name"] .'">Edit '. $rows["name"] .'</a>';
}
?>

In the html code simply write

<?php echo $links ?>

Note that you should use prepared statements instead. You should also take a look at the object oriented way to use mysqli using the mysqli class.

edited Apr 8, 2017 at 20:40

answered Apr 8, 2017 at 19:38

Aloso

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3 Comments

OldPadawan Over a year ago

your code lacks a session_start, and all links will be stuck together, not knowing which one to edit, just a link... :/

Aloso Over a year ago

No, my code does not lack a session start. I modified the code above which does not have a session_start either. Most likely, the session is started in connect.php. You're right that the links should have a more descriptive name, I corrected it.

OldPadawan Over a year ago

we now lack an answer to his last comment :)

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