Error: Display data form database to webpage

Question

I want to display my database into my webpage in tables. but it cant, i pretty sure that my database name is same. if i run, ".$no.", ".$row['nama'].", ".$row['email']." and ".$row['message']." will displayed on webpage. not data from database. please help.

here is my code:

<table class="table">
    <thead>
    <tr>
     <th>#</th>
     <th>Nama</th>
     <th>E-mail</th>
     <th>Comment</th>
    </tr>
    </thead>
     <tbody>
        <tr>
          <td>1</td>
          <td>Mark</td>
          <td>Otto</td>
          <td>@mdo</td>
        </tr>
     </tbody>
<?php 
include "conection.php";
$no = 1;
$query = mysql_query("SELECT * FROM comment");
if ($query) {
    while ($row = mysql_fetch_array($query)) {
        echo "
        <tr>
            <td>".$no."</td>
            <td>".$row['nama']."</td>
            <td>".$row['email']."</td>
            <td>".$row['message']."</td>
        </tr>
        ";

    $no++;
    }
}
?>
</table>

the error will be like this:

this is my conection.php

<?php
$db_host = "localhost";
$db_user = "root";
$db_pass = "";
$db_name = "toefl";

$koneksi = mysqli_connect($db_host, $db_user, $db_pass, $db_name);

if(mysqli_connect_error()){
    echo 'database error : '.mysqli_connect_error();
}
?>

just test this code <?php $num = 1; echo $num; ?> it will display 1 — Ramin Darvishov
– Ramin Darvishov, Commented May 1, 2016 at 13:07
i sure, because if i try that php-code in other program, it will work — Endone
– Endone, Commented May 1, 2016 at 13:12

Death-is-the-real-truth · Accepted Answer · 2016-05-01 13:33:33Z

2

You are mixing mysql_* with mysqli_*. Please don't do that. Check the below code (changes are commented):-

<table class="table">
    <thead>
    <tr>
     <th>#</th>
     <th>Nama</th>
     <th>E-mail</th>
     <th>Comment</th>
    </tr>
    </thead>
     <tbody><!-- check the change -->
<?php 
include "conection.php";
$no = 1;
$query = mysqli_query($koneksi,"SELECT * FROM comment"); // don't mix `mysql_*` with `mysqli_*`
if ($query) {
    while ($row = mysqli_fetch_array($query)) {// don't mix `mysql_*` with `mysqli_*`
        echo "<tr><td>".$no."</td><td>".$row['nama']."</td><td>".$row['email']."</td><td>".$row['message']."</td></tr>";
        $no++;
    }
}
?>
</tbody>
</table>

Note:- this code file extension must be .php.

edited May 1, 2016 at 13:33

answered May 1, 2016 at 13:20

Death-is-the-real-truth

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Comments

Death-is-the-real-truth · Accepted Answer · 2016-06-16 11:53:48Z

2

You forget a i. Change mysql_query() to mysqli_query() and mysql_fetch_array() to mysqli_fetch_array().

edited Jun 16, 2016 at 11:53

Death-is-the-real-truth

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answered May 1, 2016 at 13:19

Steffen Grahl

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Comments

Ramin Darvishov · Accepted Answer · 2016-05-01 13:25:06Z

0

Chnage conection.php. Change all mysqli methods to mysql

<?php
$db_host = "localhost";
$db_user = "root";
$db_pass = "";
$db_name = "toefl";

$koneksi = mysql_connect($db_host, $db_user, $db_pass);

if (!$koneksi) {
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully'
mysql_select_db($db_name, $koneksi) or die('Could not select database.');;
?>

answered May 1, 2016 at 13:25

Ramin Darvishov

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