CODEWITHSUNDEEP
phpmysqlhtmlsql
dead beef
dead beef

Reputation: 683

Having issues with displaying data retrieved from a MySQL database with PHP

I'm very new to PHP; and am doing a few sample projects to help me learn, this project was done for fun. There's probably some idiotic and highly unintelligibly created code in here. Considering I'm a newbie, there's probably a syntax error but I've been searching for hours without any luck.

Basically I'm making a extremely simple comment/shoutbox system in which the user who posts the data is kept anonymous. I'm doing things separate at the moment; then I plan on combining them into one. I know that the submitting part is working correctly as I've checked the MySQL database via PHPMyAdmin, however I can't seem to get the data to display right. I've searched for over 2 hours for my simple syntax mistake, and have given up deciding to bring it to a reliable and friendly community.

I've removed all of the connection information to my database, as I know all of this is correct and I'd rather not expose it on a public 'forum'.

<html>
<body>
<?php
$username="";
$password="";
$database="";

mysql_connect("localhost",$username,$password);
mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM data";
$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();
?>
<table border="2" cellspacing="2" cellpadding="2">
<tr>
<th><font face="Arial, Helvetica, sans-serif">Name(No specific order)</font></th>
</tr>

<?php
$i=0;
while ($i < $num) {

$    name=mysql_result($result,$i,"");
?>

<tr>
<td><font face="Arial, Helvetica, sans-serif"><?php echo $name; ?></font></td>
</tr>

<?php
$i++;
}
?>
</body>
</html>

I have two tables, id and name. ID is a automatic increment INT, and name is set up a text. ID is also the primary of the database. The submitting system is working perfectly fine, however the displaying isn't.

When I view the page, where there should be text($name), instead I get a blank area, and no text. This is what I received when I visited the page, with only one database entry:

<html>

<body>

<table border="2" cellspacing="2" cellpadding="2">

<tr>

<th><font face="Arial, Helvetica, sans-serif">Name(No specific order)</font></th>

</tr>





<tr>

<td><font face="Arial, Helvetica, sans-serif"></font></td>

</tr>



</body>

</html>

Any help would be appreciated, thanks.

Upvotes: 0

Views: 1468

Answers (3)

Michael
Michael

Reputation: 3141

I would recommend not using the older mysql_*() functions. You could do something like this instead (via PDO_mysql extension):

<?php
    $pdo = new PDO('mysql:host=localhost;dbname=YOURDBNAME', $username, $password);
    $query = "SELECT * FROM data";
    $stmt = $pdo->prepare($query);
    $stmt->execute(array());
    $results = $stmt->fetchAll(PDO::FETCH_ASSOC);

Then later, you would loop through with something like:

    foreach ($results as $res) {
        echo $res['column_name'];
    }

Upvotes: 3

user319198
user319198

Reputation:

Something is wrong in while loop where you trying to display names.

Replace with below : 

 <?php
    if($num >0) {
       while ($row=mysql_fetch_assoc($result)) {
         ?>
        <tr>
           <td><font face="Arial, Helvetica, sans-serif"><?php echo $row['name']; ?></font></td>
        </tr>
        <?php

        }
      }
  ?>

change :

$num=mysql_numrows($result);

to

$num=mysql_num_rows($result);

Upvotes: 2

sikander
sikander

Reputation: 2286

You have the query result from MySQL but you're not looping through it properly. Try:

$result=mysql_query($query);
if(mysql_num_rows($result) > 0)
{
    while ($row = mysql_fetch_assoc($result))
    {
        // print data
    }
}

Upvotes: 1

Related Questions