how get mysql data in php table

Question

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Browser window:
Connected to MySQL
Connected to Database

Notice: Undefined variable: link in C:\xampp\htdocs\globall\members.php on line 23

Warning: mysql_query() expects parameter 2 to be resource, null given in C:\xampp\htdocs\globall\members.php on line 23

this is line 23
$result = mysql_query($query,$link)

XML

<?php
mysql_connect("localhost", "admin", "") or die(mysql_error());

echo "Connected to MySQL<br />";

mysql_select_db("globall") or die(mysql_error());
echo "Connected to Database";

$query = " SELECT user_name,  mobile " .
        "FROM  customer ";


echo  "<br>" ;

$result = mysql_query($query,$link)
or die(mysql_error());

// this will count the total user in database
$num_users = mysql_num_rows($result);

$table_header =<<<EOD
<h2><center>GLoBALL Users Database </center></h2>

<table width="100%" border="1" cellpadding="2" cellspacing="2" align="center">
<tr>
<th>Username</th>
<th>Mobile</th>
</tr>

EOD;

$user_details = '';

while ($row = mysql_fetch_array($result)) {
$user_name = $row['user_name'];
$mobile = $row['mobile'];

$user_details .=<<<EOD
<tr>
<td>$user_name</td>
<td>$mobile</td>
</tr>
EOD;


$user_details .=<<<EOD
<tr>
<td>&nbsp;</td>
<td>Total : $num_users</td>
</tr>
EOD;
}
?>

Posted 21-Mar-12 1:56am

Peta2010

Updated 17-Oct-18 8:34am

v2

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Comments

Peta2010 21-Mar-12 8:59am

I modify my whole code according to W3schools and facing error on this line :
Parse error: syntax error, unexpected '}' in C:\xampp\htdocs\globall\members.php on line 32 which is
}
echo "</table>";

code start here-----

<tr>
<th>Username</th>
<th>Mobile</th>
</tr>";

while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row[$user_name]."</td>";
echo "<td>" . $row[$mobile]."</td>";
</tr>"
}
echo "</table>";

mysql_close($con);

?>

2 solutions

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Uday P.Singh · Accepted Answer · 2012-03-21T02:25:00

Solution 1

Try these:

Display the Result in an HTML Table[^]

How to Display MySQL Table Data[^]

hope it helps :)

Posted 21-Mar-12 2:25am

Uday P.Singh

Updated 21-Mar-12 2:45am

v2

Anand Ayyappan · Accepted Answer · 2012-03-26T04:01:00

Solution 2

Change the first line of code to

PHP

$link=mysql_connect("localhost", "admin", "") or die(mysql_error());

Thanks

Anand Ayyappan

Posted 26-Mar-12 4:01am

Anand Ayyappan