sql query output in a table.

Soumen

:bemused: 😕
I am trying to list first 2 column of a table from my database within table. Has th following HTML+PHP.

But it does not show anything in the browser.
Can anybody help me pointing where I am wrong.

Regards.

HTML+PHP Code.

<html>
<head>
<title>PHP Test 2</title>
</head>
<body>
<?php

// Connecting, selecting database
$usr = 'root';
$pwd = 'root';
$host = '155.1.99.98:3306';
$db = 'testdb';
$link = mysql_connect($host,$usr,$pwd) or die("Could not connect: " . mysql_error());
echo 'Connected successfully';
mysql_select_db($db) or die ('Could not select database');

// Performing SQL query
$query = 'SELECT * FROM usercatmst';
$result = mysql_query($query) or die('Query failed in module 3.php: '. mysql_error());
if (!$result) {
echo "Could not successfully run query ($sql) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) {
echo "No rows found, nothing to print so am exiting";
exit;
}
// Printing results in HTML
echo "<table bgcolor=\"#BBFFFF\ ">\n";
while ($row = mysql_fetch_row($result)){
$mcatid = $row[0];
echo "\n<tr>";
echo "<td>$row[0]</\td>";
echo "<td>$row[1]</\td>";
echo "\n</\tr>";
}
echo "\n</\table>";

// Free resultset
mysql_free_result($result);

// Closing connection
mysql_close($link);
?>

</body>

paulnaj

Hi there,

I fiddled about with your code and put some comments in ... could still be a typo of mine in there (but I can't fully debug at this end) ...

<?php
/*
	You script had errors in outputting the table below.
	If no error messages appeared, try putting the next 
	line at the top of your code when debugging, at least.
*/
ini_set('error_reporting', E_ALL);
?>
<html>
<head>
<title>PHP Test 2</title>
</head>
<body>
<?php

// Connecting, selecting database

/*
	I suppose these are just test data, but
	be careful not to show sensitive info on
	the forum
*/
$usr = 'root'; 
$pwd = 'root';
$host = '155.1.99.98:3306';
$db = 'testdb';

$link = mysql_connect($host,$usr,$pwd) 
	or die("Could not connect: " . mysql_error());

mysql_select_db($db) 
	or die ('Could not select database');

// Performing SQL query
$query = 'SELECT * FROM usercatmst';

$result = mysql_query($query) 
	or die('Query failed in module 3.php: '. mysql_error());

if(mysql_num_rows($result) == 0){
	echo "No rows found, nothing to print so am exiting";
	exit;
}

/*
	This is just my style of outputting HTML. You may agree
	that it's a whole lot easier to read but it's up to you.
*/
?>
<table bgcolor="#BBFFFF">
<?php	while ($row = mysql_fetch_row($result)){	?>
	<tr>
		<td><?php echo $row[0]; ?></td>
		<td><?php echo $row[1]; ?></td>
	</tr>
<?php	}	?>
</table>
<?php

// Free resultset
mysql_free_result($result);

// Closing connection
mysql_close($link);
?>
</body>

Paul 🙂

Soumen

Hi Paul,

Thanks. Great Help....

This is very easy way out like I did in ASP. But does this style of coding any declared bad effect in performance?

Regards.
Soumen.