In want to display PHP query results in a table

Question

lordtrini 0 Newbie Poster

14 Years Ago

I am trying to display the results of a query in the form of a table. this it the PHP that i am using below. The problem is that the headings of the table repeats. I really don't want that...

I am also trying to get a border around the cells of the table, but i get an error when i add table properties

Can someone help me please. i you can edit the code i posed please please do... i am new at this...

<?php

if ( isset( $_GET['id'] ) )
{

$query = "SELECT * FROM schedule WHERE course_id = '". $_GET['id'] ."' ORDER BY indexx ASC";

$numresults=mysql_query($query);
$numrows=mysql_num_rows($numresults);


if ($numrows == 0)
{
echo "This course does not have a set schedule. ";

}

$q .= " limit $s,$limit";
$result = mysql_query($query) or die("Couldn't execute query");

while ($row= mysql_fetch_array($result))
{
$indexx = $row["indexx"];
$semester = $row["semester"];
$day = $row["day"];
$time = $row["time"];
$instructor = $row["instructor"];
$location = $row["location"];


echo "<table>";
echo "<tr>";
echo " <td>Index</td>";
echo "<td>Semester</td>";
echo "<td>Day</td>";
echo "<td>Time</td>";
echo "<td>Instructor</td>";
echo "<td>Location</td>";
echo "</tr>";
echo "<tr>";
echo "<td>$indexx</td>";
echo "<td> $semester</td>";
echo "<td>$day</td>";
echo "<td>$time</td>";
echo "<td>$instructor</td>";
echo "<td>$location</td>";
echo "</tr>";
echo"</table>";




$count++ ;

} // end WHILE

echo "<br/>End test link<br/>";
} // end IF


?>

php

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Latest Post 14 Years Ago Latest Post by lordtrini

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colweb 13 Posting Whiz

14 Years Ago

You placed the complete table inside the while loop. That means there will be a new table every time the while loop executes. The way to do it is to take the <table> declaration and your table header out of the while loop.
Don't know where you use count for? To display the results you don't need to store your result in new variables first. See here:

$q .= " limit $s,$limit";
$result = mysql_query($query) or die("Couldn't execute query");

echo "<table>";
echo "<tr><td>Index</td><td>Semester</td><td>Day</td><td>Time</td><td>Instructor</td><td>Location</td></tr>";

while ($row= mysql_fetch_array($result)) {
	echo "<tr><td>".$row['indexx']."</td>
		<td>".$row['semester']."</td>
		<td>".$row['day']."</td>
		<td>".$row['time']."</td>
		<td>".$row['instructor']."</td>
		<td>".$row['location']."</td></tr>";
		$count++ ;
} // end WHILE

echo "</table><br/>End test link<br/>";

diafol

14 Years Ago

Like colweb. I'd go a bit further with the table (but just my personal preference with \n and \t - just parses 'pretty' html):

echo "<table>\n\t<thead>\n\t\t<tr>\n\t\t\t<td>Index\n\t\t\t</td>\n\t\t\t<td>Semester\n\t\t\t</td>\n\t\t\t<td>Day\n\t\t\t</td>\n\t\t\t<td>Time\n\t\t\t</td>\n\t\t\t<td>Instructor\n\t\t\t</td>\n\t\t\t<td>Location\n\t\t\t</td>\n\t\t</tr>\n\t</thead>\n\t<tbody>";

while ($row= mysql_fetch_array($result)) {
	echo "\n\t\t<tr>\n\t\t\t<td>{$row['indexx']}\n\t\t\t</td>\n\t\t\t<td>{$row['semester']}\n\t\t\t</td>\n\t\t\t<td>{$row['day']}\n\t\t\t</td>\n\t\t\t<td>{$row['time']}\n\t\t\t</td>\n\t\t\t<td>{$row['instructor']}\n\t\t\t</td>\n\t\t\t<td>{$row['location']}\n\t\t\t</td>\n\t\t</tr>";
}
echo "\n\t</tbody>\n</table>";

colweb 13 Posting Whiz

14 Years Ago

Ardav, I must agree it makes good readable html code. But the php code becomes unreadable. And it is the php code we are working with....

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lordtrini 0 Newbie Poster · Answer 1 · 2010-04-11T04:31:06+00:00

thanks guys... i am trying it now... and i will let u know how it goes

diafol · Answer 2 · 2010-04-11T04:49:19+00:00

Ardav, I must agree it makes good readable html code. But the php code becomes unreadable. And it is the php code we are working with....

Fair point. As I said - just a personal preference. Templating and heredoc syntax are much easier, but I like escaping \n\t - I'm just weird that way.:icon_rolleyes:

lordtrini 0 Newbie Poster · Answer 3 · 2010-04-11T05:06:54+00:00

I added the code as suggested but i get an error message.

this is how i places it...

<?php

if ( isset( $_GET['id'] ) )
{

$query = "SELECT * FROM schedule WHERE course_id = '". $_GET['id'] ."' ORDER BY indexx ASC";

$numresults=mysql_query($query);
$numrows=mysql_num_rows($numresults);


if ($numrows == 0)
{
echo "This course does not have a set schedule. ";

}

$q .= " limit $s,$limit";
$result = mysql_query($query) or die("Couldn't execute query");

echo "<table>";
echo "<tr><td>Index</td><td>Semester</td><td>Day</td><td>Time</td><td>Instructor</td><td>Location</td></tr>";

while ($row= mysql_fetch_array($result)) {
	echo "<tr><td>".$row['indexx']."</td>
		<td>".$row['semester']."</td>
		<td>".$row['day']."</td>
		<td>".$row['time']."</td>
		<td>".$row['instructor']."</td>
		<td>".$row['location']."</td></tr>";
		$count++ ;
} // end WHILE

echo "</table><br/>End test link<br/>";

?>

colweb 13 Posting Whiz · Answer 4 · 2010-04-11T05:25:27+00:00

Can you please tell us what the error message is?

lordtrini 0 Newbie Poster · Answer 5 · 2010-04-11T15:52:32+00:00

Sorry stupid of me to not post the error message...

Parse error: syntax error, unexpected $end in /xxx/testermax.php on line 355

line 355 is the last line of the code and contains list a plain html tag.

thanks for the help so far...

Oh and this the the php

<?php

if ( isset( $_GET['id'] ) )
{

$query = "SELECT * FROM schedule WHERE course_id = '". $_GET['id'] ."' ORDER BY indexx ASC";

$numresults=mysql_query($query);
$numrows=mysql_num_rows($numresults);


if ($numrows == 0)
{
echo "This course does not have a set schedule. ";

}

$q .= " limit $s,$limit";
$result = mysql_query($query) or die("Couldn't execute query");

echo "<table>";
echo "<tr><td>Index</td><td>Semester</td><td>Day</td><td>Time</td><td>Instructor</td><td>Location</td></tr>";

while ($row= mysql_fetch_array($result)) {
	echo "<tr><td>".$row['indexx']."</td>
		<td>".$row['semester']."</td>
		<td>".$row['day']."</td>
		<td>".$row['time']."</td>
		<td>".$row['instructor']."</td>
		<td>".$row['location']."</td></tr>";
		$count++ ;
} // end WHILE

echo "</table><br/>End test link<br/>";

?>

diafol · Answer 6 · 2010-04-11T19:31:08+00:00

You haven't closed the first 'if brace'. Add '}' just before the ?>.

lordtrini 0 Newbie Poster · Answer 7 · 2010-04-11T21:28:37+00:00

What a rookie mistake... thanks guys... i will try to see who i can help out on the forum. Everything works...

In want to display PHP query results in a table

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