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Sep 20th, 2007

Displaying query results dynamically in table

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I am trying to display MySQL query results dynamically in a table. Based on the query, the table will have a maximum of 3 columns and 15 rows. I would like to have the results display as follows:

Record 1 Record 16 Record 31
Record 2 Record 17 Record 32

Record 15 Record 30 Record 45

If the query returns less than 16 records, everything should display in the first column; if more than 15 but less than 31 - 2 columns, et al.

I am fairly new to PHP and am trying to learn all I can as I go along. Any books, websites, or online classes you can suggest to increase my knowledge of PHP would be greatly appreciated.

Thanks,

Deb

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hsmom314

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since Sep 2007

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Sep 20th, 2007

Re: Displaying query results dynamically in table

ok i will give you an example:

PHP Syntax (Toggle Plain Text)
<html>
<head>
</head>
<body>
 <?php
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name=""; // Table name
 
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");  
 
 $user_result = "select * from table;";
$qry = mysql_query($user_result) OR die(mysql_error());
$user_array = mysql_fetch_assoc($qry);
echo "<center>";
echo "<table CELLPADDING=10 border =1 >";
echo "<tr>";
echo "<td>".$user_array['field 1']."</td>";
echo "<td>".$user_array['field 2']."</td>";
echo "<td>".$user_array['field 3']."</td>";
echo "</tr>";
echo "</table>";
    }
mysql_close();
?>
 
</body>
</html>
<html>
<head>
</head>
<body>
 <?php
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name=""; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");  
    
 $user_result = "select * from table;";
$qry = mysql_query($user_result) OR die(mysql_error());
$user_array = mysql_fetch_assoc($qry);
echo "<center>";
echo "<table CELLPADDING=10 border =1 >";
echo "<tr>";
echo "<td>".$user_array['field 1']."</td>";
echo "<td>".$user_array['field 2']."</td>";
echo "<td>".$user_array['field 3']."</td>";
echo "</tr>";
echo "</table>";
    }
mysql_close();
?>

</body>
</html>

hope that helps you

Last edited by sam1; Sep 20th, 2007 at 10:52 am.

sam1

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Posting Whiz

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300 posts
since Nov 2004

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Sep 20th, 2007

Re: Displaying query results dynamically in table

Thank you for your suggestion. However, I don't think I explained what I'm trying to do well enough. I am trying to display one field from my database in this table. The results from the one field should display in a table vertically (max. 15 rows) then start a new column. I've written some code (using a while loop and count function) that will display it horizontally but everything that I try to get it to display vertically ends up duplicating the same record.

hsmom314

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Sep 20th, 2007

Re: Displaying query results dynamically in table

try using the SQL limit instruction.for example:

PHP Syntax (Toggle Plain Text)
SELECT * FROM 'users' LIMIT 0 10;
SELECT * FROM 'users' LIMIT 10 10;
SELECT * FROM 'users' LIMIT 0 10;
SELECT * FROM 'users' LIMIT 10 10;

the first number tell you the index in which the SQL DB needs to start from and the second number tell you the number of rows to be selected.
I think you can easily figure it out.
Also, try using MySQL manual if you use MySQL as your DBMS.

Good luck.

mostafadotnet

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since Jul 2006

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Sep 20th, 2007

Re: Displaying query results dynamically in table

You will of course have to implement a paging mechanism if your results exceed rows*columns and you will need to get all of the items into a single array before you generate the <tr> and <td> elements. The following example code shows how you can generate the table you want

PHP Syntax (Toggle Plain Text)
<?php
$data = array("a","b","c","d","e","f","g","h","i","j");
$rows = 4;
$itemCount=count($data);
echo "<table>";
for ($row=0; $row < $rows; $row++){
    echo "<tr>";
    for ($itemIndex=$row; $itemIndex < $itemCount; $itemIndex+=$rows){
        echo "<td>".$data[$itemIndex]."</td>";
    } 
}
echo "</table>";
?>
<?php
$data = array("a","b","c","d","e","f","g","h","i","j");
$rows = 4;
$itemCount=count($data);
echo "<table>";
for ($row=0; $row < $rows; $row++){
    echo "<tr>";
    for ($itemIndex=$row; $itemIndex < $itemCount; $itemIndex+=$rows){
        echo "<td>".$data[$itemIndex]."</td>";
    } 
}
echo "</table>";
?>

Ezzaral

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Posting Genius

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6,761 posts
since May 2007

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Sep 23rd, 2007

Re: Displaying query results dynamically in table

Here's a quick and dirty way to do it with nested "for" loops:

(Toggle Plain Text)

$strQS="select field from table order by index";
$refItems=mysql_query($strQS) or die (mysql_error());
$cols=ceil(mysql_num_rows($refItems)/15);
$tableData=array();
for($x=0;$x<$cols;$x++)
{
	for($y=0;$y<15;$y++)
	{
		$tableData[$y][$x]=@mysql_fetch_assoc($refItems);
	}
}
echo <<<TABLESTART
<table>
<tr>
<th colspan={$cols}>
Your Header Here
</th>
</tr>
TABLESTART;
for($x=0;$x<15;$x++)
{
	echo "<tr>\n";
	for($y=0;$y<$cols;$y++)
	{
		echo "<td>".$tableData[$x][$y][field]."</td>\n";
	}
	echo "</tr>\n";
}
echo "</table>"

schizoman

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No one has posted to this discussion for at least three months. Please let old threads die and do not reply to them unless you feel you have something new and valuable to contribute that absolutely must be added to make the discussion complete. Otherwise, please start a new thread in this forum instead.

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