eraxian Posted March 22, 2008 Share Posted March 22, 2008 Hello guys, I'm new to PHP and web development in general so go easy on me. Here's the sitrep. I created a database on a mysql server, it has 1 field. I want to create an html table dynamically when users click on a link, and display the data from the mysql database. I found several scripts online for doing this but none have worked. This is what I have so far: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Untitled Document</title> </head> <?php $con = mysql_connect("myservername","myusername","mypassword"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("bands", $con); $result = mysql_query("SELECT * FROM `list` WHERE name LIKE \'A%\' ORDER BY name"); echo "<table border='1'> <tr> <th>Band Name</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['name'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> <body> </body> </html> AND this is the actual output I get when I open this page, "Band Name "; while($row = mysql_fetch_array($result)) { echo ""; echo "" . $row['name'] . ""; echo ""; } echo ""; mysql_close($con); ?>" So obviously somewhere I've made a grievous error, please help! Thank you Link to comment https://forums.phpfreaks.com/topic/97419-creating-dynamic-html-table/ Share on other sites More sharing options...
Barand Posted March 22, 2008 Share Posted March 22, 2008 try this <?php $sql = "SELECT * FROM `list` WHERE name LIKE \'A%\' ORDER BY name"; echo table2Table ($sql); function table2Table($query) { $result = mysql_query($query); $str = "<TABLE border='1' cellpadding='4'>\n"; // column headings $str .= "<tr>\n"; while ($fld = mysql_fetch_field ($result)) { $str .= "<th>{$fld->name}</th>\n"; } $str .= "</tr>\n"; // list data while ($row = mysql_fetch_row($result)) { $str .= "<tr>\n"; foreach ($row as $field) { $str .= "<td>$field</td>\n"; } $str .= "</tr>\n"; } $str .= "</TABLE>\n"; return $str; } ?> Link to comment https://forums.phpfreaks.com/topic/97419-creating-dynamic-html-table/#findComment-498478 Share on other sites More sharing options...
eraxian Posted March 23, 2008 Author Share Posted March 23, 2008 still having the same type of issue, this is what I get when i inserted the code you posted, \n"; // column headings $str .= "\n"; while ($fld = mysql_fetch_field ($result)) { $str .= "{$fld->name}\n"; } $str .= "\n"; // list data while ($row = mysql_fetch_row($result)) { $str .= "\n"; foreach ($row as $field) { $str .= "$field\n"; } $str .= "\n"; } $str .= "\n"; return $str; } ?> Why is it not recognizing the rest of the code as part of the php script??? i'm baffled, plz help! Link to comment https://forums.phpfreaks.com/topic/97419-creating-dynamic-html-table/#findComment-498590 Share on other sites More sharing options...
sqlnoob Posted March 23, 2008 Share Posted March 23, 2008 you sure that you stored as a php file and that php is installed where you stored the php file? also can you show me the table where you get the info from Link to comment https://forums.phpfreaks.com/topic/97419-creating-dynamic-html-table/#findComment-498791 Share on other sites More sharing options...
eraxian Posted March 25, 2008 Author Share Posted March 25, 2008 you were right 'sqlnoob', i forgot to save it as a .php file, now i feel like an idiot, lol. it works now, thanks Link to comment https://forums.phpfreaks.com/topic/97419-creating-dynamic-html-table/#findComment-500373 Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.