This is a little more specific: I have a template in the database which looks something like this: <html> <title>$page_title</title> <body> Image: $image <p /> <img src = \"$image\"> Main Text: $mtext <br /> Text2: $text2 <br /> Text3: $text3 <br /> </body> <html> The variable name is $somecontent When I call $somecontent in the PHP file, it displays exactly like it is above. It does not replace the variables. Is there a reason for this? Here is my PHP file: <?php include "cnx.php"; include "includes/header.php"; //Include the content $select_data = "SELECT * FROM template WHERE tmpl_name = \"my_template\";"; $response = mysql_query( $select_data, $cnx ); //now print it out for the user. if ( $one_line_of_data = mysql_fetch_array( $response ) ) { extract ( $one_line_of_data ); } $file = $_POST['file']; $page_title = $_POST['page_title']; $image = $_POST['image']; $mtext = $_POST['mtext']; $text2 = $_POST['text2']; $text3 = $_POST['text3']; if (file_exists($dir . $file)) { echo "<center><p /><p class=\"NormalBold\">Success!<p /><p />"; } else { echo "The file $file does not exist"; } $filename = "$file"; //Update the table in MySQL $update_data = "UPDATE cms_pages SET page_title = '$page_title', image = '$image', mtext = '$mtext', text2 = '$text2', text3 = '$text2' WHERE filename = '$file'"; $response = mysql_query( $update_data, $cnx ); if(mysql_error()) die ('database error<br>'. mysql_error()); //Begining of Template Construction $mtext = nl2br($mtext); $text2 = nl2br($text2); $text3 = nl2br($text3); //include "content.php"; This works with an external page included, but I would like it to be in the database for easy changes. // Is the file writable? if (is_writable($dir. $filename)) { if (!$handle = fopen($dir . $filename, 'w')) { echo "Cannot open file ($filename)"; exit; } // Write $somecontent to our opened file. if (fwrite($handle, $somecontent) === FALSE) { echo "Cannot write to file ($filename)"; exit; } echo "<center><p /><p class=\"NormalText\">$filename has been updated.</p><p /></center>"; fclose($handle); } else { echo "The file $filename is not writable"; } include "includes/footer.php"; ?> </html> on 11/18/05 7:14 PM Lists (lists@xxxxxxxxxxx) wrote: > <? > $var = "time"; > $var2 = "clock"; > echo "$var$var2"; > //outputs timeclock > > $two_vars = "$var$var2"; > echo "$two_vars"; > //outputs timeclock > ?> > > Why put two variables in one field? But if you want to store them, > do: "$var, $var2". After pulling this back from the database, > explode into an array at the comma, using list to call them var and > var2. > > On Nov 18, 2005, at 7:54 PM, Marquez Design wrote: > >> Greetings. >> >> Does anyone know how to do this? >> >> I have, >> >> $var >> >> $var2 >> >> In a field called two_vars in a MySQL db. >> >> I am calling the variables inside PHP document. >> >> In that document I am saying: >> >> $var = "time" >> $var2 = "clock" >> >> <!-- I do the query in MySQL here --> >> >> echo "$two_vars"; >> >> But the what prints out is >> >> $var >> >> $var2 >> >> not "time" and "clock". I know that is what is in the database, >> but I want >> it to replace the variables when printed in the PHP file. >> >> Does this make sense to anyone? Does anyone know how to do this? >> >> -- >> Steve >> >> -- >> PHP General Mailing List (http://www.php.net/) >> To unsubscribe, visit: http://www.php.net/unsub.php >> > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php