Hi Paul
Try print $instrument."number"; jack -----Original Message----- From: Paul Evans [mailto:[EMAIL PROTECTED] Sent: Friday, October 29, 2004 9:45 AM To: [EMAIL PROTECTED] Subject: [PHP] Parsing a concatenated variable and string? Hi, I am new to PHP and cant seem to find a way to parse a concatenated string and variable. To explain: I have a database to catalogue a composers works, one of the tables has a list of instruments which I then access to dynamically display on a wep page using checkboxes - if a particular instrument is used in the work then the checkbox is checked. <?php //Select all instruments from the instrumentTbl in the correct order $resultInstrument = mysql_query("SELECT * FROM instrumentTbl ORDER BY sort_order"); //Load them into an array if ($myrowInstrument = mysql_fetch_array($resultInstrument)) { do { echo $myrowInstrument['instrument']; echo ": "; ?> //This creates a checkbox for each instrument in the database <input name="<?php echo $myrowInstrument['instrument']; ?>" type="checkbox" value="<?php echo $myrowInstrument['instrument']; ?>"> So far so good. The problem I am having is when I want to insert the number of instruments used in the work. <input name="<?php echo $myrowInstrument['instrument']; ?>number" type="text" size="3"> This snipet of code produces a dynamically created form where the user can enter the number of instruments required. (Ie 2 flutes) Here is a part of the form created from the code above for the instrument Flute: <input name="Flutenumber" type="text" size="3"> To process the information in this form I think I need to concatenate $instrument with the string 'number'. the only problem is that PHP reads this literally and the output is just $Flutenumber and not the number actually entered. Here is the code I have at the moment <?php $resultInstrument_id = mysql_query("SELECT * FROM instrumentTbl ORDER BY sort_order"); while ($myrowInstrument_id = mysql_fetch_array($resultInstrument_id)) { $instrument = $myrowInstrument_id['instrument']; if (strlen ($$instrument !='0')) { //this works fine and gives a list of all the checked instruments print $$instrument; //this just outputs the literal interpretation ie it will print $Flutenumber not the actual number entered on the form. print "$${$instrument}number"; } } ?> Any help would be greatly appreciated. Thanks, Paul. -- 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
