Siggles Posted November 23, 2007 Share Posted November 23, 2007 Hi, I have a script that updates and removes gigs from a MySQL dbase. Amongst the code that adds the gig (and is called using, for example ?nav=addnews) I have a function defined that displays some <select> tags for the day, month and year so people can specify a date. I then call that function amongst a long echo command that prints lots of html. Instead of the drop down boxes appearing amongst the form they appear at the top of the page (where the function is defined I think). What i am doing wrong? Thank you for your help. $addnews=" <form action=\"".$_SERVER['PHP_SELF']."\" method=\"post\"> ............... .............. td align=\"left\" >".DateSelector("Sample")."</td></tr> Quote Link to comment Share on other sites More sharing options...
Siggles Posted November 23, 2007 Author Share Posted November 23, 2007 This is the function. Works fine, just is in wrong place. <?php /****************************************/ /*Function:DateSelector v1.1 */ /*Code: PHP 3 */ /*Author: Leon Atkinson <leon@clearink.com> */ /*Creates three form fields for get month/day/year */ /*Input: Prefix to name of field, default date */ /*Output: HTML to define three date fields */ /****************************************/ function DateSelector($inName, $useDate=0) { /* create array so we can name months */ $monthName = array(1=> "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"); /* if date invalid or not supplied, use current time */ if($useDate == 0) { $useDate = Time(); } /* make month selector */ echo "<SELECT NAME=" . $inName . "Month>\n"; for($currentMonth = 1; $currentMonth <= 12; $currentMonth++) { echo "<OPTION VALUE=\""; echo intval($currentMonth); echo "\""; if(intval(date( "m", $useDate))==$currentMonth) { echo " SELECTED"; } echo ">" . $monthName[$currentMonth] . "\n"; } echo "</SELECT>"; /* make day selector */ echo "<SELECT NAME=" . $inName . "Day>\n"; for($currentDay=1; $currentDay <= 31; $currentDay++) { echo "<OPTION VALUE=\"$currentDay\""; if(intval(date( "d", $useDate))==$currentDay) { echo " SELECTED"; } echo ">$currentDay\n"; } echo "</SELECT>"; /* make year selector */ echo "<SELECT NAME=" . $inName . "Year>\n"; $startYear = date( "Y", $useDate); for($currentYear = $startYear - 1; $currentYear <= $startYear+3;$currentYear++) { echo "<OPTION VALUE=\"$currentYear\""; if(date( "Y", $useDate)==$currentYear) { echo " SELECTED"; } echo ">$currentYear\n"; } echo "</SELECT>"; } /****************************************/ Quote Link to comment Share on other sites More sharing options...
GingerRobot Posted November 23, 2007 Share Posted November 23, 2007 See this topic: http://www.phpfreaks.com/forums/index.php/topic,168930.0.html Either modify your DateSelector function so that it does not echo the data, but returns a string (best option) or use output buffering (explained in the reply to the above thread) Quote Link to comment Share on other sites More sharing options...
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