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[SOLVED]Adding background color to function selectboxes

Posted: Wed Jun 01, 2005 9:34 am
by mohson
Can anyone help with this simple problem, if Pimptastic is around he'll probably know what im on about as he helped me out quite a bit recently.

the function below makes creates the select boxes I require to input a date.

but al my other select boxes have a different background colour to the plain one. Im not sure how to add a background colour to the select boxes through this function.

Code: Select all

/*** Function: DateSelector** 
Input: STRING inName, INTEGER useDate** Output: ** 

Description: Creates three form fields for get month/day/year*/ function 
DateSelector($inName, $useDate)    
{  $monthName = array(1=>"January", "February", "March", "April", "May", "June", "July", "August","September", "October", "November", "December");                

if($useDate == "")        
{ 
$useDate = time();        
}        
print("<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>";               
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 ">$currentDay\n";       
}       
echo "</SELECT>";                   
echo "<SELECT NAME=".$inName."Year>\n";       
$startYear = date("Y", $useDate);        
if($startYear < 1997)        
{           
$startYear = date("Y");       
}                
for($currentYear = $startYear-1; $currentYear <= $startYear+2;$currentYear++)        {           
echo "<OPTION VALUE=\"$currentYear\"";           
if(date("Y", $useDate)==$currentYear)           
{                
echo " SELECTED";            
}           
echo ">$currentYear\n";        
}    echo "</SELECT>";}
?>

Posted: Wed Jun 01, 2005 9:42 am
by JayBird
Try this...

Code: Select all

/*** Function: DateSelector** 
Input: STRING inName, INTEGER useDate** Output: ** 
 
Description: Creates three form fields for get month/day/year*/ function 
DateSelector($inName, $useDate)    
{  $monthName = array(1=>"January", "February", "March", "April", "May", "June", "July", "August","September", "October", "November", "December");   

$style = "style=\"color: #000000; background-color: #ADD8E6\">";             
 
if($useDate == "")        
{ 
$useDate = time();        
}        
print("<SELECT NAME=\"" . $inName . "Month\" $style>\n");        
for($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>";               
echo "<SELECT NAME=".$inName."Day $style>\n";       
for($currentDay=1; $currentDay <= 31; $currentDay++)        
{           
echo "<OPTION VALUE=\"$currentDay\"";           
if(intval(date("d", $useDate))==$currentDay)            
{                
echo " SELECTED";           
}           
echo ">$currentDay\n";       
}       
echo "</SELECT>";                   
echo "<SELECT NAME=".$inName."Year $style>\n";       
$startYear = date("Y", $useDate);        
if($startYear < 1997)        
{           
$startYear = date("Y");       
}                
for($currentYear = $startYear-1; $currentYear <= $startYear+2;$currentYear++)        {           
echo "<OPTION VALUE=\"$currentYear\"";           
if(date("Y", $useDate)==$currentYear)           
{                
echo " SELECTED";            
}           
echo ">$currentYear\n";        
}    echo "</SELECT>";}
?>

Posted: Wed Jun 01, 2005 9:47 am
by mohson
one word for you - "LEGEND"

Posted: Wed Jun 01, 2005 9:49 am
by JayBird
You owe me one of these :lol:

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