your variable scope is local not the function
function name()
{
global $variable;
}
--
Chris Lee
[EMAIL PROTECTED]
""Subodh Gupta"" <[EMAIL PROTECTED]> wrote in message
002b01c0cce9$aa670b60$c834d6d2@subodhgu">news:002b01c0cce9$aa670b60$c834d6d2@subodhgu...
Can You figure out the mistake here...??
In the following code... In the function print_entry which I am calling in some other
file, I want each entry to printed in a different colour.
$colour = array("808080","800000","FF00FF","0000FF","008080");
$colouroffset=0;
function print_entry($row,$preserve="")
{
// walk through any arguments passed in after the first two
$numargs = func_num_args();
for ($i = 2; $i < $numargs; $i++)
{
$field = func_get_arg($i);
// This will transform a label string to a valid database
// field name - e.g., "Last Name" becomes "last_name"
$dbfield = str_replace(" ", "_", strtolower($field));
$dbvalue = cleanup_text($row[$dbfield],$preserve);
$name = ucwords($field);
print <<<EOQ
<tr>\n
<td valign=top align=right><b><font
color="#$colour[$colouroffset]">$name:</font></b></td>\n // The problem is here.. the
value of $colour[$colouroffset] is not <td valign=top align=left><font
color="#$colour[$colouroffset]">$dbvalue</font></td>\n // getting
substituted. Can you tell me why??
</tr>\n\n
EOQ;
}
$colouroffset++;
}
Subodh Gupta
I have learned, Joy is not in things, it is in us.
You will ultimately be known by what you give and not what you get.
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