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 Posted by BlackSpider on 10/04/06 10:07 
So I've now tried... 
 
function TableData($slot) 
{ 
 global $row_Recordset1, $colour_FREE, $colour_TOOK; 
 $var = "'"."$slot"."'"; 
 if ($row_Recordset1[$var] == "") { 
  return ($colour_FREE.'" class="timeslot"><div  
align="center">'.$row_Recordset[$var]); 
 } 
 else { 
  return ($colour_TOOK.'" class="timeslot"><div  
align="center">'.$row_Recordset1[$var]); 
 } 
 
} 
 
.... To no avail, $row_Recordset[$var] still returns nothing, even though if  
I manually input "date" instead of "$var", it works fine. (Yes, $slot is  
returning "date" correctly) 
 
Incase I misunderstood, I also tried calling the function using  
"'"."data"."'" as the argument, (Both directly, and through a secondary  
string (TableData($var)) but that didn't work either. :( 
 
 
"Vince Morgan" <vinhar@UNSPAMoptusnet.com.au> wrote in message  
news:45237850$0$22938$afc38c87@news.optusnet.com.au... 
> 
> "BlackSpider" <BlackSpider@I.Dont.Exist.com> wrote in message 
> news:v5ednayzlo_lYr_YRVnytQ@eclipse.net.uk... 
>> <--PHP 
>> function TableData($slot) 
>> { 
>>     global $row_Recordset1, $colour_FREE, $colour_TOOK; 
>> 
>>     $slotuser = $row_Recordset1[$slot]; 
>>      if ($slotuser == "") { 
>>         return ($colour_FREE.'" class="timeslot"><div 
>> align="center">'.$slotuser ); 
>>      } 
>>      else { 
>>         return ($colour_TOOK . '" class="timeslot"><div align="center">'  
>> . 
>> $row_Recordset1[ $slot ]); 
>>      } 
>> 
>> } 
>> PHP--> 
>> 
>> The $row_Recordset1 array is set by mysql_fetch_assoc to get some table 
>> data, and I call this function with the table-row name (In this case 
>> "date"), like this: TableData('date') 
>> 
>> However, when trying to get the data of the row 'date' from the Record 
> using 
>> $slot, it returns a null string, regardless of what data is actually 
> there. 
>> If I replace $slot with 'date' in the code, it returns the correct value, 
> so 
>> the connection between php/Mysql is fine, but when trying to get the data 
> by 
>> using a String as the Key, it simply doesn' twork. Any ideas? 
>> 
>> 
> I think you may need to append single quotes to get it to work. 
> 
> $var = "'"."table-row name "."'"; 
> 
> 
> HTH 
> Vince Morgan 
> 
>
 
  
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