Variables

Poofman

I'm having a problem with variable syntax. Let's say I have a variable $J2Round, which gives the rounded value of J2. Now, instead of writing 25 lines with a different $J3Round, $J4Round, I want to make a routine that increments the variable. I have a loop that starts like:

for($a = 1; $a < 26; $a++){

what is the syntax for the variable? I've tried:

$J$aRound
'$J'.$a.'Round'
$J($a)Round

none work. what's the proper syntax so that it will put the $JxRound variable together and print the actual contents of $J1Round through $J25Round?

Pests

Uhm, you can try:

$J{$a}Round

You can also just use arrays:

$JRound[$a]

Ill look into it more....

After re-reading the article I posted below, you can probabally use:

$a = 10 //For example

$var = "J".$a."Round";
$$var; //This would be the variable $J10Round
$$var = "Hello"; //Puts Hello into $J10Round
echo $$var; //Echos out Hello
echo $J10Round; //Same as above

Pests

Try here:

http://www.phpbuilder.com/columns/robert20000928.php3

Poofman

well, that created the $J10Round say, but it prints $J10Round instead of the value the variable $J10Round represents.

in other words, now I get this

score = $J10Round

instead of

score = 20

Pests

Try this then:

$a = 10 //For example 

$var = "J".$a."Round"; 
$$var; //This would be the variable $J10Round 
$$var = "Hello"; //Puts Hello into $J10Round 
echo ${$var}; //Echos out Hello 
echo $J10Round; //Same as above

Poofman

nope

look at output, just the first 25 lines which are single code instead of doing all 25.

www.squadinferno.com/jokes/pjokes.php

heres the code for first 25 lines

for($a = 1; $a < 26; $a++){
print "<tr bgcolor=#323232>\n";
print "<td width=61><a href=get.php?J$a style=text-decoration:none>$a</a></td>\n";
$var = '$J'.$a.'Round';
$$var;
print "<td width=62>$($var)</td>\n";
print "<td width=150>\n";
print showrate($a);
print "</td>\n";
print "<td width=247>\n";
print "Vote!";
print "</td></tr>\n";
}