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Old Dec 4th, 2006, 8:45 PM   #3
kworden
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If i do that i get an error saying $ga can't be found and then the #

[PHP]
for($i=1; $i<= $numberofgaquestions; $i++ ){
echo "Ga # $i:". $ga.$i ."<br>";
}
[/PHP]

Here is what i get:
Notice: Undefined variable: ga in /final_hr_attributes_7.php on line 329
Ga # 1:1
Notice: Undefined variable: ga in /final_hr_attributes_7.php on line 329
Ga # 2:2
Notice: Undefined variable: ga in /final_hr_attributes_7.php on line 329
Ga # 3:3

What i need is to form the variable $ga1 with this [php]echo "Ga # $i:". $ga.$i ."<br>";[/php]

But thanks for your help
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