php implode function not working on concatenated variables - php

Question

I have multiple arrays which I am passing to a file sales_process.php on the form submit action. The arrays are named like this : boards1 = {a , b , c}, boards2 = {b , c , d}, boards3 = {a , c , d} . . and so on (The values are not a,b,c,d). I am passing them through multiple 'multiple select boxes' in my form like this where a,b,c,d are my multiple select options :
for ($count=1;$count<10;$count++)
echo "<td>"."<select name='boards".$count."[]' multiple='multiple'>".showOptionsDrop($boards,$arr)."</select></td>";
Now when i pass them to the file sales_process.php, i want to convert these arrays into strings using implode function so that i can store them into my 'schools' table. In sales_process.php file, I am doing this:
for ($i=0;$i<$count;$i++) {
$board = implode(',',${'boards'.$i});
$query = "UPDATE schools SET board = '$board' where schoolcode = (some_no)";
$result = mysql_query($query) or die("Error in updating table :".mysql_error());
}
So in this way, every time my loop runs, the values of boardsX gets converted into string and gets stored in the table, where X is 1,2,3.... and so on.
The problem is the implode function is not working and giving error :
Warning: implode() [function.implode]: Invalid arguments passed in C:\xampp\htdocs\relationshipReport\sales_process.php on line 18
Now if you say that the variable ${'boards'.$i} is not an array, i did this and found out that its giving me an array only :
$i=1;
var_dump(${'boards'.$i});
print_r(${'boards'.$i});
which gives the output as :
array(3) { [0]=> string(4) "CBSE" [1]=> string(4) "ICSE" [2]=> string(5) "IGCSE" }
Array ( [0] => CBSE [1] => ICSE [2] => IGCSE )
I hope my question is clear. Please help me in finding out whats going wrong in the implode function. If you don't understand the question, please mention it.

Answer 1

From your examples, I believe this is your current code:
for ($i=0;$i<$count;$i++) {
$board = implode(',',${'boards'.$i});
$query = "UPDATE schools SET board = '$board' where schoolcode = (some_no)";
$result = mysql_query($query) or die("Error in updating table :".mysql_error());
}
If so, try altering it so it looks like this:
for ($i=0;$i<$count;$i++) {
if(isset($_POST['boards'.$i]))
{
$board = mysql_real_escape_string(implode(',',$_POST['boards'.$i]));
$query = "UPDATE schools SET board = '$board' where schoolcode = (some_no)";
$result = mysql_query($query) or die("Error in updating table :".mysql_error());
}
else
{
echo '$_POST[\'boards' . $i . '\'] doesn\'t exist or is null.<br>';
}
}
This contains an if condition, checking that the array entry actually exists. If it does exist, it runs the query. If it doesn't, it echos back a message, telling you which one doesn't exist.
Even if you have confirmed at the frontend with Javascript, you should always check before running code, just to be sure it won't break it. You can never be too safe, especially if your code contains sensitive information.
Try this, and let me know how you go.

Answer 2

Assume ${'boards'.$i} comes from $_POST['boards'.$i]. (If not, you are using register_globals, but that is not good.)
For multiple select, if no options is selected, then $_POST['boards'.$i] will give you null instead of an array.