checkbox array

techmeister

Hi all:

I\'m having a minor yet frustrating checkbox problem:
I have 6 checkboxes hard-coded into an HTML form:

<input type=\"checkbox\" name=\"like_to_see[0]\" value=\"statistics\">
<input type=\"checkbox\" name=\"like_to_see[1]\" value=\"statistics\">
<input type=\"checkbox\" name=\"like_to_see[3]\" value=\"statistics\">
//etc etc

The specific part of the processing script is:
while($row = mysql_fetch_array($result)) {
$date = $row[\'date\'];
$name = $row[\'name\'];
$email = $row[\'email\'];
$ext = $row[\'ext\'];
$dept = $row[\'dept\'];
$comments = $row[\'comments\'];

$like_to_see = $row[\'like_to_see\'];
while($x = each($like_to_see)) {//line 98
$disp = $disp . \"$x[0] => $x[1]<br>\"; 
}//end-2nd while

}//end 1st while

Yet I get the following error message on the procesing script:
Warning: Variable passed to each() is not an array or object in /www/httpd/html/intranet_questionnaire/questionnaire.php on line 98

I\'m sure there\'s probably something wrong with the code here somewhere but not being a programmer as such (little knowledge of loops etc) then I\'m unsure of what to do to get it to work.

My environment is php4.0.3pl1 and MySQl 3.22.32
If anyone can help me out I\'d be most grateful.
Cheers.
Russ

craigh

you're referring to $like_to_see as a plain variable instead of an array in line 97. the value of $like_to_see is "Array" so that line won't work.

techmeister

Cheers Craig - however:

I canged it to:

$like_to_see[] = $row['like_to_see'];
while($x = each($like_to_see)) {
$disp = $disp . "$x[0] => $x[1]<br>";
}//end-while

I received no error, but this is now displayed: 0 => Array

??????
Cheers.
Russ

Anon

while(list($x) = each($like_to_see))

try this 🙂

techmeister

Jeroen: Many thanx for your offer of help, but it still doesn't work and it's really p*****g me off!! It can't be THAT hard to do!

while($row = mysql_fetch_array($result)) {
$date = $row['date'];
$name = $row['name'];
$email = $row['email'];
$ext = $row['ext'];
$dept = $row['dept'];
$comments = $row

$like_to_see = $row['like_to_see'];
while(list($x) = each($like_to_see)){
$disp = $disp . "$x[0] => $x[1]<br>";
}

echo $disp;

//Results in:
Warning: Variable passed to each() is not an array or object in /www/httpd/html/intranet_questionnaire/questionnaire.php on line 98

What am I doing wrong??
Cheers again
Russ

Anon

Why are you trying to pull the 'like_to_see' from a mysql resultset?

Isn't it posted to you via a form?

techmeister

Yes it's posted via a form but I need the results to go into a DB that's the reason for the MySQL result set.

Cheers.
Russ

veddermatic

Russ, I am having this problem, and I thing you are (for lack of a better word) screwed.

I am doing the same thing... having many checkboxes with the same NAME but different VALUES and trying to access the results after positng as an array. Unfortunately, PHP only assigns ONE value to the $name as a string... no array of all checked values like perl does. =(

Workaround is to have your names for your checkboes numbered, then pass the number of moxes along as well:

========

Then parse through the boxes one by one to see if they are set and make an array out of the values that are.

HTH.