?
Solved

Check Boxes

Posted on 2002-04-06
7
Medium Priority
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Last Modified: 2008-03-06
i have two files.
this is the first one:
authority_process.php
-------
<html>
<table border="1" width="100%" bordercolor="#FFFFFF">
<form action="authority_send_emails.php" method="POST">
<?
   // link to the database
   $linkID = mysql_connect("localhost","","");
   $result = mysql_select_db("test_toaster",$linkID);

   $condition = $Submit && ($Subjects != "Choose one");
   if ($condition) {
   $selected = mysql_query("SELECT id FROM subject WHERE name='$Subjects'", $linkID);
   $selected_subject_id = mysql_fetch_array($selected);
   $selected_subject_id = $selected_subject_id["id"];
   $table_teaches = mysql_query("SELECT instructor_id FROM teaches WHERE subject_id='$selected_subject_id'", $linkID);
   if ($table_teaches != "")
   $teaches_rows_num = mysql_num_rows($table_teaches);
   if ($teaches_rows_num == 0)
       echo "NO instructors teach this subject";
   else{
     for ($num=0 ; $num < $teaches_rows_num ; $num++)
    {
    $teaches = mysql_fetch_array($table_teaches);
?>
    <tr>
    <td width="100%">
    <INPUT NAME="Check[<? echo $num ?>]" TYPE="CHECKBOX" VALUE="<? echo $num  ?>">
<?
    $instID = $teaches["instructor_id"];
    $table_instructor = mysql_query("SELECT fullname FROM instructor WHERE id='$instID'", $linkID);
    $name = mysql_fetch_array($table_instructor);
    $name[$num] = $name["fullname"];
    echo $name[$num] ?>
    </td>
    </tr>
<?
}//end for
?>
  <tr>
    <td width="100%">
        <p><input type="submit" value="Submit" name="B1"></p>
      </form>
    </td>
  </tr>
</table>
<?
}//end else
}//end if ($condition)
?>
</html>
------------------
------------------
and this is the 2nd one:
authority_send_emails.php
------------------
<?php
     for ($num=0 ; $num < $teaches_rows_num ; $num++){
     if ($Check[$num] == "$num")
     echo $name[$num];
}
?>
------------------

the problem that the variable $name is not recognizied!
0
Comment
Question by:almaha
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7 Comments
 
LVL 32

Expert Comment

by:Batalf
ID: 6922967
The $name is not recognized in authority_send_emails ??

I can't find any form-fields with the name="name[]" in authority_process.php. That's probably your problem.

The only field I could find is Check[]

Maybe you need a hidden-field in authority_process.php?

<input type="hidden" Name="name[<? echo $num ?>]" value="<? echo $name[$num]; ?>">

 
0
 

Author Comment

by:almaha
ID: 6923053
the following is more easy to understand and know what's i'm stuck in:
--------------------------------
<html>
<table border="1" width="100%" bordercolor="#FFFFFF">
<form action="authority_send_emails.php" method="POST">
<?
   // link to the database
   $linkID = mysql_connect("localhost","","");
   $result = mysql_select_db("test_toaster",$linkID);

   $condition = $Submit && ($Subjects != "Choose one");
   if ($condition) {
   $selected = mysql_query("SELECT id FROM subject WHERE name='$Subjects'", $linkID);
   $selected_subject_id = mysql_fetch_array($selected);
   $selected_subject_id = $selected_subject_id["id"];
   $table_teaches = mysql_query("SELECT instructor_id FROM teaches WHERE subject_id='$selected_subject_id'", $linkID);
   if ($table_teaches != "")
   $teaches_rows_num = mysql_num_rows($table_teaches);
   if ($teaches_rows_num == 0)
       echo "NO instructors teach this subject";
   else{
     for ($num=0 ; $num < $teaches_rows_num ; $num++)
    {
    $teaches = mysql_fetch_array($table_teaches);
?>
    <tr>
    <td width="100%">
    <INPUT NAME="Check[<? echo $num; ?>]" TYPE="CHECKBOX" VALUE="<? echo $num;  ?>">
<?
    $instID = $teaches["instructor_id"];
    $table_instructor = mysql_query("SELECT fullname FROM instructor WHERE id='$instID'", $linkID);
    $name = mysql_fetch_array($table_instructor);
    $name[$num] = $name["fullname"];
    echo $name[$num];
    //if ($Check[$num] == "$num")
    //echo $name[$num];
    echo $Check[$num];   <<<<<<<<<<<< THIS LINE
?>
    </td>
    </tr>
<?
}//end for
?>
  <tr>
    <td width="100%">
        <p><input type="submit" value="Submit" name="Submit"></p>
      </form>
    </td>
  </tr>
</table>
<?
}//end else
}//end if ($condition)
?>
</html>
-------------------------------
i have a problem with this line >> echo $Check[$num];
it doesn't recognize the variable $Check !
0
 
LVL 32

Expert Comment

by:Batalf
ID: 6923076
Instead of

   $name[$num] = $name["fullname"];
   echo $name[$num] ?>

why don't simply type

echo $name["fullname"];

?
0
WordPress Tutorial 2: Terminology

An important part of learning any new piece of software is understanding the terminology it uses. Thankfully WordPress uses fairly simple names for everything that make it easy to start using the software.

 
LVL 5

Expert Comment

by:harwantgrewal
ID: 6923560
I think you are mistaking somewhere instead try to use of array like check[<?echo $num?>] will not create an array it will create a variable like check[1] but this is not a array try to use like this

check<?echo $num?> this will generate variables check1, check2 etc and make a hidden field which calculates the number of check boxes and the in for loop you can retive the the variables like
$chk="check".$num;
echo $$chk;

Harry
0
 
LVL 5

Accepted Solution

by:
harwantgrewal earned 46 total points
ID: 6927936
did your problem solved
0
 
LVL 32

Expert Comment

by:Batalf
ID: 6927944
Harry:

About:
"I think you are mistaking somewhere instead try to use of array like check[<?echo $num?>] will not create
an array it will create a variable like check[1] but this is not a array"

Are you sure about this? I don't think so, I have used this a lot of times, and it's easy to loop through the array afterwards. check[<? echo $num; ?>] WILL CREATE A ARRAY.

Batalf
0
 

Author Comment

by:almaha
ID: 6933755
Problem unsolved :| EOS i'm movin to another procesture!
0

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