please help me in drop down list

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Views: 977 | Replies: 21

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Join Date: May 2008

Posts: 24

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almualim

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Re: please help me in drop down list

#11

May 8th, 2008

hi, ok i tried to print this value: $firstdropdownlistvalue = $_POST['col_name'];
by echo($firstdropdownlistvalue);
but it's doesn't appeared in the browser. why?

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Join Date: Nov 2007

Location: Bangalore, India

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nav33n

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Re: please help me in drop down list

#12

May 8th, 2008

Post your latest code.

Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.

*PM asking for help will be ignored*

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Join Date: May 2008

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almualim

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Re: please help me in drop down list

#13

May 8th, 2008

okay:

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
<?php
            mysql_connect ("localhost","root","");
            mysql_select_db ("University");
 
            $firstdropdownlistvalue = $_POST['col_name'];
 
            $get_list_result = mysql_query("select Col_Name from College");
 
 echo '<form name="delete" method="POST" action="delete_dept2.php">';
 
echo'<select name=col_name onchange=javascript: document.delete.submit(); style="position:absolute;left:311px;top:146px;width:303px;z-index:2">';
while ($recs = mysql_fetch_array($get_list_result)){
 
            $display_list =  $recs['Col_Name'];
if($display_list==$firstdropdownlistvalue){
 $selected = "selected";
}
else
{ $selected = ""; 
}     
echo "<option value='".$display_list."' $selected>" .$display_list."</option>";
           }
echo "</select>";
echo ($firstdropdownlistvalue);
 
$get_list_result2 = mysql_query("select * from College where Col_Name='$firstdropdownlistvalue'");
$get_list_result3=  mysql_query("select D_Name from Department where Col_No='$get_list_result2'");
 
echo'<select name="formselect2" style="position:absolute;left:310px;top:223px;width:308px;z-index:4">';
while ($recs2 = mysql_fetch_array($get_list_result3)){
 
            $display_list2 =  $recs2['D_Name'];
 
           echo "<option value='".$display_list2."'> $display_list2 </option>";
           }
echo '</select>';
?>
<?php
            mysql_connect ("localhost","root","");
            mysql_select_db ("University");
             
            $firstdropdownlistvalue = $_POST['col_name'];

            $get_list_result = mysql_query("select Col_Name from College");

 echo '<form name="delete" method="POST" action="delete_dept2.php">';

echo'<select name=col_name onchange=javascript<b></b>: document.delete.submit(); style="position:absolute;left:311px;top:146px;width:303px;z-index:2">';
while ($recs = mysql_fetch_array($get_list_result)){
           
            $display_list =  $recs['Col_Name'];
if($display_list==$firstdropdownlistvalue){
 $selected = "selected";
}
else
{ $selected = ""; 
}     
echo "<option value='".$display_list."' $selected>" .$display_list."</option>";
           }
echo "</select>";
echo ($firstdropdownlistvalue);

$get_list_result2 = mysql_query("select * from College where Col_Name='$firstdropdownlistvalue'");
$get_list_result3=  mysql_query("select D_Name from Department where Col_No='$get_list_result2'");

echo'<select name="formselect2" style="position:absolute;left:310px;top:223px;width:308px;z-index:4">';
while ($recs2 = mysql_fetch_array($get_list_result3)){
           
            $display_list2 =  $recs2['D_Name'];
     
           echo "<option value='".$display_list2."'> $display_list2 </option>";
           }
echo '</select>';
?>

Last edited by peter_budo : May 11th, 2008 at 10:19 am. Reason: Keep It Organized - please use [code] tags

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Join Date: Nov 2007

Location: Bangalore, India

Posts: 3,098

Reputation: nav33n has a spectacular aura about

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nav33n

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Posting Sensei

Re: please help me in drop down list

#14

May 8th, 2008

onchange="javascript: document.delete.submit();" instead of
onchange=javascript: document.delete.submit();

Secondly,

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$get_list_result2 = mysql_query("select * from College where Col_Name='$firstdropdownlistvalue'");
$get_list_result3= mysql_query("select D_Name from Department where Col_No='$get_list_result2'");

is wrong. What are you trying to do ?

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Join Date: May 2008

Posts: 24

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almualim

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Re: please help me in drop down list

#15

May 8th, 2008

see the last update on the code:

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
<?php
            mysql_connect ("localhost","root","");
            mysql_select_db ("University");
 
            $firstdropdownlistvalue = $_POST['col_name'];
 
            $get_list_result = mysql_query("select Col_Name from College");
 
 echo '<form name="delete" method="POST" action="delete_dept2.php">';
 
echo'<select name=col_name onchange="javascript: document.delete.submit();" style="position:absolute;left:311px;top:146px;width:303px;z-index:2">';
while ($recs = mysql_fetch_array($get_list_result)){
 
            $display_list =  $recs['Col_Name'];
if($display_list==$firstdropdownlistvalue){
 $selected = "selected";
}
else
{ $selected = ""; 
}     
echo "<option value='".$display_list."' $selected>" .$display_list."</option>";
           }
echo "</select>";
echo ($firstdropdownlistvalue);
 
$get_list_result2 = mysql_query("select col_No from College where Col_Name='$firstdropdownlistvalue'");
while ($recs2 = mysql_fetch_array($get_list_result2)){
 
            $display_list2 =  $recs2['col_No'];
}
 
 
 
$get_list_result3=  mysql_query("select D_Name from Department where Col_No='$$display_list2'");
 
echo'<select name="formselect2" style="position:absolute;left:310px;top:223px;width:308px;z-index:4">';
while ($recs3 = mysql_fetch_array($get_list_result3)){
 
            $display_list3 =  $recs3['D_Name'];
 
           echo "<option value='".$display_list3."'> $display_list3 </option>";
           }
echo '</select>';
?>
<?php
            mysql_connect ("localhost","root","");
            mysql_select_db ("University");
             
            $firstdropdownlistvalue = $_POST['col_name'];

            $get_list_result = mysql_query("select Col_Name from College");

 echo '<form name="delete" method="POST" action="delete_dept2.php">';

echo'<select name=col_name onchange="javascript<b></b>: document.delete.submit();" style="position:absolute;left:311px;top:146px;width:303px;z-index:2">';
while ($recs = mysql_fetch_array($get_list_result)){
           
            $display_list =  $recs['Col_Name'];
if($display_list==$firstdropdownlistvalue){
 $selected = "selected";
}
else
{ $selected = ""; 
}     
echo "<option value='".$display_list."' $selected>" .$display_list."</option>";
           }
echo "</select>";
echo ($firstdropdownlistvalue);

$get_list_result2 = mysql_query("select col_No from College where Col_Name='$firstdropdownlistvalue'");
while ($recs2 = mysql_fetch_array($get_list_result2)){
           
            $display_list2 =  $recs2['col_No'];
}



$get_list_result3=  mysql_query("select D_Name from Department where Col_No='$$display_list2'");

echo'<select name="formselect2" style="position:absolute;left:310px;top:223px;width:308px;z-index:4">';
while ($recs3 = mysql_fetch_array($get_list_result3)){
           
            $display_list3 =  $recs3['D_Name'];
     
           echo "<option value='".$display_list3."'> $display_list3 </option>";
           }
echo '</select>';
?>

Help with Code Tags

(Toggle Plain Text)

$get_list_result2 = mysql_query("select col_No from College where Col_Name='$firstdropdownlistvalue'");

here i want to get college_no(col_No) from college table, if the college name(col_Name) equals the value came from 1st dropdown list($firstdropdownlistvalue)

$get_list_result3= mysql_query("select D_Name from Department where col_No='$display_list2'");

and here i want to get department name(D_Name) from Department table, if college_no(col_No) equals $display_list2

Last edited by peter_budo : May 11th, 2008 at 10:20 am. Reason: Keep It Organized - please use [code] tags

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Join Date: Nov 2007

Location: Bangalore, India

Posts: 3,098

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nav33n

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Posting Sensei

Re: please help me in drop down list

#16

May 8th, 2008

Then use a sub query.

Help with Code Tags

php Syntax (Toggle Plain Text)

$query = "select D_Name from Department where col_No = (select col_No from College where col_name = '$firstdropdownlistvalue')";

This will return the D_name of the selected col_name.

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Join Date: May 2008

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almualim

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Re: please help me in drop down list

#17

May 8th, 2008

Tired with me, but the problem still exist and I do not know what to do...

Last edited by almualim : May 8th, 2008 at 8:37 am.

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Join Date: Nov 2007

Location: Bangalore, India

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nav33n

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Posting Sensei

Re: please help me in drop down list

#18

May 8th, 2008

Use firefox to execute your scripts. Make use of error console to see any javascript error.

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Join Date: May 2008

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almualim

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Re: please help me in drop down list

#19

May 8th, 2008

i tried with firefox, but when i was select value from 1st dropdown list Transmission occurs for the second page, but i want the values appear in the same page.