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Need help, displaying database content not working

Posted: Mon Sep 22, 2008 12:20 am
by wolfwood16
hi all,

im making a program in which a student could view the subject curriculum with his decided curriculum year,...I've created a new table named curr_subjs with the ff fields:

Code: Select all

 
id INT  
subj_code VARCHAR,
subj_desc VARCHAR,
lec_hrs INT,
lab_hrs INT,
subj_units INT,
subj_preq VARCHAR,
subj_sem INT,
subj_level INT
subj_curr VARCHAR
 
now, if the admin created a new curriculum, the name(for example, The New Curriculum) of that curriculum will be stored in curr_names.When he add new subjects, those subjects will be stored in curr_subjs but the value for subj_curr field will be the modified one(thenewcurriculum).

the code is working properly...

but i have a situation which i cant handle regarding on retrieving the data from the table curr_subjs. The code returns no error, but ain't working

on the admin page, there's an option to view the added curriculum, here's the code

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include_once 'conn.php';
$getCurriculum = "SELECT * FROM curr_names";
$getCurriculumNow = mysql_query($getCurriculum);
while($crclm = mysql_fetch_array($getCurriculumNow)){
         print<<<HERE
                 <tr>
              <td width="493" height="27" valign="middle" style="background:#ffffff; color:#333333; padding-left: 10px;">$crclm[currname]</td>
          <td width="80" valign="middle" style="background:#ffffff; color:#333333; padding-left: 10px;">$crclm[currcourse]</td>
          <td width="71" valign="middle"><form name="form2" method="post" action="[color=#80FF00]viewcurr.php?mode=View&viewCurrIs=[/color][color=#800000]$crclm[/color][color=#40FF00][id][/color]"><input type="submit" class="qpost_button" name="LetsViewCurr" value="View" /></form></td>
         </tr>
HERE;
 
}
 
Upon selecting his decided curriculum he want to view, the page will bring him to viewcurr.php

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//on the top of the page
if(!isset($_SESSION)){
    session_start();
}
 
 
if ($_REQUEST['mode'] == "View") {
    include_once '../admin_/conn.php';
    if ($_GET['id']) {
        $id = (int)$_GET['id'];
        
        $getCurriculumNow = mysql_query("SELECT currname FROM curr_names WHERE id = '$id'");
        
        $strippedCurrName = strtolower(strip_tags(str_replace(' ', '', $getCurriculumNow)));
        echo ($getCurriculumNow." and ".$strippedCurrName) ;
    }
}
 
get the id of the selected curriculum, and returns the unmodified name(The New Curriculum), and modify it with the $strippedCurrName variable so the name now will be (thenewcurriculum).


and on the result section

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[color=#800000]$GetSubjsF[/color] = "SELECT * FROM curr_subjs WHERE subj_curr = '$strippedCurrName' AND subj_level = 1 AND subj_desc != '' AND subj_sem = 1";
 
[color=#400000]$GetSubjsNowF[/color] = mysql_query([color=#800000]$GetSubjsF[/color]);
 
while ($subjSecondSemView = mysql_fetch_array($GetSubjsNowS)) {
echo "$subjSecondSemView[subj_code] ";
echo "$subjSecondSemView[subj_desc] ";
echo "$subjSecondSemView[lec_hrs] ";
echo "$subjSecondSemView[lab_hrs] ";
echo "$subjSecondSemView[subj_units] ";
echo "$subjSecondSemView[subj_preq]";
}
 

again, the code did not return any error, but not functioning...

Please help me solve this....Many thanks in advance and God bless us always...

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 12:32 am
by pcoder
First of all you make sure your returned query is perfect and try to run it on PHPMYADMIN.

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 12:44 am
by wolfwood16
when i try running a query to phpmyadmin, it works....i duno, it seems that i'm having problem on getting the id from the fetched array on the view section of admin page...

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 12:46 am
by pcoder
Did you try to view your fetched array, I mean like

Code: Select all

 
print_r($subjSecondSemView);
 

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 12:57 am
by wolfwood16
i run the program and view a curriculum with an id=7 (/admin_/viewcurr.php?mode=View&viewCurrIs=7)

print_r(); did not return anything,

so i assumed it's on the fetched array id of that curriculum so

print_r($getCurriculumNow); returns nothing,
print_r($strippedCurrName); returns nothing,
print_r($GetSubjsNowF); returns "Resource id #4 "

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 1:02 am
by pcoder
On your result section note down the line 4 and line 6.
In line 6, i think $GetSubjsNowS should be $GetSubjsNowF.
And then view the array $subjSecondSemView.

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 1:07 am
by wolfwood16
pcoder wrote:On your result section note down the line 4 and line 6.
In line 6, i think $GetSubjsNowS should be $GetSubjsNowF.
And then view the array $subjSecondSemView.

Sorry bout that but its ok because there's a two results need to be done, first for the First Semester ($GetSubjsNowF) and for the Second Semester ($GetSubjsNowS)

but still the print_r($subjSecondSemView) or print_r($subjFirstSemView) returns nothing...

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 3:12 am
by wolfwood16
*bump*....anyone?

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 3:55 am
by pcoder
Your view Section.

Code: Select all

 
$GetSubjsF = "SELECT * FROM curr_subjs WHERE subj_curr = '$strippedCurrName' AND subj_level = 1 AND subj_desc != '' AND subj_sem = 1"; 
$GetSubjsNowF = mysql_query($GetSubjsF);
while ($subjSecondSemView = mysql_fetch_array($GetSubjsNowS)) {
echo "$subjSecondSemView[subj_code] ";
echo "$subjSecondSemView[subj_desc] ";
echo "$subjSecondSemView[lec_hrs] ";
echo "$subjSecondSemView[lab_hrs] ";
echo "$subjSecondSemView[subj_units] ";
echo "$subjSecondSemView[subj_preq]";
}
 
In this section, the resource is returned in $GetSubjsNowF if Success and you are passing $GetSubjsNowS in mysql_fetch_array. Then how do you get the result?? :)

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 4:09 am
by wolfwood16

Code: Select all

 
<?php 
        
            $GetSubjsF = "SELECT * FROM curr_subjs WHERE subj_curr = '$strippedCurrName' AND subj_level = 1 AND subj_desc != '' AND subj_sem = 1";
            $GetSubjsNowF = mysql_query($GetSubjsF);
            
            
            
            
            while ($subjFirstSemView = mysql_fetch_array($GetSubjsNowF)) {
                echo "<tr>";
                echo "  <td width='8%' style=\"background:#ffffff; color:#333333; padding-left: 5px;\"> $subjFirstSemView[subj_code] </td>";
                echo "  <td width='20%' style=\"background:#ffffff; color:#333333; padding-left: 5px;\"> $subjFirstSemView[subj_desc] </td>";
                echo "  <td width='8%' style=\"background:#ffffff; color:#333333; padding-left: 5px;\"> $subjFirstSemView[lec_hrs] </td>";
                echo "  <td width='8%' style=\"background:#ffffff; color:#333333; padding-left: 5px;\"> $subjFirstSemView[lab_hrs] </td>";
                echo "  <td width='8%' style=\"background:#ffffff; color:#333333; padding-left: 5px;\"> $subjFirstSemView[subj_units] </td>";
                echo "  <td width='8%' style=\"background:#ffffff; color:#333333; padding-left: 5px;\"> $subjFirstSemView[subj_preq] </td>";                
                echo "</tr>";
            }
            
 

still no luck sir...:(

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 4:10 am
by pcoder
Try this:

Code: Select all

 
$GetSubjsF = "SELECT * FROM curr_subjs WHERE 1=1";
$GetSubjsNowF = mysql_query($GetSubjsF);
while ($subjSecondSemView = mysql_fetch_array($GetSubjsNowF)) {
//get the content
}
print_r($subjSecondSemView);
 
Does this return any value?? :)

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 4:17 am
by wolfwood16
still not returning anything sir... :cry: :cry:

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 4:26 am
by pcoder
I am assuming your connection is ok.
First of all, try this:

Code: Select all

 
var_dump($GetSubjsNowF);
 
I mean if $GetSubjsNowF returns some resource id and the table curr_subjs is not empty, then this query should return some value.

Re: Need help, displaying database content not working

Posted: Mon Sep 22, 2008 10:26 am
by wolfwood16
hi sir pcoder, accept my apology for my very late reply. About the program..


var_dump() now return these results

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var_dump($GetSubjsNowF)    > resource(4) of type (mysql result) 
var_dump($GetSubjsNowS)    > resource(5) of type (mysql result) 
 
yes my connection is ok, but still no luck sir... :(