sql code not displaying table data

Question

Hello I am trying to understand why this isn't working on my page. I am using the php block below to use the variable I created at the beginning of my page $sel_subj (I used $_GET to get the id of what I clicked on on the previous page. I want the new page to reflect the data of the link I clicked on) I got the url to work to show the correct number from the database but I cannot get my page to display the name of what link was pressed; aka the data in the column labeled 'subject_name' from the 'subjects' table.

<?php
$query = "SELECT * FROM subjects WHERE id = \"$sel_subj\"";
$result_set = mysql_query($query, $connection);
if(!$result_set) {
    die("Database query failed: " . mysql_error());
$subject = mysql_fetch_array($result_set);
return $subject;
?>

into this h2 tag right here.

<h2><?php echo $subject['subject_name']; ?>Hello</h2>

I can post the whole page if it will help. I appreciate everyone's input.
Thank you.

edit:new problems

Here is the bottom half of my code. I have a navigation div above this code which separates the links to the pages that relate to the database tables from the content that I'd like to pull from the db and display in the div for the page table.

However with the code I've provided nothing is showing up on the page when I open it in firefox. In my html when I "view source" while previewing on the testing server there is nothing underneath ...

<td id="page">

<?php
$query = "SELECT * FROM subjects WHERE id ='$sel_subj'";
$result_set = mysql_query($query, $connection);
if(!$result_set) 
    die("Database query failed: " . mysql_error());
$subject = mysql_fetch_assoc($result_set);
return $subject;

?>
            <h2><?php echo $subject['subject_name']; ?>Hello</h2>
            <br />
            <?php echo $sel_page; ?><br />
            <?php echo $sel_subj; ?><br />
            <?php echo $subject; ?><br />
            <?php echo $subject['id']; ?>
        </td>

I have tried what you all said and its now giving me an error I had earlier which I'm not sure what it means. You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 if I add LIMIT 1 to the sql query the error changes, saying : near 'LIMIT 1' at line 1 at the end of the error instead... "SELECT * FROM subjects WHERE id ={$sel_subj} LIMIT 1"; — 1337475
– 1337475, Commented May 1, 2012 at 7:51
And just to clarify. This is the links on my page which relate to the database with the table "subject have bulleted list underneath them that relate to a table called "pages". This error arrises when I click the link relating to the pages table, not the subjects' table's links.... I'm sure I'll figure that problem out as soon as I get some help regarding my new problem in the question in the comment above; however I figured I'd post this information regardless as it may help someone help me. Thank you so much. — 1337475
– 1337475, Commented May 1, 2012 at 7:59

Radu Murzea · Accepted Answer · 2012-05-01 08:47:55Z

3

Change this:

$query = "SELECT * FROM subjects WHERE id = \"$sel_subj\"";

to this:

$query = "SELECT * FROM subjects WHERE id = '$sel_subj'";

and this:

$subject = mysql_fetch_array($result_set);

to this:

$subject = mysql_fetch_assoc($result_set);

PS: Try not to use the mysql class of functions anymore, they're not too good. Instead, use mysqli or PDO.

EDIT

If the column id is of numeric type, remove the apostrophes from the query. Like this:

$query = "SELECT * FROM subjects WHERE id = $sel_subj";

edited May 1, 2012 at 8:47

answered May 1, 2012 at 6:16

Radu Murzea

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1 Comment

1337475 Over a year ago

I have changed my code and nothing is showing up under my main content section. I have updated my question a little to further explain.

vvr · Accepted Answer · 2012-05-01 06:19:00Z

1

Try This

$query = "SELECT * FROM subjects WHERE id = '$sel_subj' ";

edited May 1, 2012 at 6:19

answered May 1, 2012 at 6:12

vvr

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Comments

Moyed Ansari · Accepted Answer · 2012-05-01 06:19:23Z

1

Try this

$query = "SELECT * FROM subjects WHERE id = {$sel_subj} ";

answered May 1, 2012 at 6:19

Moyed Ansari

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