not pulling down from database properly

[cass27]

Hi,

I have been trying to create a small CMs with a Staff page, content page and index page. I have created a navigation system and this appears to work fine. The database has fields id, menu_name, position, visible, and content.

the content page is a trial page. A lot of it is html and css. I am trying out the php in the first main box in the centre.

However when i try to draw down the contents of the table, when I click on home I get the home contents "this is home". But when i click on other menu items the same content comes up - "this is home".

http://www.books3333.com/content.php
The functions are as follows:

 function confirm_query($result){
    global $connection;
    if(!$result){
    die("Database query failed".mysqli_error($connection));
}
}
/*---------------------------------------get_menu_names----------------------------------------*/
function get_menu_names(){
    global $connection; 
    $query="Select * From pages ";
    $result=mysqli_query($connection,$query);
    confirm_query($result);
    return $result;
}
/*--------------------------------------Read database--------------------------------------------*/
function get_subjects_by_id($subject_id){
    global $connection;
    $query = "SELECT * FROM pages ";
    $result=mysqli_query($connection,$query);
    confirm_query($result);
    if($subject = mysqli_fetch_array($result)){
    return $subject;
    }else{
    return NULL;
}
} 

The content page is as follows:

 <?php include("includes/connect.php");
include("includes/functions.php");
if(isset($_GET['entry'])){
    $sel_name = $_GET['entry'];
    }else{
    $sel_name = "";
}
$allpages = get_subjects_by_id($sel_name);
include("includes/header.php");
?>
<link href="temps1.css" rel="stylesheet" type="text/css">


<div class="menu">
<ul >
<?php
//get-menu-names() draws down the data. The while loop puts the data in the right place in the list.
$result = get_menu_names();
    while($row=mysqli_fetch_array($result )){
    echo "<li><a href=\"content.php?entry="
    .urlencode($row['id'])."\">"
    .$row['menu_name']."</a></li>";
    }
    ?>
</ul>

</div><!--end menu-->
</div><!--end outerbanner-->
<div  class="maincontent"><div class="newsbox"><h3><b><u>Corporate News</u></b></h3><p>Lorem ipsum dolor sit amet, consectetur adipiscing elit. Quisque ornare, nibh a posuere viverra, leo leo varius felis, nec molestie nisi arcu quis lorem.</p><h4 align="center" class="style1"><u>New Policies</u></h4>
<p> Fusce dignissim, tortor et ullamcorper eleifend, diam felis auctor mauris, mollis blandit est lacus vel nulla. Vestibulum ante ipsum primis. Fusce dignissim, tortor et ullamcorper eleifend, diam felis auctor mauris, mollis blandit est lacus vel nulla.</p><h4 align="center" class="style1"><u>Wind of Change</u></h4> 
<p>Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere cubilia Curae; Fusce porttitor nisl ac tellus tincidunt adipiscing. </p></div><!--endof newsbox-->
<div class="splash"><div class="servicepics1" >
<div class="servicepic"><img src="images/offices1.jpg" /></div><!--pic1-->
<div class="servicepic"><img src="images/officemeeting1.jpg" /></div><!--pic2-->
<div class="servicepic"><img src="images/officeadvice1.jpg" /></div><!--pic3-->
<div class="servicepic"><img src="images/officeGirl1.jpg" /></div><!--pic4-->
</div><!--end of servicepics wrap-->
</div><!--end of splash-->
<div class="databox">
  <div class="data1"><?php echo $allpages['content'];   ?> </div>
    <!--end of data1-->

I have not put all the html and css for this page in.

Can any one see where I am going wrong?

Thanks for your help

cass27

[Dan Grossman]

Your get_subjects_by_id() function does not actually use the subject_id passed into it. You need to add a WHERE clause to the query to only get the row for that ID.

$query = "SELECT * FROM pages WHERE id = $subject_id"; 

[cass27]

hi,

I did as you suggested but I got this error message:

Database query failedYou 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

The function now looks like this:

 function get_subjects_by_id($subject_id){
    global $connection;    
    $query = "SELECT * FROM pages WHERE id = $subject_id ";    
    $result=mysqli_query($connection,$query);
    confirm_query($result);
    if($subject = mysqli_fetch_array($result)){
    return $subject;
    }else{
    return NULL; 

Any ideas?

[Dan Grossman]

Does $subject_id have a value in it?

[captainccs]

Your get_subjects_by_id() function does not actually use the subject_id passed into it. You need to add a WHERE clause to the query to only get the row for that ID.

$query = "SELECT * FROM pages WHERE id = $subject_id"; 

You might need to add single quotes in the WHERE clause:

$query = "SELECT * FROM pages WHERE id = '$subject_id'"; 

[Dan Grossman]

I'm assuming his id column is numeric since all the values in his links are numbers. Numeric columns aren't quoted.

[captainccs]

I'm assuming his id column is numeric since all the values in his links are numbers. Numeric columns aren't quoted.

But if empty it will throw the error he reported.

[Dan Grossman]

I'm guessing the error is originating here:

if(isset($_GET['entry'])){ 
    $sel_name = $_GET['entry']; 
    }else{ 
    $sel_name = ""; 
}

The default should be the ID of the default page, not a blank.

[cass27]

hi, Thanks for all your input.

The solution was the commas around '$subject_id'

I adapted this script from one on the web as i am still a bit of a newbie with php. What I am still trying to get my head around is when you need to pass an argument and when you dont need to. And what that argument should be.


cheers

cass27

[captainccs]

What I am still trying to get my head around is when you need to pass an argument and when you dont need to. And what that argument should be.


cheers

cass27

Cass:

Think of a function as a car body shop. If you want them to work on your car, you have to drive your car to the shop. If you want them to work on your wife's car, then that is the car or 'argument' or 'parameter' you pass to the body_shop($car) function.

function body_shop($car) {
  work on $car;
  return fixed $car;
}

$myCar = 'VW';
// to fix my car
body_shop($myCar);

$wifeCar = 'Mercedes';
// to fix wife's car
body_shop($wifeCar); 

[darkdrgn2k]

i find the fastest way to debug SQL is to dump the SQL string before each "query" and run them through mysql manualy

that way i see WHERE the SQL string looks wierd, or the result is not what i expected!

[cass27]

Denny wrote:

Cass:

Think of a function as a car body shop. If you want them to work on your car, you have to drive your car to the shop. If you want them to work on your wife's car, then that is the car or 'argument' or 'parameter' you pass to the body_shop($car) function.



PHP Code:
function body_shop($car) { work on $car; return fixed $car;}$myCar = 'VW';// to fix my carbody_shop($myCar);$wifeCar = 'Mercedes';// to fix wife's carbody_shop($wifeCar);


This makes sense. $car although an empty variable in the function is a kind of class in a way.

So why are thier times when no argument is passed?

cass

p.s. Surely i should have the mercedes and my wife should have the VW....?

[captainccs]

So why are thier times when no argument is passed?

cass

I suppose the answer to that depends on how the function is written. The SQL query function that started this thread expected a non-empty number and it didn't bother to check if the incoming data was valid. Since it wasn't, the function blew up. Let's call it "less than optimal coding." By adding the single quotes, I prevented the function from blowing up from bad or missing data but it probably still gives bad results when passed an empty argument.

Since functions work on data it makes sense to pass them arguments. The only time you would not do so is if there are default values the function can use. For example, the function time() takes a void argument because it uses "now" as the default argument.