query not going on inside a function

har_dick

hi. I have this function in a file:

<?php
// set of previous functions...

function comment()
{
	$conn = db_connect();

$s_sql = "select * from imagesdb where id = '$imageid'";
$s_result = mysql_query($s_sql, $conn);
if (mysql_num_rows($s_result))
{
	$s = mysql_fetch_array($s_result);
}
echo "hola";
echo $imageid;
echo $s[poster];
?>
// some html
<?php
}
?>

my browser returns no errors to me but the query doesn't happen, i mean it returns no result (blank page)

because I am new in php, I think my mistake is with $imageid maybe not making it as global or something... I tried so but I got parse errors.

the function that displays content including the initial $imageid is:

<?php

function display_content()
{

echo "<p align='center'><font face='Arial, Helvetica, sans-serif' size=4><b>Image gallery</b></font></p>";

$conn = db_connect();

$s_sql = "select * from imagesdb order by image asc LIMIT 0 , 5";
$s_result = mysql_query($s_sql, $conn);
if (mysql_num_rows($s_result)) 
{
	$s = mysql_fetch_array($s_result);
}
?>
<p>
<table width='99%' border='1' cellspacing='0' cellpadding='5'>


<p><a href="javascript:popUp('comment.php?imageid=<? echo $s[id]; ?>')"><img src="images/images/<? echo $s[poster]; ?>" width="94" height="140"></a></p>    
<p><font face="Arial, Helvetica, sans-serif" size=2><b><? echo $s[image]; ?></b></font></p>
<p><font face="Arial, Helvetica, sans-serif" size=2><b>Genre:</b> <? echo $s[genre]; ?></font></p>   


<input type=hidden name="imageid" value="<?php echo $s[id]; ?>">
<?php

// etc...........

?>

where as you can see I included:

to send $imageid to the comment() function's query, but I don't know what's wrong with it.

Can anyone help me on this ? thanks.

AQHost

Hi,

Ok, first things first: the hidden input field that you've added won't do anything if it isn't part of an HTML form, so it serves no purpose within your code. The link to...

comment.php?imageid=<? echo $s[id]; ?>

...should be fine. In your comment.php file, which I assume calls the comment() function (right?) you need to make the call like this:

comment($_GET['imageid']);

Then you should modify the comment() function to look like this:

<?php 
// set of previous functions... 

function comment($myimageid) 
{ 
    $conn = db_connect(); 

$s_sql = "select * from imagesdb where id = '$myimageid'"; 
$s_result = mysql_query($s_sql, $conn); 
if (mysql_num_rows($s_result)) 
{ 
    $s = mysql_fetch_array($s_result); 
} 
echo "hola"; 
echo $myimageid; 
echo $s[poster]; 
?> 
// some html 
<?php 
} 
?>

Best wishes,
Simon.

har_dick

I did as you indicated but it didn't work at all 🙁

AQHost

Need a little more information please. Do you get an error message? How are you calling the comment() function and from what file? Does the echo $myimageid; line in that function return anything? Etc etc. The more details you can provide the better.

Best wishes,
Simon.

har_dick

it's ok now.

I had a misstype someplace prior the comment() function..

thanks a lot..

can i ask you something else ?

i have heard that using $variable is non secure, and that's better to use $ .... but since i'm new at php, i would like to have a brief explanation on these $ variables.... why and where to use them.

rheroux

Take a gander here.

har_dick

ok another ....question:

do I have to pass a parameter to every function in the file that involves working with a query and a submit form ?

I think I may have the same problem with the submit_image() function listed below:

<?php

function submit_final()
{
	$date = date("M d, Y");

$conn = db_connect();

$time = time();

$sql = "insert into imagesdb (filename, image, stars, plot, poster) values ('$filename', '$image', '$stars', '$plot', '$poster')";

$result = mysql_query($sql,$conn);

if (!$result) {
print "There was an error on dbase executing.<PRE>$sql</PRE>";
print mysql_error();
exit;
}

if ($poster_name != "")
{    
	move_uploaded_file("$poster", "/home/mywebbsite/public_html/images/images/$poster_name") or die("Couldn't Upload Your File.");

} else {

	die("Didnt upload any file at all!");
}

if (!$story)
{
	$story = mysql_insert_id();

$sql = "update imagesdb set poster = '$poster_name' where id = $story";
$result = mysql_query($sql,$conn);

echo "<center><font face='arial' size='2'><b>Well done !</b></font></center>"; 
	}
}
?>

I call my submitfile.php like this:

<?php
include functions.php

// bla bal bal 
submit_final();
?>

thanks for answering 🙂