PHP Program help please.

[SpeedCrazy]

I am making myself an image gallery from scratch and as i am planning to build an image uploader into it so i have to has a dynamic image loader to make sure all images are loaded. So i have a working mysql db that stores image # album # comments title and date uploaded.

So far my program was able to load all that info and display it, so now i am trying to make it take the album and img numbers and form the <img /> src url.

I need some help getting it to work as right now it is not working, i get an "unexpected T_string parse error"

PHP Code

<?php

$conn = mysql_connect("localhost","portfolio","portfolio") or die(mysql_error());
mysql_select_db("portfolio");

$sqlA = "SELECT album FROM pics";
$resultA = mysql_query($sqlA, $conn) or die(mysql_error());

while($row=mysql_fetch_assoc($resultA)){
foreach($resultA as $album){
	$sqlI = "SELECT img FROM pics WHERE album $resultA";
	$resultB = mysql_query($sqlI, $conn) or die(mysql_error());

	print "<div>"<img src="http://localhost/PortfolioSite/img/content/full/$resultB.jpg"/>"</div>"
} //end foreach
}// end while

?>

Any ideas?

Thanks

 

EDIT: I could just add the url to the db. Will that work ya think?

Edited May 26, 2011 by SpeedCrazy

[flareback]

print "<div>"<img src="http://localhost/PortfolioSite/img/content/full/$resultB.jpg"/>"</div>"

A quick look makes it seem your print statement has the quotes messed up. try

 

print '<div><img src="http://localhost/PortfolioSite/img/content/full/'.$resultB.'.jpg"/></div>'

 

actually I'm not sure it likes the php variable in the middle of the url either.

Edited May 26, 2011 by flareback

[SpeedCrazy]

oh i forgot the single quotes tried everything else and then decided the error was else where. It still might be easier to just store the url in the db.

[SpeedCrazy]

The foreach doesn't work. It tells me it should be an array but it is just a single value. How would i fix that?Is there an alternative to foreach that deals with single values?

[flareback]

Okay, on a closer look you're using the result from the queries wrong. Those are resources being returned not text. I would try something that looks more like the following. I took out the foreach and added a while statement for your second query

 

<?php

$conn = mysql_connect("localhost","portfolio","portfolio") or die(mysql_error());
mysql_select_db("portfolio");

$sqlA = "SELECT album FROM pics";
$resultA = mysql_query($sqlA, $conn) or die(mysql_error());

while($row=mysql_fetch_assoc($resultA)){
	$sqlI = "SELECT img FROM pics WHERE album = ". $row['album'];
	$resultB = mysql_query($sqlI, $conn) or die(mysql_error());
	while($rowB = mysql_fetch_assoc($resultB)){
	   print '<div><img src="http://localhost/PortfolioSite/img/content/full/'.$rowB['img'].'.jpg"/></div>';
               }
}// end while

?>

[SpeedCrazy]

Okay, that makes more sense. I will try it.

LOL just realized that the password is in the code i gave. Stupid mistake.

[SpeedCrazy]

Thanks that works.

Now i will have a go at adding title and comments to that.

[SpeedCrazy]

Okay, i think this should add titles but it throws a very unusual error.

<?php

$conn = mysql_connect("localhost","portfolio","portfolio") or die(mysql_error());
mysql_select_db("portfolio");

$sqlA = "SELECT album FROM pics";
$resultA = mysql_query($sqlA, $conn) or die(mysql_error());

while($row=mysql_fetch_assoc($resultA)){
               $sqlI = "SELECT img FROM pics WHERE album = ". $row['album'];
               $resultB = mysql_query($sqlI, $conn) or die(mysql_error());

               $sqlT = "SELECT title FROM pics WHERE album =". $row['album']."AND img=" .$resultB['img'];
			$resultC = mysql_query($sqlT, $conn) or die(mysql_error());

			$rowC=mysql_fetch_assoc($resultC);

               while($rowB = mysql_fetch_assoc($resultB)){
                  print '<div id="album'.$row['album'].'"><img src="http://localhost/PortfolioSite/img/content/album'.$row['album'].'/full/'.$rowB['img'].'.jpg"/><h3>'. $rowC['title'] .'</h3></div>';
               }
}// end while


 ?>

it says

div #main 4 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 'img=' at line 1

This program is quite literally the only thing on the page so i haven't a clue what it means.

[flareback]

you can select more than one item from a table at a time. try this

 

$sqlI = "SELECT img, title FROM pics WHERE album = ". $row['album'];

then in your while loop you can access it with $rowB['title']. That will help cut down on querying the database. Try checking out some tutorials on using the select statement. Here is a pretty simple example that should get you what you want.

[SpeedCrazy]

But i need to separate them.

I want to have the image and then below it the title and comments. Though i have decided it might be easier to have the <img src="htt://etc." /> all in the database requiring just 1 query.

Edited May 27, 2011 by SpeedCrazy

[flareback]

But i need to separate them.

I want to have the image and then below it the title and comments. Though i have decided it might be easier to have the <img src="htt://etc." /> all in the database requiring just 1 query.

You can separate it.

$sqlI = "SELECT img, title FROM pics WHERE album = ". $row['album'];
$resultB = mysql_query($sqlI, $conn) or die(mysql_error());

while($rowB = mysql_fetch_assoc($resultB)){
  print '<div id="album'.$row['album'].'"><img src="http://localhost/PortfolioSite/img/content/album'.$row['album'].'/full/'.$rowB['img'].'.jpg"/><h3>'. $rowB['title'] .'</h3></div>';
}

[flareback]

have a look at this page. Be sure to read about the returned value.