Passing array from one page to another

Jack31081

What I want to do is simple, and I'm sure it gets done often enough, so here goes.

I want to send an array from one page to another page, for the purposes of displaying a slideshow of images. The first page sets up the array of images to be shown and when the user clicks the link, the slideshow page opens and displays the images in order.

Ideally, I'd like to be able to send it directly through the URL, so I don't need to resort to sessions or cookies, in case some user's security settings are really high. So if it's possible to send the array through the URL query, how would I do this? A note: the array can be of varying size.

If I can't do it that way, can someone show me how to do it by using sessions?

thanks.

patmcv

There are a few ways to do it. You could pass it by the URL, but you will need to use serialize() on it and unserialize() in the receiving page, and probably urlencode() and urldecode() as well.
Another way, which I would prefer as long as its simple would be to pass the elements of the array in hidden form fields. Like this:

foreach($array as $key=>$value)
{
echo '<input type="hidden" name="array['. $key .']" value="'. $value .'">';
}

When this is received, php will reassemble your array for you.

Jack31081

Ok. So let's say my goal is to have an array on the second page that looks like this:

$path = array(0=>1,2,3,4,5,6,7,8,9);

In the first page, I'd write:

$temp = array(0=>1,2,3,4,5,6,7,8,9);

foreach($temp as $key=>$value)
{
echo '<input type="hidden" name="array['. $key .']" value="'. $value .'">';
}

Is that it? What would I need to write on the second page, if anything? Do I need anything more than:

$path = $temp;

Also, there is no form on the first page...does that affect whether these hidden inputs get from page1 to page2?

Jack31081

On second thought, I don't think that will work.

There will be a number of links on the page, and the link clicked on will determine which and how many images are shown on the slideshow page. For instance, think of a page that has a bunch of different slideshows you can watch, but all the links go to the same php page, but each page sends a different array of numbers, which are the IDs of the pictures to be seen.

patmcv

What I wrote assumed your array is called $array, so in your case:
foreach($temp as $key=>$value)
{
echo '<input type="hidden" name="temp['. $key .']" value="'. $value .'">';
}

Secondly, yes you will have to have a form and submit the form. So instead of a link to take you to the next page, use a submit button. And echo out your form tag before the 'foreach' and the submit button and closing form tag after it.

$path = $temp will work if register_globals is on your php.ini,if not, then $path = $POST['temp']. (this assumes the method of the form is set to POST, otherwise $GET['temp']).

patmcv

On second thought, I don't think that will work.

It should still work fine, you would just have several forms, depending on which one is submitted, a different array of values would be passed, so a different group of images would be shown.

But the simpler way to do it would be to store the image identifiers in a database, and associate each with a slideshow. Then you could have slideshow number 1, your sql query would select the image identifiers associated with this id, and you would only have to pass the slideshow id in the url.

Jack31081

Damn, why didn't I think of that.

Ok, I like this idea, storing the slideshows in a sql database and referring to them by their ID.

So I can do

What would I store in the database though?

The only info I'd need to store would be the images to show for that slideshow, and maybe the title of the show.

would I store 'array(0=>1,2,3,4,5,6,7,8,9)' in the database and then write:

$sql = "SELECT * from slideshowDB where ID=$slideID";
$result = mysql_query($sql,$conn);

$path = mysql_result($result,0,'array');
$title = mysql_result($result,0,'showTitle');

That should work, no?

patmcv

I would do it by having a slideshow table:

slideshow_id int unsigned not null auto_increment, slideshow_title varchar(128)

then a table called slides: slide_id int (etc), slide_title, slide_path
(slide_path is the path to the image file)

and then have a join table which just contains slide_id and slideshow_id (call it slides_to_slideshows)

then if there is an entry in that table like 1,12 it means that slide 1 belongs to slideshow 12. This way, you can also have 1, 2 (slide 1 belongs to slideshow 2), and 3, 12 (slide 3 belongs to slideshow 12) so any slide can belong to any slideshow and vice versa.

then your sql will be something like:

SELECT slide_id, slide_title, slide_path FROM slides AS s, slides_to_slideshows AS ss, slideshows as ssh WHERE s.slide_id = ss.slide_id AND ss.slideshow_id = ssh.slideshow_id AND ssh.slideshow_id = '". $_GET['id'] ."'";

This might seem complicated, but it is worth doing it properly.

Jack31081

Ok. So I took what you wrote and put together this simple script to print out the IDs of the slides in a given slideshow:

<?php

$conn = mysql_connect(...);
mysql_select_db(...,$conn);

$sql = "SELECT slideID FROM slides AS s, slides_to_shows AS ss, 
slideshows AS ssh WHERE s.slideID = ss.slideID AND ssh.slideshowID = ".$ssID." AND ss.slideshowID = ssh.slideshowID";
$result = mysql_query($sql, $conn) or die(mysql_error());
while($newArray = mysql_fetch_array($result)) {
$sID = $newArray['slideID']
print "<p>$sID</p>";
}

?>

But it doesn't work. I just get a blank page.

any idea why?

Jack31081

I did some muckin' around, made the script even simpler, and I get the error:

Column: 'slide_id' in field list is ambiguous

Here's the script

<?php

$conn = mysql_connect("...");
mysql_select_db("...",$conn);

$sql = "SELECT slide_id FROM slides AS s, slides_to_shows AS ss, slideshows AS ssh WHERE s.slide_id = ss.slide_id AND ssh.slideshowID = ".$ssID." AND ss.slideshowID = ssh.slideshowID";
$result = mysql_query($sql, $conn) or die(mysql_error());
$rows = mysql_num_rows($result);
print "Number of rows: $rows";
?>

patmcv

Column: 'slide_id' in field list is ambiguous

because there is a slide_id in more than one of the tables in your join, you have to specify which one you want, e.g. SELECT s.slide_id FROM slides AS s....

In this case, you are only selecting where they are equal, so it wouldn't matter which slide_id you selected

I just get a blank page.

any idea why?

Have you tried running the query from the command line or through phpmyadmin to see if you get any results? The syntax of the query looks correct to me. In your new version you use mysql_num_rows, so that will tell you whether or not any rows are returned.

Jack31081

thanks a ton. I replaced "SELECT slide_id" with "SELECT s.slide_id" and although it doesn't make a lot of sense, it worked. Now I can get a table of the slides that belong to any slideshow.

I may drop by later today to ask about pulling info from this table in PHP, but until then, that's all.

khurramijaz

Hi, Just saw your discussion. you can simply collect array input from a form as <input name='slides[]' .....

I dont think nothing else is required. Used it millions of times in many projects. Will work.

Jack31081

Well, I thought I had it nailed, but I'm still running into a problem. When the page is opened, it endlessly loads, never getting past 20% or so? What does this mean? I think it may be linked to the following chunk of code...wanna check it out for me?

//Set up connection
$conn = mysql_connect("...");
mysql_select_db("...",$conn);

//Retrieve image path and title path
$sql = "SELECT s.slide_id, s.slidename FROM slides as s, slides_to_shows as ss, slideshows as ssh WHERE s.slide_id = ss.slide_id AND ss.slideshowID = ssh.slideshowID AND ssh.slideshowID = ".$ssID."";
$result = mysql_query ($sql, $conn) or die(mysql_error());
$inc = 0;
while($newArray = mysql_fetch_array($result)) {
$path[] = $newArray['slideID'];
$pathtitles[] = $newArray['slidename'];
}

$pathsize = count($path);
//Set $curr to the current point in the path
$curr = 0;
while($id != $path[$curr]) {$curr++;}

//Get all info for current slide
$sql = "SELECT * from slides where slide_id=$id";
$result = mysql_query($sql, $conn) or die(mysql_error());

$newArray = mysql_fetch_array($result);
$testField = $newArray['slidename'];
$imgsrc = $newArray['image'];
$notes = $newArray['notes'];
$FAQ = $newArray['faq'];

Jack31081

I hate it when one tiny thing kills your code!!!

I still had a 'slideID' that I hadn't converted to 'slide_id'....:mad:

I've got it all working great now...for the time being at least.

Having a 'slides_to_shows' table is going to make my life much easier. Thanks a lot, patmcv.

patmcv

Glad it works!

Just one thing though, I can't see why you need to repeatedly query the database in a loop to get all the info for your slides - you could do this in the original query, just by listing all the fields instead of just 2 of them:

$sql = "SELECT s.slide_id, s.slidename, s.next_field, s.next_field ... FROM slides ...

Jack31081

there's only two loops. One to get the arrays for the slide names and slide IDs, and one to increment the $curr variable until it's equal to the location of the current slide in the array.

neither loop has a sql query in it...

patmcv

Sorry, I didn't read it properly. Still, you could make the whole array in one query and save a bit of code. Then again, if it ain't broke...