Arrays within variables

Question

I am trying to produce an array of values to be used by a variable, but I can't get it to work. I don't want to "see" the results. Instead, I want the variable to "use" the results. Here's my code:

$rss_results = mysql_query("SELECT feed_id, feed_url, feed_title, feed_order, feed_page_id FROM user_feeds  WHERE ((feed_owner = '" . $_SESSION['UserId'] . "'));");

    if ($rss_results) {

    while($row = mysql_fetch_array($rss_results))
    {
    $feed->set_feed_url(print "'".$row[feed_url]."',");

    }

    }
    else {
      // something went wrong.
      echo mysql_error();

The problem is with the line:

$feed->set_feed_url(print "'".$row[feed_url]."',");

EDIT 1) Yes. I know print doesn't belong there. I forgot to remove it after doing some troubleshooting earlier.

2) I am trying to use this code with an RSS parser calleld SimplePie.

3) Everyone seems to agree that the following code works

   $feed->set_feed_url(array("'" . $row['feed_url'] . "',"));

...but it doesn't. If I weren't accessing my db, the code would look something like this....

$feed->set_feed_url(array('http://www.tmz.com/rss.xml', 'feed://rss.cbc.ca/lineup/topstories.xml'));

...which works just fine. I am trying to reproduce that effect using values stored in a mysql db. If the code worked, it would be working. So there must be a problem with the code that is "supposed" to be working.

It's not helpful to your question, but it should be noted that you should stop using `mysql_*` functions. They're being deprecated. Instead use [PDO](http://php.net/manual/en/book.pdo.php) (supported as of PHP 5.1) or [mysqli](http://php.net/manual/en/book.mysqli.php) (supported as of PHP 4.1). If you're not sure which one to use, [read this SO article](http://stackoverflow.com/questions/13569/mysqli-or-pdo-what-are-the-pros-and-cons). — Matt, Aug 07 '12 at 15:43
Why are you setting a method parameter as 'print'? This makes no sense, you pass variables to methods.... not printed characters... — David Barker, Aug 07 '12 at 15:45
@DavidBarker - Actually, `print()` always returns 1, so you're not passing printed characters, you're always passing 1. — nickb, Aug 07 '12 at 15:48
I'm pretty sure no one would agree that `$feed->set_feed_url(array("'" . $row['feed_url'] . "',"));` works, because that's not how dynamically populating arrays works. — Matt, Aug 07 '12 at 16:28

score 1 · Answer 1 · edited May 23 '17 at 12:04

1

When accessing an associative array, you must put the index name inside quotes:

Instead of

$row[feed_url]

use

$row['feed_url']

Also,

$feed->set_feed_url(print "'".$row[feed_url]."',");

makes no sense. If you want to pass a value to a method, pass the variable, don't print it to the screen.

$feed->set_feed_url("'" . $row['feed_url'] . "',");

If you don't want the string you pass to $feed->set_feed_url() to be enclosed in 's, just pass it like a regular variable:

$feed->set_feed_url($row['feed_url']);

UPDATE:

Since you mention you're trying to pass an array, you have to first initialize it, then populate it, then send it to the method:

$urls = array();
while ($row = mysql_fetch_array($rss_results)) {
    $urls[] = $row['feed_url'];
}
$feed->set_feed_url($urls);

Doing this:

$feed->set_feed_url(array("'" . $row['feed_url'] . "',"));

Will simply pass the following to $feed->set_feed_url():

array(
    [0] => String() "'[value of $row['feed_url'] ]',"
)

If you want to call $feed->set_feed_url() once for each iteration of the loop, this will work fine:

while($row = mysql_fetch_array($rss_results)) {
    $feed->set_feed_url(array($row['feed_url']));
}

Please note: You should stop using mysql_* functions. They're being deprecated. Instead use PDO (supported as of PHP 5.1) or mysqli (supported as of PHP 4.1). If you're not sure which one to use, read this SO article.

edited May 23 '17 at 12:04

Community

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answered Aug 07 '12 at 15:44

Matt

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Exactly what I was thinking :) – David Barker Aug 07 '12 at 15:47
I forgot to remove print from when I was testing something earlier. Also I tried the code you suggested, and it doesn't seem to be working. – Richard Aug 07 '12 at 15:56
@Richard What does `var_dump($row['feed_url']); exit();` inside the loop produce? – Matt Aug 07 '12 at 15:57
@Matt cURL error 6: Couldn't resolve host 'http'string(39) "http://rss.cbc.ca/lineup/topstories.xml" – Richard Aug 07 '12 at 16:01
@Richard I'm assuming you added that AFTER `$feed->set_feed_url()`? – Matt Aug 07 '12 at 16:03
Yes. I added after $feed->set_feed_url... Also http://rss.cbc.ca/lineup/topstories.xml is only one of the values I am trying to access. – Richard Aug 07 '12 at 16:03
@Richard Well you're passing the variable to your method correctly. Now the error lies in "why can't cURL resolve the host?" That's a new question. – Matt Aug 07 '12 at 16:05
@Richard OH OH OH You're passing **'rss.cbc.ca/lineup/topstories.xml'** instead of **rss.cbc.ca/lineup/topstories.xml** See the difference? – Matt Aug 07 '12 at 16:10
actually CBC government owned radio/tv broadcast in Canada – Richard Aug 07 '12 at 16:13
@Richard sorry, that was a typo. My point was that you were surrounding your URL with single-quotes. – Matt Aug 07 '12 at 16:14
The RSS parser I am using requires single quotes. I have updated my original post with more details. – Richard Aug 07 '12 at 16:24
@Richard, the quotes aren't necessary except to define a `string literal`. I'll update my answer. I didn't realize you needed an `array` to pass to the method. – Matt Aug 07 '12 at 16:26

Arrays within variables

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