从回显中的字符串中删除空格和BR

Question

I got the below code but at the moment it generates a string of results but with about 40+ empty spaces.

$user_  = JFactory::getUser();
$db     = JFactory::getDBO();
$levels = JAccess::getAuthorisedViewLevels($user->id);
foreach($levels as $key => $level)
{
  $query  = 'SELECT title FROM #__pf_projects';
  $query .= ' WHERE access = ' . $level . " AND TRIM(title) != ''";
  $db->setQuery($query);
  $projectlist = $db->loadResult($query).'<br>';
  echo $projectlist;
}

At first I thought that array_filter() would be good here but as PatrickQ points out it is a string so the array filter won't work. Then I adapted the code according to the answer from Don't Panic. This adapted code is what you can see above.

It returns now a list like this.

<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
http://www.domain1.com
<br>
<br>
<br>
<br>
http://www.domain5.com
http://www.domain23.com
http://www.domain65.com
http://www.domain213.com
<br>
<br>
<br>
<br>
<br>
<br>

So how to adapt the code to just get a list like this:

http://www.domain1.com
http://www.domain5.com
http://www.domain23.com
http://www.domain65.com
http://www.domain213.com

When you change the <br> into a , then the list becomes ,,,,,,,,,,,http,,,,,,httphttphttphttp,,,,,,, <= I wrote it down a bit shorter.

Answer 1

If array_filter isn't filtering out empty values, then they probably aren't really empty. Assuming there is some sort of whitespace there rather than nulls or empty strings, you can probably modify your query to trim the title and only return results where there's still something there.

SELECT title FROM #__pf_projects
WHERE access = ? AND title IS NOT NULL AND TRIM (title) != ''

Or, in terms of your original PHP code:

$query  = 'SELECT title FROM #__pf_projects';
$query .= ' WHERE access = ' . $level . " AND TITLE IS NOT NULL AND TRIM(title) != ''";

It is best to avoid concatenating variables into your SQL like this, though. If the framework you're using has some way to utilize prepared statements, you should go that route instead.

---

If this still doesn't work, I don't really know what else to try with the query, but you should be able to just check for an empty result in PHP and only echo if there's something to show.

$projectlist = $db->loadResult($query);
if (trim($projectlist)) echo $projectlist.'<br>';

Answer 2

First thing, array_filter, if no callback was passed, will remove only falsy elements. String with empty spaces is evaluated to true and therefore will not be remove from array. You can do something like:

$filteredArray = array_filter($projectList, function($val) {
    return trim($val);
});

print_r($filteredArray);

Also, you can't echo an array. You can use print_r or var_dump.

Answer 3

Thanks to everyone's answers I finally figured out a way to get the results. My way is not necessarily the right way for everyone. And someone with more knowledge then me in this area would probably do it different.

Essentially my answer is outputting the entire string with each result inside a div. This will create a lot of empty divs but those are not generated by HTML. They do however show up in the Inspector.

The final code that I use in my script.

$user_  = JFactory::getUser();
$db     = JFactory::getDBO();
$levels = JAccess::getAuthorisedViewLevels($user->id);
foreach($levels as $key => $level)
{
  $query  = 'SELECT title FROM #__pf_projects';
  $query .= ' WHERE access = ' . $level;
  $db->setQuery($query);
  $projectlist = '<div class="project">'.$db->loadResult($query).'</div>';
  echo $projectlist;
}

This is now giving me a list like this:

http://www.domain1.com
http://www.domain5.com
http://www.domain23.com
http://www.domain65.com
http://www.domain213.com