How to Remove PHP Array Implode Character from Beginning of String

Question

I am working on a project that requires imploding an array with character separation. I have successfully used join and implode interchangeably in other parts of the project, but I can't get it to work in this section.

$dbQuery = "SELECT ftc.*, fc.name
            FROM facilities f
            LEFT JOIN facility_to_category ftc ON f.fid = ftc.fid
            LEFT JOIN facility_categories fc ON ftc.cid = fc.cid
            WHERE f.listing_year = '2011'
                AND fc.parent_cid = '2'
              AND f.fid = ('".$listing_fid."')";
    $dbResult = $dbc->query($dbQuery,__FILE__,__LINE__);
    $num_results = $dbc->num_rows($dbResult);
    echo '<h3>Demographics</h3>
    <div>';
    while($catdata = $dbc->fetch($dbResult)) {
        $demographics = array();
        $demographic_names = array('',trim($catdata->name));
        $demographics = implode('  '.chr(149).'   ',$demographic_names);   
        print $demographics; 
    }

The result is this:

Demographics
• Affluent • Children • Hard-to-Reach • Parents

instead of

Demographics
Affluent • Children • Hard-to-Reach • Parents

I've tried using double quotes instead of single quotes around '.chr(149).'. I've tried using commas or bars or just spaces. I've tried different ways of trimming and not trimming $catdata->name.

I also thought about trying string concatenation, but then I'll end up with an extra character at the end instead of the beginning. Implode or join seem the better way to go.

What am I missing?

Is $demographic_names[0] blank? Also, why are you setting $demographics = array(); then almost immediately turning it into a string? — Matt H
– Matt H, Commented Jan 24, 2012 at 21:01

Jerome WAGNER · Accepted Answer · 2012-01-24 21:03:20Z

1

I am not sure I understand what you are trying to do, but why not use

$demographics_names = array();
while($catdata = $dbc->fetch($dbResult)) {
    array_push($demographics_names, trim($catdata->name));
}
$demographics = implode('  '.chr(149).'   ',$demographic_names);   
print $demographics;

answered Jan 24, 2012 at 21:03

Jerome WAGNER

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1 Comment

deewilcox Over a year ago

Moving the implode outside of the while statement helped!

deewilcox · Accepted Answer · 2012-01-26 18:44:31Z

Thank you, everyone, for your help! I ended up reformatting the query and array. Here is the final code:

$dbQuery = "SELECT ftc.*, fc.name
            FROM facilities f
            LEFT JOIN facility_to_category ftc ON f.fid = ftc.fid
            LEFT JOIN facility_categories fc ON ftc.cid = fc.cid
            WHERE f.listing_year = '2011'
                AND fc.parent_cid = '2'
              AND f.fid = ('".$listing_fid."')";
    $dbResult = $dbc->query($dbQuery,__FILE__,__LINE__);
    $num_results = $dbc->num_rows($dbResult);
    echo '<h3>Demographics</h3>
    <div>';
    while($catdata = $dbc->fetch($dbResult)) {
        $demographics = array();
        if (is_array($demographics)) {
        $demographic_names[] = trim($catdata->name);
        }
    }
    $demographics = join('  '.chr(149).'   ',$demographic_names);           
    print $demographics;

The result is:

Demographics
Affluent • Children • Hard-to-Reach • Parents

Hopefully this will be helpful to someone else!

FtDRbwLXw6 · Accepted Answer · 2012-01-24 21:04:24Z

0

You're inserting a blank element into the array that you're imploding:

$demographic_names = array('',trim($catdata->name));
                           ^--- here

answered Jan 24, 2012 at 21:04

FtDRbwLXw6

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2 Comments

deewilcox Over a year ago

Thank you -- I tried removing that, but it somehow prevented the join character from working at all. The result was AffluentChildrenHard-to-ReachParents

FtDRbwLXw6 Over a year ago

Well this is because you only have 1 element in your $demographic_names array. implode() will only put the join character between 2 elements. Use @JeromeWAGNER's answer.

Explosion Pills · Accepted Answer · 2012-01-24 21:04:56Z

0

Seems like you have a blank demographic_name. Use array_filter on the resulting array.

answered Jan 24, 2012 at 21:04

Explosion Pills

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Comments

sathishkumar · Accepted Answer · 2012-01-24 21:05:02Z

0

Check the first array is empty.

while($catdata = $dbc->fetch($dbResult)) {
    $demographics = array();
    $demographic_names = array('',trim($catdata->name));
    if(!empty($demographic_names))
    {
      $demographics = implode('  '.chr(149).'   ',$demographic_names);   
      print $demographics; 
    }
}

answered Jan 24, 2012 at 21:05

sathishkumar

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