PHP Break Apart Array Variables

Question

I am extracting XML from a feed. There is a tag entitled:

<georss:point>34.234 -34.435</georss:point>

which contains two variables I am inserting into MySQL When I run this code, the variable is 'array'. I then place an extract to break out the variables, unsure of next

$xmlString = str_replace('georss:point','point',$xmlString);  
$xml = new SimpleXMLElement($xmlString); 
$items = $xml->xpath('channel/item'); 
$closeItems = array(); 
foreach($items as $item) 
{     
    $latlng = explode(' ',trim($item->point));
    array(
            'lat'=>$latlng[0],
            'lng'=>$latlng[1]
    );

    $lat = array($latlng[0]);
    $lng = array($latlng[1]);

    echo $lat;
    echo $lng;  
}

When I place those echo statements (in the last two lines of code), the variables get echod to the screen. However when I place these varaibles outside of the array, the values do not get echod. I am attempting to get these variables outside of the array, so that I can insert these variables into the database. I have tried extract on the varaiables, but this prints back to the screen as 'Array'... unsure of what I need to do to extract these variables from the array. Thanks,

is there only one lat and long, or many?

Prisoner
– Prisoner

2011-08-15 18:52:11 +00:00
Commented Aug 15, 2011 at 18:52 — Prisoner
– Prisoner, Commented Aug 15, 2011 at 18:52

jondavidjohn · Accepted Answer · 2011-08-15 18:51:39Z

1

Replace your foreach with a version like this...

$new_array = array();
foreach($items as $item) 
{     
    $latlng = explode(' ',trim($item->point));
    $new_array[] = array('lat'=>$latlng[0],'lng'=>$latlng[1]); 
}

var_dump($new_array);

this should give you a array of your lat/long values.

answered Aug 15, 2011 at 18:51

jondavidjohn

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Comments

brian_d · Accepted Answer · 2011-08-15 19:36:33Z

0

Some observations.

1) you can register xpath namespaces instead of replacing in the string
2) array('lat'=>$latlng[0],'lng'=>$latlng[1]); does nothing. It is a dead store.
3) Instead of

$lat = array($latlng[0]);
$lng = array($latlng[1]);
echo $lat;
echo $lng;

You can do: echo $latlng[0].'-'.$latlng[1];
If you try to echo an array, you will always get Array.
Use a specific index to echo an individual element.

4) Alternatively you could use list to store the variables directly
list($lat, $long) = explode(' ',trim($item->point));

Edit To insert multiple results into your database, store the results in an array:

$lat_long = array();
foreach($items as $item){
    $lat_long[] = explode(' ', $item);
}

Then later iterate through this array and do your database insert.

foreach($lat_long as $point){   
  //do SQL injection prevention as well
  //INSERT INTO your_table (`lat`, `long`) VALUES ($point[0], $point[1]);
}

edited Aug 15, 2011 at 19:36

answered Aug 15, 2011 at 18:53

brian_d

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2 Comments

matt colley Over a year ago

Hi Brian, Your suggestion makes more sense to me... however when I implement, the values are the same. list($lat, $long) = explode(' ',trim($item->point)); the lat and lng are always the same on all 23 items.

matt colley Over a year ago

Ahhh... 'I see' said the blind man! Thank you very much Brian!

Headshota · Accepted Answer · 2011-08-15 18:50:24Z

0

I think this:

$lat = array($latlng[0]);
$lng = array($latlng[1]);

should be:

$lat = $latlng[0];
$lng = $latlng[1];

answered Aug 15, 2011 at 18:50

Headshota

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Comments

Mark Biesheuvel · Accepted Answer · 2011-08-15 18:51:11Z

0

Obmitting the array(..) around $latlng[..] would do the trick, I think.

$lat = $latlng[0];
$lng = $latlng[1];

answered Aug 15, 2011 at 18:51

Mark Biesheuvel

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