How to Display Element of Array that has specific value/character?

Question

I Have A json Array from hxxp://best1st.info/Moviedb/json.php?m=tt2015381&o=json

Here is sample of the array output :

. and so on
.
. 
[STORYLINE] => On planet Earth in 1988, young Peter Quill ( ) sits in the waiting room of a hospital...
[ALSO_KNOWN_AS] => Array
    (
        [0] => Guardianes de la galaxia = Argentina
        [1] => Qalaktikanin MÃ¼hafizeÃ§ileri = Azerbaijan
        [2] => ÐŸÐ°Ð·Ð¸Ñ‚ÐµÐ»Ð¸Ñ‚Ðµ Ð½Ð° Ð“Ð°Ð»Ð°ÐºÑ‚Ð¸ÐºÐ°Ñ‚Ð° = Bulgaria (Bulgarian title)
        [3] => GuardiÃµes da GalÃ¡xia = Brazil
    )

[RELEASE_DATES] => Array
    (
.
.
. and so on

I want to check if in "ALSO_KNOWN_AS" Element, have "Argentina" word , If "ALSO_KNOWN_AS" Element have the "Argentina" word, then display it (the value).

I have try to do it (search with google and here on stackoverflow), but seem my code dont work, Can some one here help me to fix it , here is my code

$url   = 'http://best1st.info/Moviedb/json.php?m=tt2015381&o=json';
$newdata = json_decode(file_get_contents($url));

$alsoKnownAs = $newdata->ALSO_KNOWN_AS;

if (in_array('Argentina', $alsoKnownAs)) {
    echo "Match found";
    // echo the-array-value
}

else
{
echo "Match not found";
return false;
}

Thanks You

Satish Sharma · Accepted Answer · 2014-09-01 05:29:14Z

1

try this

$url   = 'http://best1st.info/Moviedb/json.php?m=tt2015381&o=json';
$newdata = json_decode(file_get_contents($url));
$found = false;

foreach($newdata->ALSO_KNOWN_AS as $value)
{
    if(strpos($value, "Argentina") !==false)
    {
        echo $value;
        echo "<br/>";
        $found = true;
    }   
}   

if($found===false)
{
    echo "Not found";
}

answered Sep 1, 2014 at 5:29

Satish Sharma

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Comments

gvgvgvijayan · Accepted Answer · 2014-09-01 05:36:35Z

0

function searchArray($search, $array)
{
    foreach($array as $key => $value)
    {
        if (stristr($value, $search))
        {
            return true;
        }
    }
    return false;
}

Here in the above function first argument is the string you like to search and second argument is the array and it will iterate through entire array with the function stristr whose work is to return portion on found else false.

Note: if you need case sensitive search instead of stristr use strstr or strchr

if (searchArray('Argentina', $alsoKnownAs)) {
    echo "Match found";
    // echo the-array-value
}

else
{
echo "Match not found";
return false;
}

edited Sep 1, 2014 at 5:36

answered Sep 1, 2014 at 5:26

gvgvgvijayan

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Comments

peter_the_oak · Accepted Answer · 2014-09-01 05:46:35Z

0

Your entries in the ALSO_KNOWN_AS array are not single words like "Argentina" but rather comparisons with country names and other text. This is why in_array() won't find it. in_array() needs precise matches. So you have to check all items.

I suggest you replace the if statement by a foreach loop:

...
...
$found = false;
foreach ($alsoKnownAs as $item) {
   if (strpos($item, 'Argentina') !== false) {
      $found = true;
      break;
   }
}

if (!$found) {
  echo "Match not found";
  return false;
}

Hope this helps :-)

edited Sep 1, 2014 at 5:46

answered Sep 1, 2014 at 5:30

peter_the_oak

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2 Comments

esi0411 Over a year ago

this give me "Parse error: syntax error, unexpected 'as' (T_AS), expecting ';' in D:\XAMPP\htdocs..." massage, but the above answer have solve my problem thaks q anyway peter_the_oak :)

peter_the_oak Over a year ago

Yes, it must be a foreach loop :-) I've corrected the answer.

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