PHP - How to ignore numbers/numeric on string using explode() or preg_split() function

Question

I want to make explode or preg_split function to ignore numbers/numeric.

$string = "my name is numbre 9000 900 1";

$dictionary = array("my", "name", "is", "number");

$words = explode(' ',$string);

foreach($words as $wrd):

if(in_Array($wrd, $dictionary)){
  echo $wrd;
}
elseif(in_Array($wrd, $dictionary) == FALSE){
  echo $wrd."->wrong";
}

the output I want should be:

my
name
is
numbre<-wrong
9000
900
1

not:

my
name
is
numbre<-wrong
9000<-wrong
900<-wrong
1<-wrong

any idea how do I do this?

what is the difference between you expected and actual <-wrong? your title doesn't really match the example you gave and that made your question a bit confusing. — Ghostff
– Ghostff, Commented Jun 16, 2019 at 18:36

The fourth bird · Accepted Answer · 2019-06-16 18:10:02Z

2

As mentioned in the comment, you could use an or || to check if the string is_numeric.

Note that you might write your code a bit shorter:

$string = "my name is numbre 9000 900 1";
$dictionary = array("my", "name", "is", "number");
foreach(explode(' ',$string) as $wrd){
    if(in_array($wrd, $dictionary) || is_numeric($wrd)){
        echo $wrd . PHP_EOL;
    } else {
        echo $wrd."->wrong" . PHP_EOL;
    }
}

Result:

my
name
is
numbre->wrong
9000
900
1

See a php demo

answered Jun 16, 2019 at 18:10

The fourth bird

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Comments

Emma Marcier · Accepted Answer · 2019-06-16 19:26:00Z

2

Your original method is just fine, we'd slightly modify that and apply the preg_split. Here, we check first for is_numeric, if TRUE we continue, then we array_search our dictionary, if FALSE we append ->wrong, otherwise we continue.

Test

$str = "my name is numbre 9000 900 1 and some other undesired words";
$dictionary = array("my", "name", "is", "number");
$arr = preg_split('/\s/', $str);

foreach ($arr as $key => $value) {
    if (is_numeric($value)) {
        continue;
    } elseif (array_search($value, $dictionary) === false) {
        $arr[$key] = $value . "->wrong";
    } else {
        continue;
    }
}

var_dump($arr);

Output

array(12) {
  [0]=>
  string(2) "my"
  [1]=>
  string(4) "name"
  [2]=>
  string(2) "is"
  [3]=>
  string(13) "numbre->wrong"
  [4]=>
  string(4) "9000"
  [5]=>
  string(3) "900"
  [6]=>
  string(1) "1"
  [7]=>
  string(10) "and->wrong"
  [8]=>
  string(11) "some->wrong"
  [9]=>
  string(12) "other->wrong"
  [10]=>
  string(16) "undesired->wrong"
  [11]=>
  string(12) "words->wrong"
}

RegEx Demo

edited Jun 16, 2019 at 19:26

answered Jun 16, 2019 at 18:07

Emma Marcier

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1 Comment

Nigel Ren Over a year ago

Hardcoding if ($value === 'numbre') seems to miss a main point of the code which is to check the words against the dictionary.

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