Question

The following code takes in an $input like: "a banana is a fruit"

$t = sscanf($input,"a %s is a %s",$child,$parent);
echo $t.",".$child.",".$parent;

and it outputs: 2,banana,fruit, which is perfect!

However, if $input = "a cow is an animal" we get the output: 2,cow,n. This is not ideal.

Ideally, for $input = "a cow is an animal" it would return 1,cow, and I will use a separate formatting string: "a %s is an %s" for this input which would return 2,cow,animal.

So: why doesn’t the space after the second a in "a %s is a %s" stop this problem from occurring? Is there a way to explicitly declare that a space MUST occur after that a for it to find a correct match?

Hopefully that makes sense. Any help appreciated!

EDIT: here is the actual code:

$teachArray = array("a %s is a %s","an %s is a %s","a %s is an %s","an %s is an %s");
$i=0;
$t = -1;
while( (t <= 2) && (i < count($teachArray)) ) {
    $t = sscanf($input,$teachArray[i],$child,$parent);
    $i++;
}

So it keeps looping until it either finds a matching sentence structure, or it runs out of sentence templates (stored in $teachArray).

Answer 1

Instead of scanf, you should use a regular expression:

$input = "a cow is an animal";
preg_match("/^an? (.+?) is an? (.+?)$/", $input, $matches);
echo (count($matches)-1) . ',' . $matches[1] . ',' . $matches[2];