Is there any simple way of checking if all elements of an array are instances of a specific type without loopin开发者_JAVA技巧g all elements? Or at least an easy way to get all elements of type X from an array.
$s = array("abd","10","10.1");
$s = array_map( gettype , $s);
$t = array_unique($s) ;
if ( count($t) == 1 && $t[0]=="string" ){
print "ok\n";
}
You cannot achieve this without checking all the array elements, but you can use built-in array functions to help you.
You can use array_filter
to return an array. You need to supply your own callback function as the second argument to check for a specific type. This will check if the numbers of the array are even.
function even($var){
return(!($var & 1));
}
// assuming $yourArr is an array containing integers.
$newArray = array_filter($yourArr, "even");
// will return an array with only even integers.
As per VolkerK's comment, as of PHP 5.3+ you can also pass in an anonymous function as your second argument. This is the equivalent as to the example above.
$newArray = array_filter($yourArr, function($x) { return 0===$x%2; } );
Is there any simple way of checking if all elements of an array [something something something] without looping all elements?
No. You can't check all the elements of an array without checking all the elements of the array.
Though you can use array_walk
to save yourself writing the boilerplate yourself.
You can also combine array_walk
with create_function
and use an anonymous function to filter the array. Something alon the lines of:
$filtered_array = array_filter($array, create_function('$e', 'return is_int($e)'))
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