How can I check if one input have integer value?
For example:
if ($request->input('public') == ¿int?){
}
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2php.net/manual/en/function.is-int.phpMubashar Iqbal– Mubashar Iqbal2017-08-21 14:27:41 +00:00Commented Aug 21, 2017 at 14:27
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you can use is_int().Saquib Lari– Saquib Lari2017-08-21 15:28:23 +00:00Commented Aug 21, 2017 at 15:28
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6 Answers
You should try this:
if (is_int($request->input('public'))){
}
OR
if (is_numeric($request->input('public'))){
}
answered Aug 21, 2017 at 14:25
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1 Comment
dodancs
is_int() does not work in Lumen 6+ for inputs. It always says false, only is_numeric() works, but that does not have the same functionality
filter_var($request->input('public'), FILTER_VALIDATE_INT) !== false
because:
- is_int() is true for 12 but isn't for "12"
- is_numerric() is true for 12.12312323
Comments
You can use ctype_digit
<?php
$cadenas = array('1820.20', '10002', 'wsl!12');
foreach ($cadenas as $caso_prueba) {
if (ctype_digit($caso_prueba)) {
echo "La cadena $caso_prueba consiste completamente de dígitos.\n";
} else {
echo "La cadena $caso_prueba no consiste completamente de dígitos.\n";
}
}
?>
Comments
Use is_int:
$number = $request->input('public');
if (is_int($number)) {
dd('number is an integer');
} else {
dd('number is not an integer');
}
Comments
Data taken from the input is always a string.
Use Laravel validation to check if the data from the input is integer or number.
Comments
You can also use shorthand if:
$isInteger = (is_int($request->input('public'))) ? true : false;
1 Comment
salty_splundh
Using the ternary operator like this is redundant because is_int() already only resolves to true or false.
