Use numbers starting with 0 in a variable in php - php

Hi i need to save a 010 number in $number and if i do like this php will remove the starting 0
$number = 010
And echo of this will return 10 how can i make it not to remove the initial 0
BR
Martin

Use it as a String:
$number = '010';

Use str_pad() function.
echo str_pad('10',3,'0',STR_PAD_LEFT)
http://php.net/manual/en/function.str-pad.php

Do remember that numbers starting with 0 can also be treated as octal number notation by the PHP compiler, hence if you want to work with decimal numbers, simply use:
$num = '010';
This way the number is saved, can be stored in the database and manipulated like any other number. (Thx to the fact that PHP is very loosely typed language.)
Another method to use would be:
Save number as $num = 10;
Later while printing the value you can use sprintf, like:
sprintf("%03d", $i);
This will print your number in 3 digit format, hence 0 will be added automatically.
Another method:
<?php
$num = 10;
$zerofill = 3;
echo str_pad($num, $zerofill, "0", STR_PAD_LEFT);
/* Returns the wanted result of '010' */
?>
You can have a look at the various options available to you and make a decision. Each of the method given above will give you a correct output.

Related

generate a random string with with exactly 6 digits (show first 0)

I am trying to generate a string with exactly 6 random numbers in it. My current code:
$test = sprintf('%6d', rand(1, 1000000));
With this code I get a string that sometimes has an empty value at the beginning like " 53280". I would want to have it produce "053280" in that case. How to achieve this?
You should add a 0 in your conversion specification to indicate that you want zero-padding:
$test = sprintf('%06d', rand(1, 1000000));
// ^-- here
The conversion specifications are documented on the sprintf manual page.
If you don't want to use sprintf (some dont!), an alternative way to do it would be:
$test = str_pad(mt_rand(1, 999999),6,0,STR_PAD_LEFT);
Example output:
736523
024132
003145
Using mt_rand here because its a better random number function (not perfect, but better than just rand). Also adjusted to 999999 since 1000000 could possibly produce a 7 digit number.
Doing a benchmark of 10000 iterations on the three answers provided (Sean, Mine, Aslan), these are the results in speed:
Sean's Method: 0.005
My Method: 0.006
Aslan's Method: 0.009
So you would be better off going with Sean's method.
You can just replace the empty character with 0.
$test = str_replace(" ", "0", sprintf('%6d', rand(1, 1000000)));

How to calculate using a POST variable?

I need to multiply this POST variable by 12. As an example, if the amount was 10, the result should say:
Amount: 120
Here's my code so far:
Amount :'.$_POST['my_amount'].'<br/>
I tried to run the calculation in another variable, but this doesn't seem to work:
$result = ($_POST['my_amount'])*12;
or maybe it works and my output code is not working:
$vl_text='';
Amount :'.$_POST['my_amount'].'<br/>'.;
If you want your output to resemble your first example.,.. Amount:120 your missing chunks in each of the following 3 examples. first ensure that your $_POST variable is a valid one and set it to a new variable so you can print out the variable if you need to ...
// if you only expect $_POST['my_amount'] to contain integers...
if(is_int(intval($_POST['my_amount']))){
$my_amount = intval($_POST['my_amount']) * 12;
// or if you expect $_POST['my_amount'] to possibly contain a decimal
if(is_float(floatval($_POST['my_amount']))){
$my_amount = floatval($_POST['my_amount']) * 12;
intval ensures that a variable is cast as an integer if it can be, while not entirely necessary as multiplying in php will do this...its good practice to check any variables that you are using for and math functionality.
floatval does the same for for numbers with decimal. as an integer has to be a whole number if your variable could numbers that could contain decimals... use floatval
all of your examples then need to specify to print/echo the string....so
// your second line
echo 'Amount :'.$my_amount .'<br/>';
// your fourth line...
$vl_text='Amount: '.$my_amount;
echo $vl_text;
}
The most logical explanation is that you get string from POST. A good way to achieve what you want is to convert the POST value to int but keep in mind that it could not be numerical.
$int = (is_numeric($_POST['my_amount']) ? (int)$_POST['my_amount'] : 0); //If POST value is numeric then convert to int. If it's not numeric then convert it to 0
$_POST['my_amount'] = 150;
$data = $_POST['my_amount'] * 12;
echo $data;
Result will be 1800

How to convert a "decimal string" into an integer without the period in PHP

If I have, say, 8.1 saved as a string/plaintext, how can I change that into the integer (that I can do addition with) 81? (I've got to remove the period and change it into an integer. I can't seem to figure it out even though I know it should be simple. Everything I try simply outputs 1.)
You can also try this
$str = '8.1';
$int = filter_var($str, FILTER_SANITIZE_NUMBER_INT);
echo $int; // 81
echo $int+1; // 82
DEMO.
If you're dealing with whole numbers (as you said), you could use the intval function that is built into PHP.
http://php.net/manual/en/function.intval.php
So basically, once you have your string parsed and setup as a whole number you can do something like:
intval("81");
And get back the integer 81.
Example:
$strNum = "81";
$intNum = intval($strNum);
echo $intNum;
// "81"
echo getType($intNum);
// "integer"
Since php does auto-casting, this should work:
<?php
$str="8432.145522";
$val = str_replace('.','', $str);
print $str." : ".$val;
?>
Output:
8432.145522 : 8432145522
Not sure if this will work. But if you always have something.something,(like 1.1 or 4.2), you can multiply by 10 and do intval('string here'). But if you have something.somethingsomething or with more somethings(like 1.42 and 5.234267, etc.), I don't know what to say. Maybe a function to keep multiplying by ten until it's an integer with is_int()?
Sources:
http://php.net/manual/en/function.intval.php
http://php.net/manual/en/function.is-int.php
Convert a string to a double - is this possible?

How can I separate a number and get the first two digits in PHP?

How can I separate a number and get the first two digits in PHP?
For example: 1345 -> I want this output=> 13 or 1542 I want 15.
one possibility would be to use substr:
echo substr($mynumber, 0, 2);
EDIT:
please not that, like hakre said, this will break for negative numbers or small numbers with decimal places. his solution is the better one, as he's doing some checks to avoid this.
First of all you need to normalize your number, because not all numbers in PHP consist of digits only. You might be looking for an integer number:
$number = (int) $number;
Problems you can run in here is the range of integer numbers in PHP or rounding issues, see Integers Docs, INF comes to mind as well.
As the number now is an integer, you can use it in string context and extract the first two characters which will be the first two digits if the number is not negative. If the number is negative, the sign needs to be preserved:
$twoDigits = substr($number, 0, $number < 0 ? 3 : 2);
See the Demo.
Shouldn't be too hard? A simple substring should do the trick (you can treat numbers as strings in a loosely typed language like PHP).
See the PHP manual page for the substr() function.
Something like this:
$output = substr($input, 0, 2); //get first two characters (digits)
You can get the string value of your number then get the part you want using
substr.
this should do what you want
$length = 2;
$newstr = substr($string, $lenght);
With strong type-hinting in new version of PHP (> PHP 7.3) you can't use substr on a function if you have integer or float. Yes, you can cast as string but it's not a good solution.
You can divide by some ten factor and recast to int.
$number = 1345;
$mynumber = (int)($number/100);
echo $mynumber;
Display: 13
If you don't want to use substr you can divide your number by 10 until it has 2 digits:
<?php
function foo($i) {
$i = abs((int)$i);
while ($i > 99)
$i = $i / 10;
return $i;
}
will give you first two digits

Php set value as a number

How do I output a value as a number in php? I suspect I have a php value but it is outputting as text and not as a number.
Thanks
Here is the code - Updated for David from question below
<?php
if (preg_match('/\-(\d+)\.asp$/', $pagename1, $a))
{
$pageNumber = $a[1];}
else
{ // failed to match number from URL}
}
?>
If I call it in: This code it does not seem to work.
$maxRows_rs_datareviews = 10;
$pageNum_rs_datareviews = $pagename1; <<<<<------ This is where I want to use it.
if (isset($_GET['pageNum_rs_datareviews'])) {
$pageNum_rs_datareviews = $_GET['pageNum_rs_datareviews'];
}
If I make page name a static number like 3 the code works, if I use $pagename1 it does not, this gives me the idea $pagename1 is not seen as a number?
My stupidity!!!! - I used $pagename1 instead of pageNumber
What kind of number? An integer, decimal, float, something else?
Probably the easiest method is to use printf(), eg
printf('The number %d is an integer', $number);
printf('The number %0.2f has two decimal places', $number);
This might be blindingly obvious but it looks like you want to use
$pageNum_rs_datareviews = $pageNumber;
and not
$pageNum_rs_datareviews = $pagename1;
echo (int)$number; // integer 123
echo (float)$number; // float 123.45
would be the easiest
I prefer to use number_format:
echo number_format(56.30124355436,2).'%'; // 56.30%
echo number_format(56.30124355436,0).'%'; // 56%
$num = 5;
echo $num;
Any output is text, since it's output. It doesn't matter what the type of what you're outputting is, since the human eye will see it as text. It's how you actually treat is in the code is what matters.
Converting (casting) a string to a number is different. You can do stuff like:
$num = (int) $string;
$num = intval($string);
Googling php string to number should give you a beautiful array of choices.
Edit: To scrape a number from something, you can use preg_match('/\d+/', $string, $number). $number will now contain all numbers in $string.

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