I am learning PHP and I just read about Assignment Operators and I saw this
$a .= 5 which means $a equals $a concatenated with 5. To test this I coded a simple script
<?php
$a = 12345;
$a .=6;
$b = 12345;
$b .=006;
$c = 12345;
$c .=678;
echo " a=$a and b=$b c=$c" ;
?>
the output was a=123456 and b=123456 c=12345678.My question is why the b isn't equal with 12345006? Is it because treats 6 == 006?
Because 006 is treated as the octal number 6, which is the converted to the string "6", and concatenated to "12345" (which is the number 12345 converted to a string).
Use $b .= "006", and the result will be 12345006.
Because numbers with leading zeroes are treated as octals. if the concatenation would've included the leading zeroes, then 006 would've been interpretted as a string, which makes no sense, because it hasn't got quotes around it.
If you want 006 to be treated as a string, write it as one: '006'. Leave it as is, and it'll be interpreted as an octal:
$b = $a = 123;
$a .= 8;
$b .= 010;//octal
echo $a, ' === ', $b, '!';//echoes 1238 === 1238!
Just as an asside: yes, those are comma's/. echo is a language construct to which you can push multiple values, separated by comma's. The upshot of using comma's is that the values aren't concatenated into a single string before being pushed to the output stream.
This means that it is (marginally faster). The downsides: there aren't any AFAIK.
In case you're interested in the internals of PHP, I've explain this a bit more detailed here...
Just try b as a string:
$b.= '006';
Then you will get output 12345006.
Related
Simple maths:
$a=$b/$c; echo $a;
if $b equals to 123.00 and $c equals to 1 then $a becomes 123.
If $b is 123.50, $c is 1, $a is 123.50. But in the former case , I want $a to be 123.00.
It is possible to test whether $a has any non-zero fraction part or not, and then add the trailing zeros as necessary.
But I am looking for php functions to do the same thing. Possible?
EDIT :
What if I do not want the commas from number_format there ?
Use sprintf("%0.2f",$a);. docs
Use the number_format function. If you don't want comma separators, set all four parameters of the function like so (the thousands separator is the fourth parameter):
$number = number_format(1234, 2, '.', '');
Yup, using https://www.php.net/manual/en/function.number-format.php function like this:
$a = 123;
$answer = number_format($a,"2");
echo $answer;
I need to add numbers in php without changing the number format like below
$a = "001";
$b = "5";
$c = $a+$b;
Now the result comes like "6" but I need "006" if $a is "01" then the result should be "06".
Thanks
Technically speaking, the $a and $b in your example are strings - when you use the addition operator on them they converted to integers which can't retain leading zeroes. More details on string-to-number conversion are in the manual
Something like this would do it (assuming positive integer strings with leading zeros)
#figure out how long the result should be
$len=max(strlen($a), strlen($b));
#pad the sum to match that length
$c=str_pad($a+$b, $len, '0', STR_PAD_LEFT);
If you always know how long the string has to be, you could use sprintf, e.g.
$c=sprintf('%03d', $a+$b);
Here, % introduces a placeholder, 03 tells it we want zero padded to fill at least 3 digits, and d tells it we're formatting an integer.
Hope this would help you:
<?php
$a="001";
$b="5";
$l=max(strlen($a),strlen($b));
$c=str_pad($a+$b, $l,"0", STR_PAD_LEFT);
echo $c;
?>
For common case. Your code should looks like this.
$a = someFormat($original_a);
$b = someFormat2($original_b); // $b has different format.
$c = someFormat($a + $b);
Or, you need write formatRecognition function.
$a = getValueA();
$b = getValueB();
$c = someFormat(formatRecognition($a), $a + $b);
$a = "3dollars";
$b = 20;
echo $a += $b;
print($a += $b);
Result:
23
43
I have a question from this calculation.$a is a string and $b is number.I am adding both and print using echo its print 23 and print using print return 43.How is it
It casts '3dollars' as a number, getting $a = 3.
When you echo, you add 20, to $a, so it prints 23 and $a = 23.
Then, when you print, you again add 20, so now $a = 43.
The right way to add (which is technically concatenating) strings is
$a = 7;
$b = "3 dollars";
print ($a . $b); // 73 dollars
The + operator in php automatically converts string into numbers, which explains why your code carried out arimethic instead of concatenation
PHP automatically associates a data type to the variable, depending on its value. Since the data types are not set in a strict sense, you can do things like adding a string to an integer without causing an error.
In PHP 7, type declarations were added. This gives us an option to specify the expected data type when declaring a function, and by adding the strict declaration, it will throw a "Fatal Error" if the data type mismatches.
To specify strict we need to set declare(strict_types=1);. This must be on the very first line of the PHP file. Then it will show fatal error and if you didn't declare this strict then it convert string into integer.
If you need both the values, return them in an array
PHP treats '3dollars' as a integer 3 because string starting with integer and participating in arithmetic operation, so
$a = "3dollars";
$b = 20;
echo $a += $b;
it echo 23; //$a=$a+$b;
now $a = 23 + 20;
print($a += $b); //$a=$a+$b;
it print 43;
Since You have created a variable for the two, it stores the result of each, so when you added $a to 20 it will echo 23 which stores in the system, them when you print $a which is now 23 in addition to $b which is 20. You will get 43.
<?php
$a = array("ABC","DEF");
foreach($a as &$b){
$b++;
echo $b . "<br />";
}
?>
Can any body explain me why the above code snippet generate the following output?
output:
ABD
DEG
Thanks in advance
It is explained in php manual
PHP follows Perl's convention when dealing with arithmetic operations on character variables and not C's. For example, in Perl 'Z'+1 turns into 'AA', while in C 'Z'+1 turns into '[' ( ord('Z') == 90, ord('[') == 91 ). Note that character variables can be incremented but not decremented and even so only plain ASCII characters (a-z and A-Z) are supported.
And if you mean why $b has changed:
You pass reference to $b in foreach loop, so assignments, increases etc. are done on reference to $b not on value that $b represents.
If you use ++ on a string it is "counted up": the next char in the alphabet is used. Remove the $b++; and your output will be "ABC DEF"
Good explanations here
hint:
If you call $a[0][1], it would return "B".
Watch following code:
$a = 'Test';
echo ++$a;
This will output:
Tesu
Question is, why ?
I know "u" is after "t", but why it doesn't print "1" ???
PHP Documentation:
Also, the variable being incremented
or decremented will be converted to
the appropriate numeric data
type—thus, the following code will
return 1, because the string Test is
first converted to the integer number
0, and then incremented.
PHP follows Perl's convention when dealing with arithmetic operations on character variables and not C's. For example, in Perl 'Z'+1 turns into 'AA', while in C 'Z'+1 turns into '[' ( ord('Z') == 90, ord('[') == 91 ). Note that character variables can be incremented but not decremented and even so only plain ASCII characters (a-z and A-Z) are supported.
Source: http://php.net/operators.increment
In PHP you can increment strings (but you cannot "increase" strings using the addition operator, since the addition operator will cause a string to be cast to an int, you can only use the increment operator to "increase" strings!... see the last example):
So "a" + 1 is "b" after "z" comes "aa" and so on.
So after "Test" comes "Tesu"
You have to watch out for the above when making use of PHP's automatic type coercion.
Automatic type coercion:
<?php
$a="+10.5";
echo ++$a;
// Output: 11.5
// Automatic type coercion worked "intuitively"
?>
No automatic type coercion! (incrementing a string):
<?php
$a="$10.5";
echo ++$a;
// Output: $10.6
// $a was dealt with as a string!
?>
You have to do some extra work if you want to deal with the ASCII ordinals of letters.
If you want to convert letters to their ASCII ordinals use ord(), but this will only work on one letter at a time.
<?php
$a="Test";
foreach(str_split($a) as $value)
{
$a += ord($value); // string + number = number
// strings can only handle the increment operator
// not the addition operator (the addition operator
// will cast the string to an int).
}
echo ++$a;
?>
live example
The above makes use of the fact that strings can only be incremented in PHP. They cannot be increased using the addition operator. Using an addition operator on a string will cause it to be cast to an int, so:
Strings cannot be "increased" using the addition operator:
<?php
$a = 'Test';
$a = $a + 1;
echo $a;
// Output: 1
// Strings cannot be "added to", they can only be incremented using ++
// The above performs $a = ( (int) $a ) + 1;
?>
The above will try to cast "Test" to an (int) before adding 1. Casting "Test" to an (int) results in 0.
Note: You cannot decrement strings:
Note that character variables can be incremented but not decremented and even so only plain ASCII characters (a-z and A-Z) are supported.
The previous means that echo --$a; will actually print Test without changing the string at all.
The increment operator in PHP works against strings' ordinal values internally. The strings aren't cast to integers before incrementing.