Converting an integer to a string in PHP

Question

Is there a way to convert an integer to a string in PHP?

Merritt

PHP is loosely typed. What was an integer once, can be a string as well, e.g. when you echo it (used in so called string context). – Coxalgia

@Coxalgia yes php is loosely typed and echoing would print the value in string context . But this does not change the datatype of a variable internally . Hence , strval($variable) is correct . – Bellbella

@Vivek: strval() doesn't change the $variable internally neither. – Coxalgia

I meant $variable = strval($variable); – Bellbella

Answer 1

You can use the strval() function to convert a number to a string.

From a maintenance perspective its obvious what you are trying to do rather than some of the other more esoteric answers. Of course, it depends on your context.

$var = 5;

// Inline variable parsing
echo "I'd like {$var} waffles"; // = I'd like 5 waffles

// String concatenation 
echo "I'd like ".$var." waffles"; // I'd like 5 waffles

// The two examples above have the same end value...
// ... And so do the two below

// Explicit cast 
$items = (string)$var; // $items === "5";

// Function call
$items = strval($var); // $items === "5";

Shilohshim

Good point. I threw in some mysql_escape_string functions in there to clean it up. – Shilohshim

edited since mysql_escape_string is deprecated since it ignores charset. – Quint

Here is a live version - link – Hildebrand

if we have those values, then will it be 55 or 10? – Yoheaveho

@freaky: $var + $var = 10, $var . $var = "55". "." is the string concatenation operator while "+" is the addition operator. – Shilohshim

@Quint for anyone who looks at this example to try and avoid injection attacks, I'd say abandon your attempts at escaping strings and use prepared statements – Previse

After 5 years I finally removed the SQL example as it was unnecessary to answer the question and introduced confusion. – Shilohshim

Here is a benchmark, for the examples in the answer: leifw.wickland.net/2009/08/… – Perse

How is about 0 as string ? $var =0; – Fibroma

Warning!! If you pass in certain numbers php will round them when converting to string: strval(0.999999997) returns 1. – Americanist

Can anyone provide a reason to choose a cast over a function call or vice versa? – Lupelupee

@Lupelupee cast is theoretically faster because of the overhead of making a function call. – Phyllisphylloclade

Answer 2

There's many ways to do this.

Two examples:

 $str = (string) $int;
 $str = "$int";     

See the PHP Manual on Types Juggling for more.

Ottoottoman

Answer 3

$foo = 5;

$foo = $foo . "";

Now $foo is a string.

But, you may want to get used to casting. As casting is the proper way to accomplish something of that sort:

$foo = 5;
$foo = (string)$foo;

Another way is to encapsulate in quotes:

$foo = 5;
$foo = "$foo"

Eldridgeeldritch

So incredibly strange that we both picked 5. Also, + is java. – Serna

I think you're mixing up your concatenation syntax between languages there. – Ullage

I'd lean toward casting, simply because it makes your intentions abundantly clear. Casting has one and only one purpose: to change types. The other examples may almost look like mistakes, in some contexts. – Madagascar

Answer 4

There are a number of ways to "convert" an integer to a string in PHP.

The traditional computer science way would be to cast the variable as a string:

$int = 5;
$int_as_string = (string) $int;
echo $int . ' is a '. gettype($int) . "\n";
echo $int_as_string . ' is a ' . gettype($int_as_string) . "\n";

You could also take advantage of PHP's implicit type conversion and string interpolation:

$int = 5;
echo $int . ' is a '. gettype($int) . "\n";

$int_as_string = "$int";
echo $int_as_string . ' is a ' . gettype($int_as_string) . "\n";

$string_int = $int.'';
echo $int_as_string . ' is a ' . gettype($int_as_string) . "\n";

Finally, similar to the above, any function that accepts and returns a string could be used to convert and integer. Consider the following:

$int = 5;
echo $int . ' is a '. gettype($int) . "\n";

$int_as_string = trim($int);
echo $int_as_string . ' is a ' . gettype($int_as_string) . "\n";

I wouldn't recommend the final option, but I've seen code in the wild that relied on this behavior, so thought I'd pass it along.

Necrolatry

Answer 5

Use:

$intValue = 1;
$string = sprintf('%d', $intValue);

Or it could be:

$string = (string)$intValue;

Or:

settype($intValue, 'string');

Damask

Answer 6

Warning: the below answer is based on the wrong premise. Casting 0 number to string always returns string "0", making the code provided redundant.

All these answers are great, but they all return you an empty string if the value is zero.

Try the following:

    $v = 0;

    $s = (string)$v ? (string)$v : "0";

Particolored

Note that the question specifically says 'converting an integer', not a float. :) – Particolored

Correct the mistake, please. The correct form should be: $s = (string)$v ? (string)$v : "0"; – Bonnibelle

var_dump((string)0); prints string(1) "0" on PHP 5.5.9 (Released: 6 Feb 2014) for me. Which version of PHP did you get "an empty string if the value is zero"? – Lefthand

Answer 7

There are many possible conversion ways:

$input => 123
sprintf('%d',$input) => 123
(string)$input => 123
strval($input) => 123
settype($input, "string") => 123

Tobias

Answer 8

You can either use the period operator and concatenate a string to it (and it will be type casted to a string):

$integer = 93;
$stringedInt = $integer . "";

Or, more correctly, you can just type cast the integer to a string:

$integer = 93;
$stringedInt = (string) $integer;

Gallinacean

Answer 9

As the answers here demonstrates nicely, yes, there are several ways. However, in PHP you rarely actually need to do that. The "dogmatic way" to write PHP is to rely on the language's loose typing system, which will transparently coerce the type as needed. For integer values, this is usually without trouble. You should be very careful with floating point values, though.

Astronaut

I don't think that PHP will coerce an integer to string all the time. It will do it for strlen(12345); but not for $x = 12345; echo $x[2]; All these casting functions are quite useful and lots of programmers are checking their types more and more. – Promotion

Answer 10

I would say it depends on the context. strval() or the casting operator (string) could be used. However, in most cases PHP will decide what's good for you if, for example, you use it with echo or printf...

One small note: die() needs a string and won't show any int :)

Metaplasia

"in most cases PHP will decide whats good for you" amazing :,) – Gooseflesh

Answer 11

$amount = 2351.25;
$str_amount = "2351.25";

$strCorrectAmount = "$amount";
echo gettype($strCorrectAmount);    //string

So the echo will be return string.

Chartism

Re "string": literally or the type? Or something else? Can you make it more clear (by editing your answer)? – Nuts

Thanks for the comment@ Peter Mortensen That is the return type "STRING" means your value should be converted as a string in a new variable. so that's why I edited string – Chartism

Answer 12

My situation :

echo strval("12"); => 12
echo strval("0"); => "0"

I'm working ...

$a = "12";
$b = "0";
echo $a * 1; => 12
echo $b * 1; => 0

Woolery

As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Libove

It doesn't seem like an answer but rather a question. – Barrettbarrette

Answer 13

I tried all the methods above yet I got "array to string conversion" error when I embedded the value in another string. If you have the same problem with me try the implode() function. example:

$integer = 0;    
$id = implode($integer);    
$text = "Your user ID is: ".$id ;

Galleon

This code makes no sense and returns an error, Argument #1 ($pieces) must be of type array, string given – Barrettbarrette

Answer 14

You can simply use the following:

$intVal = 5;
$strVal = trim($intVal);

Tadzhik

The trim() function removes whitespace and other predefined characters from both sides of a string Although it returns string it's a bad practice. It's better to use strval() or number_format() perhaps. – Forbearance

Answer 15

$integer = 93;
$stringedInt = $integer.'';

is faster than

$integer = 93;
$stringedInt = $integer."";

Araarab

$foo = 5; $foo = "$foo" extremely waistfull for memory in PHP use ''. – Araarab

What do you mean by wasteful? ``` $foo = 5; $foo = "$foo"; ``` This code is just overwriting int value with string, of course, it's not the answer but what's wasteful? – Ezara