Converting an integer to a string in PHP - php

Question

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

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";

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.

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"

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.

Answer 5

Use:
$intValue = 1;
$string = sprintf('%d', $intValue);
Or it could be:
$string = (string)$intValue;
Or:
settype($intValue, 'string');

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";

Answer 7

There are many possible conversion ways:
$input => 123
sprintf('%d',$input) => 123
(string)$input => 123
strval($input) => 123
settype($input, "string") => 123

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;

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.

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 :)

Answer 11

$amount = 2351.25;
$str_amount = "2351.25";
$strCorrectAmount = "$amount";
echo gettype($strCorrectAmount); //string
So the echo will be return string.

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

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 ;

Answer 14

You can simply use the following:
$intVal = 5;
$strVal = trim($intVal);

Answer 15

$integer = 93;
$stringedInt = $integer.'';
is faster than
$integer = 93;
$stringedInt = $integer."";