Converting an integer to a string in PHP
Converting an integer to a string in PHP
Accepted Answer
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";
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There's many ways to do this.
Two examples:
$str = (string) $int;
$str = "$int";
See the PHP Manual on Types Juggling for more.
$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"
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.
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";
Use:
$intValue = 1;
$string = sprintf('%d', $intValue);
Or it could be:
$string = (string)$intValue;
Or:
settype(&$intValue, 'string');