Convert float to string in php?

Question

Like:

float(1.2345678901235E+19) => string(20) "12345678901234567890"

Can it be done?

(it's for json_decode...)

Answer 1

echo number_format($float,0,'.','');

note: this is for integers, increase 0 for extra fractional digits

• Anthony: This is a good solution if you want to have a maximum number of decimal places. Based on the question, I'd change the above to echo number_format($float,10,'.',''); giving it a max of 10 decimal places. (Arbitrary, but I'm pretty sure it should be higher than 0).
• tweety: The json value I get is not float, it's a really big integer, like 23453453245324532453253425. But it gets converted to float by json_decode. Will your solution always get me the original json value? :) (it seems to work though for the data I have now)
• Karoly Horvath: floats have a limited precision, once your integers are large enough it won't work. should work for approx 14 digits. beyond that it might work, but could be sheer luck.
• tweety: The value has 17 numbers right now. Do you know when it will stop to precisely convert the float?
• Anthony: You can change the setting in the function call, see my answer.

Answer 2

$float = 0.123;
$string = sprintf("%.3f", $float); // $string = "0.123";

Answer 3

It turns out json_decode by default casts large integers as floats. This option can be overwritten in the function call:

$json_array = json_decode($json_string, , , 1);

I'm basing this only on the main documentation, so please test and let me know if it works.

• tweety: that's the first thing I tried - JSON_BIGINT_AS_STRING, and I get: json_decode expects at most 3 parameters bla bla... and a undefined constant message
• Anthony: Ha! I just tested and I got the same error, but with "expects at most 2 parameters". It looks like the 3rd parameter was added in 5.3 and the 4th in 5.4. So it would work if we upgraded our php.
• netcoder: Anyway, it's not even valid code, you can't omit arguments like that.
• Anthony: can't you? It says I passed in 4 parameters... Let me do a test.

Answer 4

One funny way is to use json_encode()

json_encode($float, JSON_PRESERVE_ZERO_FRACTION)

Answer 5

I solved this issue by passing the argument JSON_BIGINT_AS_STRING for the options parameter.

json_decode($json, false, 512, JSON_BIGINT_AS_STRING)

See example #5 in the json_decode documentation

Answer 6

The only way to decode a float without losing precision is to go through the json and frame all the floats in quotation marks. By making strings of numbers.

PHP json_decode integers and floats to string

Answer 7

I often use:

$v = float(1.2345678901235E+19);
$s = "{$v}";

Not sure if there may be any "gotcha's" associated but it generally works for me.

Answer 8

A double precision floating point number can only contain around 15 significant digits. The best you could do is pad the extra digits out with zeroes.