Convert float to string in php?

Question

Like:

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

Can it be done?

(it's for json_decode...)

score 55 · Accepted Answer · answered Jul 29 '11 at 17:21

55

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

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

answered Jul 29 '11 at 17:21

Karoly Horvath

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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). – Anthony Jul 29 '11 at 17:26
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) – tweety Jul 29 '11 at 17:27
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. – Karoly Horvath Jul 29 '11 at 17:29
The value has 17 numbers right now. Do you know when it will stop to precisely convert the float? – tweety Jul 29 '11 at 17:32
You can change the setting in the function call, see my answer. – Anthony Jul 29 '11 at 17:33
are these originally floats? because then they already lost precision. if they are ints try to send them as strings. – Karoly Horvath Jul 29 '11 at 17:34
[`number_format` docs](https://www.php.net/manual/en/function.number-format.php) – Tomás Cot Jul 04 '22 at 17:19

score 18 · Answer 2 · answered Jan 11 '14 at 15:05

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

answered Jan 11 '14 at 15:05

lubart

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score 2 · Answer 3 · answered Jul 29 '11 at 17:32

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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.

answered Jul 29 '11 at 17:32

Anthony

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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 – tweety Jul 29 '11 at 17:34
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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. – Anthony Jul 29 '11 at 17:53
Anyway, it's not even valid code, you can't omit arguments like that. – netcoder Jul 29 '11 at 17:58
can't you? It says I passed in 4 parameters... Let me do a test. – Anthony Jul 29 '11 at 23:01

score 0 · Answer 4 · answered Apr 14 '14 at 06:10

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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

answered Apr 14 '14 at 06:10

Astrotim

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score 0 · Answer 5 · answered Nov 29 '20 at 05:55

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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

answered Nov 29 '20 at 05:55

Андрей Глызин

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score 0 · Answer 6 · answered Jul 29 '11 at 17:21

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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.

answered Jul 29 '11 at 17:21

Alnitak

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Convert float to string in php?

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