Like:
float(1.2345678901235E+19) => string(20) "12345678901234567890"
Can it be done?
(it's for json_decode...)
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8 Answers
echo number_format($float,0,'.','');
note: this is for integers, increase 0 for extra fractional digits
7 Comments
Anthony
This is a good solution if you want to have a maximum number of decimal places. Based on the question, I'd change the above to
echo number_format($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.
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$float = 0.123;
$string = sprintf("%.3f", $float); // $string = "0.123";
Comments
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.
4 Comments
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.
One funny way is to use json_encode()
json_encode($float, JSON_PRESERVE_ZERO_FRACTION)
Comments
I solved this issue by passing the argument JSON_BIGINT_AS_STRING for the options parameter.
json_decode($json, false, 512, JSON_BIGINT_AS_STRING)
See example #5 in the json_decode documentation
Comments
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.
Comments
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.
Comments
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.
