当我使用json_encode时为什么PHP会绕过我的号码？

Question

I need execute json_encode() and convert my original number from:

50610101800060384093800100001010000000056199999999

to

"50610101800060384093800100001010000000056199999999" 

But it return

5.061010180006E+49

I tried this:

ini_set('precision', 30); //With 1, 30, 50, 100, 1000
ini_set('serialize_precision', -1);
'content' => json_encode($params, JSON_NUMERIC_CHECK)

but doesn't work. Can you help me?

Answer 1

50610101800060384093800100001010000000056199999999 exceeds the value of the maximum integer in PHP and so it is promoted to a float and expressed in scientific notation. The float result may be problematic for various reasons as the Manual explains in warning about floating point precision.

If you wish to express the value as if it were an integer you must encapsulate it in a string. That string you may add zero to it but when you do so the result in scientific notation will refer to a float, as follows:

<?php

$s = "50610101800060384093800100001010000000056199999999";

echo $s,"
";

$x = $s + 0;

echo $x, "
",is_float($x); 

See here.

For more info in re PHP and floats, see here.

On the other hand, if there were an array of numbers whose digits corresponded to the numerical display in the OP's post, you could write code as follows:

<?php
    $a = [5,0,6,1,0,1,0,1,8,0,0,0,6,0,3,8,4,0,9,3,8,0,0,1,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,5,6,1,9,9,9,9,9,9,9,9];
    foreach($a as $e) {
       $e = (string) $e;
    }
    $foo = join($a);
    var_dump($foo);
    $foo = bcadd($foo, 1);
    var_dump($foo);

See live code.

The reason this example works is because each array value is converted to a numerical string and then the individual elements are joined to form one very long numerical string. BC Math is an extension in PHP which supports arbitrary precision. In this case, the bcadd() function adds one to the numerical string which results in the display of an incremented numerical string value.