[FIX] web: remove human_number limitations

The formatting of very big numbers (like
1.0045e+22) was not handled adequately by
the function 'human_number' in web.utils.
For those numbers, it seems more appropriate
to use the scientific format. The only
difference introduced here is the following.
Beyond a magnitude of e+21, we represent the
numbers by themselves but we keep the
number of decimals provided as parameter
(or zero by default) and we remove useless
zero decimals like in the other cases.
For instance, 1.0045e+22 is replaced by 1e+22
instead of 1.00e+22 if ask 2 decimals of
precision (and by 1.005e+22 for 3 decimals).

We have also fixed a small bug occuring in the
previous version of the function: the symbol
'E' was never used. So by the past the above
number would have be represented as
10045000P and, even worse, 1.01e+45 by
1.012e+28P.

closes odoo/odoo#31140

1 files changed, 11 insertions, 1 deletions

diff --git a/addons/web/static/src/js/core/utils.js b/addons/web/static/src/js/core/utils.js
index f4031fccec9..a21c4241864 100644
--- a/addons/web/static/src/js/core/utils.js
+++ b/addons/web/static/src/js/core/utils.js
@@ -112,7 +112,17 @@ var utils = {
         var d2 = Math.pow(10, decimals);
         var val = _t('kMGTPE');
         var symbol = '';
-        for (var i = val.length - 1 ; i > 0 ; i--) {
+        var numberMagnitude = number.toExponential().split('e')[1];
+        // the case numberMagnitude >= 21 corresponds to a number
+        // better expressed in the scientific format.
+        if (numberMagnitude >= 21) {
+            // we do not use number.toExponential(decimals) because we want to
+            // avoid the possible useless O decimals: 1e.+24 prefered to 1.0e+24
+            number = Math.round(number * Math.pow(10, decimals - numberMagnitude)) / d2;
+            // formatterCallback seems useless here.
+            return number + 'e' + numberMagnitude;
+        }
+        for (var i = val.length; i > 0 ; i--) {
             var s = Math.pow(10, i * 3);
             if (s <= number / Math.pow(10, minDigits - 1)) {
                 number = Math.round(number * d2 / s) / d2;