[kamailio/kamailio] s.substr transformation with negative offset and length greater than 1 not working (#1426)
4 Feb
2018
4 Feb
'18
3:58 a.m.
According to https://www.kamailio.org/wiki/cookbooks/5.1.x/transformations (and I'm
using 5.1), I should be able to do something like `$var(num)=$(fU{s.substr,-1,10})` and
`$var(num)` should be `123456789` if `$fU` is `0123456789` (only the last 10 characters
are displayed). However, `$var(num)` ends up with only the last character of the
substring, in this case, `9`.
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4 Feb
4 Feb
7:58 a.m.
Closed #1426.
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7:58 a.m.
Not sure if the docs are confusing, but the second parameter is the position, which is
counted from end backwards when negative value. The length, the third parameter, is always
counted forward (never backwards). So if you want to get the last 10 chars:
```
$var(num)=$(fU{s.substr,-10,10})
```
Feel free to improve the docs.
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7:30 p.m.
Thanks @miconda. When you say it that way and I go back to read the docs, they make
sense. Sorry for the noise.
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Anthony Messina -
Daniel-Constantin Mierla