4 ms·
There is a parser and producer for yaml in all the major languages. I mean, even perl -MYAML -E 'say YAML::LoadFile("filename.yaml")->{foo}{bar}' what le
by nrdvana 4y ago
There is a parser and producer for yaml in all the major languages. I mean, even
perl -MYAML -E 'say YAML::LoadFile("filename.yaml")->{foo}{bar}'
what level of tooling do you mean?
- planede 4y agoAnd by YAML, some libraries implement YAML 1.1, some others implement YAML 1.2, and by YAML 1.2 they actually mean YAML 1.2 core schema. And often they implement that poorly. So it doesn't help that: * there is YAML 1.1 and YAML 1.2, and you need to know which one you want to use * there is this thing called "core schema" and it's possible to use other schemas, but nobody seems to care * the spec itself is complex, and hard to implement well
- deleted 4y ago[deleted]