graphql: return correct logs for tx #25612
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graphql/graphql.go
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| ret := make([]*Log, 0) | ||
| for _, log := range logs { | ||
| if uint64(log.TxIndex) != t.index { | ||
| continue |
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logs is sorted by TxIndex, so this could be improved to skip most iteration:
ix := sort.Search(len(logs), func (i int) bool { return logs[i].TxIndex == t.index })
for ix < len(logs) && logs[ix].TxIndex == t.index {
ret = append(ret, logs[ix])
ix++
}
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Also, it's kind of weird to use a filter here. We could just call t.r.backend.GetLogs to get the block logs.
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Hmm, just realized that the filter is used here exactly because GetLogs doesn't have access to the cache.
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I kind of like the simplicity even tho it's slightly less fast, if you don't have a strong feeling about it.
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Fix tested working on my personal node.
Would deploy 1.10.23 + your_fix on prod if no 1.10.24 official version before the merge (currently using 1.10.21 + TTD set in CLI params)
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I'm slightly in favor of @fjl's suggestion.
graphql/graphql.go
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| ret := make([]*Log, 0) | ||
| for _, log := range logs { | ||
| if uint64(log.TxIndex) != t.index { | ||
| continue |
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I'm slightly in favor of @fjl's suggestion.
graphql/graphql.go
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| return nil, err | ||
| } | ||
| ret := make([]*Log, 0, len(logs)) | ||
| ret := make([]*Log, 0) |
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| ret := make([]*Log, 0) | |
| var ret []*Log |
As part of #25459 I made a change so that graphql's tx log query goes through the usual filter system so it benefits from the newly added cache. However the filter system only handles logs on the block level, not tx level and I didn't select logs based on txindex which caused a regression.
Fixes #25583