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Querylog -- Read from querylog files when answering to /querylog API, it now survives restarts.
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parent
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commit
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2 changed files with 92 additions and 0 deletions
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@ -88,6 +88,11 @@ func handleQueryLog(w http.ResponseWriter, r *http.Request) {
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logBufferLock.RLock()
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values := logBuffer
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logBufferLock.RUnlock()
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if len(values) < queryLogAPI {
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values = appendFromLogFile(values, queryLogAPI, time.Hour*24)
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}
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var data = []map[string]interface{}{}
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for _, entry := range values {
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var q *dns.Msg
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@ -144,3 +144,90 @@ func periodicQueryLogRotate(t time.Duration) {
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}
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}
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}
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func appendFromLogFile(values []logEntry, maxLen int, timeWindow time.Duration) []logEntry {
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now := time.Now()
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// read from querylog files, try newest file first
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files := []string{
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queryLogFileName + ".gz",
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queryLogFileName + ".gz.1",
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}
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a := []logEntry{}
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// read from all files
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for _, file := range files {
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if len(a) >= maxLen {
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// previous file filled us with enough fresh entries
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break
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}
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if _, err := os.Stat(file); os.IsNotExist(err) {
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// do nothing, file doesn't exist
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continue
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}
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trace("Opening file %s", file)
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f, err := os.Open(file)
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if err != nil {
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log.Printf("Failed to open file \"%s\": %s", file, err)
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// try next file
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continue
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}
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defer f.Close()
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trace("Creating gzip reader")
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zr, err := gzip.NewReader(f)
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if err != nil {
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log.Printf("Failed to create gzip reader: %s", err)
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continue
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}
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trace("Creating json decoder")
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d := json.NewDecoder(zr)
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i := 0
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// entries on file are in oldest->newest order
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// we want maxLen newest
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for d.More() {
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var entry logEntry
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err := d.Decode(&entry)
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if err != nil {
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log.Printf("Failed to decode: %s", err)
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// next entry can be fine, try more
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continue
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}
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if now.Sub(entry.Time) > timeWindow {
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trace("skipping entry")
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continue
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}
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i++
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a = append(a, entry)
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if len(a) > maxLen {
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toskip := len(a) - maxLen
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a = a[toskip:]
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}
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}
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err = zr.Close()
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if err != nil {
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log.Printf("Encountered error while closing gzip reader: %s", err)
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}
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log.Printf("file \"%s\": read %d entries", file, i)
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}
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// now that we've read all eligible entries, reverse the slice to make it go from newest->oldest
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for left, right := 0, len(a)-1; left < right; left, right = left+1, right-1 {
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a[left], a[right] = a[right], a[left]
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}
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// append it to values
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values = append(values, a...)
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// then cut off of it is bigger than maxLen
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if len(values) > maxLen {
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values = values[:maxLen]
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}
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return values
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}
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