148 lines
6.4 KiB
Kotlin
148 lines
6.4 KiB
Kotlin
package com.gzzn.omms.msgexchange.ingress
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import com.gzzn.omms.msgexchange.config.PipelineProps
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import com.gzzn.omms.msgexchange.domain.ErrorClass
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import com.gzzn.omms.msgexchange.domain.ProcStatus
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import com.gzzn.omms.msgexchange.infra.stub.StubInbox
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import com.gzzn.omms.msgexchange.infra.stub.StubInboxCursor
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import com.gzzn.omms.msgexchange.infra.stub.StubPipelineTx
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import com.gzzn.omms.msgexchange.infra.stub.StubProcState
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import org.junit.jupiter.api.Assertions.assertEquals
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import org.junit.jupiter.api.Assertions.assertFalse
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import org.junit.jupiter.api.Assertions.assertNotNull
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import org.junit.jupiter.api.Assertions.assertNull
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import org.junit.jupiter.api.BeforeEach
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import org.junit.jupiter.api.Test
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import java.time.Instant
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/**
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* 收报环节最要紧的几条规矩：
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* - 取新消息只看 ID，不看处理标记。处理完却没能回填的行（尤其是永远不回填的死信）
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* 不允许占住批次，也不允许挡住后面的新消息——这是曾经的线上隐患；
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* - 水位只在成功登记后才推进，而且和登记写在同一个事务里，中断后重扫就能补齐；
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* - 遇到 ID 缺口先停下来（可能有更小的消息还没到），缺口等太久则跳过（否则水位永远卡住）；
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* - 这一层不碰信箱的处理标记，标记留给回填环节写。
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*/
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class InboxPollerTest {
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private val t0: Instant = Instant.parse("2026-09-08T03:00:00Z")
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private val props = PipelineProps()
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private lateinit var inbox: StubInbox
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private lateinit var proc: StubProcState
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private lateinit var cursor: StubInboxCursor
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private lateinit var poller: InboxPoller
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@BeforeEach
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fun setUp() {
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inbox = StubInbox().apply { clear() }
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proc = StubProcState().apply { clear() }
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cursor = StubInboxCursor().apply { clear() }
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poller = InboxPoller(inbox, proc, cursor, StubPipelineTx(), props)
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}
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@Test
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fun `external rows are enqueued in id order and advance the watermark without marking the mailbox`() {
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val first = inbox.simulateExternalWrite("<MSG/>")
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val second = inbox.simulateExternalWrite("<MSG/>")
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assertEquals(2, poller.pollOnce(t0))
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assertEquals(ProcStatus.PENDING, proc.find(first)!!.state)
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assertEquals(ProcStatus.PENDING, proc.find(second)!!.state)
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assertEquals(second, cursor.cursor.committedUpTo)
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assertNull(cursor.cursor.holeSince)
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// 收报只写自有库，不碰信箱的处理标记
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assertFalse(inbox.isMarked(first))
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assertEquals(0, poller.pollOnce(t0)) // 重复扫描幂等
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}
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/**
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* 回归用例：处理完却永远不会回填的行（典型是解码失败的死信）曾经占满每一批的名额，
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* 导致收报整体停摆。取新消息这件事必须和"有没有处理标记"彻底分开。
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*/
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@Test
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fun `terminal rows without a mailbox mark do not block discovery of later messages`() {
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props.pipeline.claimBatch = 3
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val dead = (1..3).map { inbox.simulateExternalWrite("<MSG/>") }
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assertEquals(3, poller.pollOnce(t0))
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dead.forEach {
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proc.markTerminal(it, ProcStatus.DEAD, errorClass = ErrorClass.MALFORMED, lastError = "raw-missing")
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}
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val fresh = inbox.simulateExternalWrite("<MSG/>")
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assertEquals(1, poller.pollOnce(t0))
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assertEquals(ProcStatus.PENDING, proc.find(fresh)!!.state)
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assertEquals(fresh, cursor.cursor.committedUpTo)
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}
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@Test
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fun `watermark stops at a hole so later ids cannot overtake a missing smaller id`() {
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val first = inbox.simulateExternalWrite("<MSG/>")
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val hole = inbox.simulateExternalWrite("<MSG/>")
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val afterHole = inbox.simulateExternalWrite("<MSG/>")
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inbox.removeRow(hole)
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assertEquals(1, poller.pollOnce(t0))
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assertEquals(first, cursor.cursor.committedUpTo)
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assertNotNull(cursor.cursor.holeSince)
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assertNull(proc.find(afterHole)) // 不得越过空洞入队（FIFO）
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assertEquals(0, poller.pollOnce(t0.plusSeconds(60))) // 宽限期内水位不推进
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assertEquals(first, cursor.cursor.committedUpTo)
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}
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@Test
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fun `an aged hole is released and later ids resume enqueuing`() {
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val hole = inbox.simulateExternalWrite("<MSG/>")
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val afterHole = inbox.simulateExternalWrite("<MSG/>")
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inbox.removeRow(hole)
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poller.pollOnce(t0) // 记录空洞观测时刻
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val agedOut = t0.plus(props.pipeline.maxCommitDelay)
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assertEquals(0, poller.pollOnce(agedOut)) // 空洞判永久：推进水位但不越过入队
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assertNull(cursor.cursor.holeSince)
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assertEquals(afterHole - 1, cursor.cursor.committedUpTo)
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assertEquals(1, poller.pollOnce(agedOut)) // 下一轮恢复发现
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assertEquals(afterHole, cursor.cursor.committedUpTo)
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assertNotNull(proc.find(afterHole))
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}
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@Test
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fun `compat http path and poller do not double enqueue the same message`() {
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val receipt = InboxService(inbox, proc).accept("<MSG/>")
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assertEquals(0, poller.pollOnce(t0))
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assertEquals(receipt.msgId, cursor.cursor.committedUpTo) // 已在 PG：读取进度照常推进
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assertNotNull(proc.find(receipt.msgId))
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}
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@Test
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fun `a newly exposed hole does not inherit the age of the previous hole`() {
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val first = inbox.simulateExternalWrite("<MSG/>")
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val oldHole = inbox.simulateExternalWrite("<MSG/>")
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val third = inbox.simulateExternalWrite("<MSG/>")
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val newHole = inbox.simulateExternalWrite("<MSG/>")
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val fifth = inbox.simulateExternalWrite("<MSG/>")
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inbox.removeRow(oldHole)
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inbox.removeRow(newHole)
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assertEquals(1, poller.pollOnce(t0))
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val almostAged = t0.plus(props.pipeline.maxCommitDelay).minusSeconds(1)
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inbox.restoreRow(oldHole, "<MSG/>", almostAged)
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assertEquals(2, poller.pollOnce(almostAged))
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assertEquals(third, cursor.cursor.committedUpTo)
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assertEquals(almostAged, cursor.cursor.holeSince)
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assertNull(proc.find(fifth))
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// 旧空洞的期限已到，但新空洞必须获得完整等待窗口。
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assertEquals(0, poller.pollOnce(t0.plus(props.pipeline.maxCommitDelay)))
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assertEquals(third, cursor.cursor.committedUpTo)
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assertNull(proc.find(fifth))
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assertEquals(first + 2, third)
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}
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}
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