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fix: 完善轨迹追踪页右侧时间线显示第1条时间以及合并时间的处理

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刘洪洪
2024-03-06 15:13:15 +08:00
parent 973a9aef0b
commit d93f732a18
1 changed files with 16 additions and 14 deletions
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30
src/views/location/Index.vue
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@@ -685,28 +685,30 @@ export default {
this.trackingSubscribers.forEach(s => {
if (s.trackRecords && s.trackRecords.length > 0) {
for (let i = 0; i < s.trackRecords.length; i++) {
// 最新的位置点不计算停留时间
if (i === 0) {
s.trackRecords[i].stayTime = '-'
} else {
if ((s.trackRecords[i - 1].subscriberLongitude === s.trackRecords[i].subscriberLongitude) && (s.trackRecords[i - 1].subscriberLatitude === s.trackRecords[i].subscriberLatitude)) {
// 如果连续两条地址重复，则将时间累加，并将上一条删除，键值-1继续循环
if (s.trackRecords[i - 2]) {
const stayTime = unitConvert(s.trackRecords[i - 2].time - s.trackRecords[i].time, unitTypes.time, 's')
if (Number(stayTime[0]) === Number(Number(stayTime[0]).toFixed(0))) {
stayTime[0] = Number(stayTime[0]).toFixed(0)
}
s.trackRecords[i].stayTime = stayTime.join(' ')
s.trackRecords.splice(i - 1, 1)
i = i - 1
if ((s.trackRecords[i - 1].subscriberLongitude === s.trackRecords[i].subscriberLongitude) && (s.trackRecords[i - 1].subscriberLatitude === s.trackRecords[i].subscriberLatitude)) {
// 如果连续两条地址重复，则将时间累加，并将上一条删除，键值-1继续循环
if (s.trackRecords[i - 2]) {
const stayTime = unitConvert(s.trackRecords[i - 2].time - s.trackRecords[i].time, unitTypes.time, 's')
if (Number(stayTime[0]) === Number(Number(stayTime[0]).toFixed(0))) {
stayTime[0] = Number(stayTime[0]).toFixed(0)
}
s.trackRecords[i].stayTime = stayTime.join(' ')
} else {
// 数据只有2条，或者第1条和第2条地点重复，删除1，合并时间
const stayTime = unitConvert(s.trackRecords[i - 1].time - s.trackRecords[i].time, unitTypes.time, 's')
if (Number(stayTime[0]) === Number(Number(stayTime[0]).toFixed(0))) {
stayTime[0] = Number(stayTime[0]).toFixed(0)
}
s.trackRecords[i].stayTime = stayTime.join(' ')
}
s.trackRecords.splice(i - 1, 1)
i = i - 1
} else {
const stayTime = unitConvert(s.trackRecords[i - 1].time - s.trackRecords[i].time, unitTypes.time, 's')
if (Number(stayTime[0]) === Number(Number(stayTime[0]).toFixed(0))) {
stayTime[0] = Number(stayTime[0]).toFixed(0)
}
s.trackRecords[i].stayTime = stayTime.join(' ')
}
}
}