提交 2c105cc5 编写于 作者: 杨广俊's avatar 杨广俊

排序问题

上级 a9f70b43
...@@ -163,28 +163,46 @@ export default { ...@@ -163,28 +163,46 @@ export default {
let orderData = this.oldParmDataDoctList.slice() let orderData = this.oldParmDataDoctList.slice()
// let orderData = parmDataDoctList // let orderData = parmDataDoctList
orderData.sort((a, b)=> { orderData.sort((a, b)=> {
if (!a[orderName] || !b[orderName]) {
return 0
}
if (orderName === 'doctorFirstNameChar') { if (orderName === 'doctorFirstNameChar') {
if (a[orderName].charCodeAt(0) > b[orderName].charCodeAt(0)) { if (a[orderName] > b[orderName]) {
return 1 * orderType return 1 * orderType
} else if (a[orderName].charCodeAt(0) < b[orderName].charCodeAt(0)) { } else if (a[orderName] < b[orderName]) {
return -1 * orderType return -1 * orderType
} else { } else {
return 0 return 0
} }
} else if(orderName === 'newTime') { } else if(orderName === 'newTime') {
if ((new Date(a[orderName]).getTime())/1000 > (new Date(b[orderName]).getTime())/1000) { // if ((new Date(a[orderName].replace(/-/g, "/")).getTime())/1000 > (new Date(b[orderName].replace(/-/g, "/")).getTime())/1000) {
return 1 * orderType // alert('1: ' + (new Date(a[orderName].replace(/-/g, "/")).getTime())/1000)
} else if ( (new Date(a[orderName]).getTime())/1000 < (new Date(b[orderName]).getTime())/1000) { // return 1 * orderType
return -1 * orderType // } else if ( (new Date(a[orderName].replace(/-/g, "/")).getTime())/1000 < (new Date(b[orderName].replace(/-/g, "/")).getTime())/1000) {
} else { // alert('2: ' + (new Date(a[orderName].replace(/-/g, "/")).getTime())/1000)
return 0 // return -1 * orderType
}
// if(isSortUp) {
// return ((new Date(a[orderName]).getTime())/1000 > (new Date(b[orderName]).getTime())/1000) ? 1 : -1
// } else { // } else {
// return ((new Date(a[orderName]).getTime())/1000 < (new Date(b[orderName]).getTime())/1000) ? 1 : -1 // alert('3: ' + (new Date(a[orderName].replace(/-/g, "/")).getTime())/1000)
// return 0
// } // }
// return
// alert('1: ' + (new Date(a[orderName].replace(/-/g, "/")).getTime())/1000 (new Date(b[orderName].replace(/-/g, "/")).getTime())/1000)
// alert(JSON.stringify(a[orderName].replace(/-/g, "/")))
let aTime = (new Date((a[orderName]).replace(/-/g, "/")).getTime())/1000
let bTime = (new Date((b[orderName]).replace(/-/g, "/")).getTime())/1000
if(isSortUp) {
if(aTime == bTime) {
return 0
} else {
return aTime > bTime ? 1 : -1
}
} else {
if(aTime == bTime) {
return 0
} else {
return aTime > bTime ? -1 : 1
}
}
} else { } else {
if (a[orderName] > b[orderName]) { if (a[orderName] > b[orderName]) {
return 1 * orderType return 1 * orderType
......
Markdown 格式
0% or
您添加了 0 到此讨论。请谨慎行事。
先完成此消息的编辑!
想要评论请 注册