JavaScript splice()方法是否不删除数组中超过半小时的时间项?
我有3个数组,其中包含的时间段是字符串。。。像这样:JavaScript splice()方法是否不删除数组中超过半小时的时间项?,javascript,arrays,Javascript,Arrays,我有3个数组,其中包含的时间段是字符串。。。像这样: $scope.shiftMorning = ['09:00', '09:30', '10:00', '10:30', '11:00', '11:30']; $scope.shiftAfternoon = ['12:00', '12:30', '13:00', '13:30', '14:00', '14:30', '15:00', '15:30', '16:00', '16:30', '17:00', '17:30']; $scope.shif
$scope.shiftMorning = ['09:00', '09:30', '10:00', '10:30', '11:00', '11:30'];
$scope.shiftAfternoon = ['12:00', '12:30', '13:00', '13:30', '14:00', '14:30', '15:00', '15:30', '16:00', '16:30', '17:00', '17:30'];
$scope.shiftEvening = ['18:00', '18:30', '19:00', '19:30', '20:00', '20:30', '21:00', '21:30', '22:00', '22:30', '23:00', '23:30'];
现在我有一系列函数来确定接下来30分钟的当前时间,然后将当前时间与数组中的值进行比较。如果当前时间在数组值之后,我们将删除数组值。。。像这样:
function dateCompare(time1, time2) {
//console.log("time 1 = " + time1 + " time 2 = " +time2);
var t1 = new Date();
var parts = time1.split(":");
t1.setHours(parts[0], parts[1], "00", 0);
var t2 = new Date();
parts = time2.split(":");
t2.setHours(parts[0], parts[1], "00", 0);
// returns 1 if greater, -1 if less and 0 if the same
if (t1.getTime() > t2.getTime()) {
return 1;
} else if (t1.getTime() < t2.getTime()) {
return -1;
} else {
return 0;
}
}
function getNearestHalfHourTimeString() {
var now = new Date();
var hour = now.getHours();
var minutes = now.getMinutes();
if (minutes < 15) {
minutes = "30";
} else if (minutes < 45) {
minutes = "00";
hour = hour + 1;
} else {
minutes = "00";
++hour;
}
return(hour + ":" + minutes);
}
var currentTime = getNearestHalfHourTimeString();
function removeUnavailableTimes(timeArray) {
for (var i = 0; i < timeArray.length; i++) {
console.log(timeArray[i], currentTime, dateCompare(timeArray[i], currentTime));
if (dateCompare(timeArray[i], currentTime) === -1) {
// remove the offending item from the array
timeArray.splice(i, 1);
}
}
//console.log(timeArray);
return timeArray;
}
removeUnavailableTimes($scope.shiftMorning);
removeUnavailableTimes($scope.shiftAfternoon);
removeUnavailableTimes($scope.shiftEvening);
函数日期比较(时间1,时间2){
//console.log(“time 1=“+time1+”time 2=“+time2”);
var t1=新日期();
var parts=time1.split(“:”);
t1.设定小时数(第[0]部分、第[1]部分、“00”、0);
var t2=新日期();
parts=time2.拆分(“:”);
t2.设定小时数(第[0]部分、第[1]部分、“00”、0);
//如果较大,则返回1;如果较小,则返回1;如果相同,则返回0
if(t1.getTime()>t2.getTime()){
返回1;
}else if(t1.getTime()
现在一切似乎都很好,但是过去的项目(我们的当前时间更大)具有半小时值('09:30')似乎没有从数组中删除(使用.splice()),但是我注意到我的比较是正确的。有人能看出我做错了什么吗?在迭代时,您正在使用
.splice()
删除数组元素。例如,当您删除timeArray[0]
时,数组的其余部分将移位,因此当您检查timeArray[1]
时,实际上跳过了一个元素
您需要使用其他系统,如:
- 在原始数组上迭代,并用要保留的元素填充另一个数组李>
- 迭代原始数组并从副本中删除元素李>
- 在原始数组上向后迭代
- 代码就在那里。运行
splice()
run时,timeArray
会更改,因为splice()
pop变量和索引在该运行时会更改。请检查我的密码
function removeUnavailableTimes(timeArray) {
var i = 0;
while( i < timeArray.length ){
console.log(timeArray[i], currentTime, dateCompare(timeArray[i], currentTime));
if (dateCompare(timeArray[i], currentTime) === -1) {
// remove the offending item from the array
timeArray.splice(i, 1);
}else{
i++;
}
}
//console.log(timeArray);
return timeArray;
}
函数removeUnavailableTimes(timeArray){
var i=0;
while(i
这里的简单修复方法是向后迭代:for(var i=timeArray.length-1;i