编写一个功能，将格式为 M / D / YYYY 的用户输入日期转换为API所需的格式（ YYYYMMDD ）。参数“userDate”和返回值是字符串。
例如，它应将用户输入的日期“12/31/2014”转换为适用于API的“20141231”。
function formatDate(userDate)
{
userDate = new Date();
y = userDate.getFullYear();
m = userDate.getMonth();
d = userDate.getDate();
return y + m + d;
}
我的代码有什么问题吗？
无法通过在线测试。
答案 0 :(得分：8)
代码有五个问题。
Date()）。getMonth方法返回月份索引，而不是月份编号，因此您必须为其添加一个。忽略日期格式问题，可以使用以下方法修复其他格式：
function formatDate(userDate) {
userDate = new Date(userDate);
y = userDate.getFullYear().toString();
m = (userDate.getMonth() + 1).toString();
d = userDate.getDate().toString();
if (m.length == 1) m = '0' + m;
if (d.length == 1) d = '0' + d;
return y + m + d;
}
您可以重新排列字符串中的字符，而不是将字符串解析为日期，并将其格式化为字符串。这避免了日期格式问题：
function formatDate(userDate) {
var parts = userDate.split('/');
if (parts[0].length == 1) parts[0] = '0' + parts[0];
if (parts[1].length == 1) parts[1] = '0' + parts[1];
return parts[2] + parts[0] + parts[1];
}
答案 1 :(得分：2)
function formatDate(userDate) {
// format from M/D/YYYY to YYYYMMDD
let array = userDate.split("/");
while(array[0].length < 2) {
array[0] = "0" + array[0];
}
while(array[1].length < 2) {
array[1] = "0" + array[1];
}
let arrayAnswer = array[2]+ array[0]+ array[1];
return arrayAnswer;
}
console.log(formatDate("1/3/2014"));
//output must equal 20140103
作为初学者，这是我能够做到这一点的最简单方法。我将它拆分，添加0然后将其分类到正确的位置对我来说更有意义。
答案 2 :(得分：1)
按照代码中的注释进行操作 - 逐步显示解决问题的单向。
// Function shell. Accepts a parameter userDate, returns a value
function formatDate(userDate) {
// Step 1: attempt to convert parameter to a date!
var returnDate = new Date(userDate);
// Step 2: now that this is a date, we can grab the day, month and year
// portions with ease!
var y = returnDate.getFullYear();
var m = returnDate.getMonth() + 1; // Step 6
var d = returnDate.getDay();
// Step 3: The bit we did above returned integer values. Because we are
// *formatting*, we should really use strings
y = y.toString();
m = m.toString();
d = d.toString();
// Step 4: The value of our month and day variables can be either 1 or 2
// digits long. We need to force them to always be 2 digits.
// There are lots of ways to achieve this. Here's just one option:
if (m.length == 1) {
m = '0' + m;
}
if (d.length == 1) {
d = '0' + d;
}
// Step 5: combine our new string values back together!
returnDate = y + m + d;
// Step 6: did you notice a problem with the output value?
// The month is wrong! This is because getMonth() returns a value
// between 0 and 11 i.e. it is offset by 1 each time!
// Look back up at Step 2 and see the extra piece of code
// Step 7: Looks pretty good, huh? Well, it might pass you your quiz
// question, but it's still not perfect.
// Do you know why?
// Well, it assumes that the parameter value is
// a) always an actual date (e.g. not "dave")
// b) our Step1 correctly converts the value (e.g. the client, where
// the JS is run, uses the date format m/d/y).
// I am in the UK, which doesn't like m/d/y, so my results will
// be different to yours!
// I'm not going to solve this here, but is more food for thought for you.
// Consider it extra credit!
return returnDate;
}
// Display result on page -->
document.getElementById("result").innerText += formatDate("1/1/2015");<div id="result">Result: </div>
答案 3 :(得分：0)
您正在使用带有整数的+，您必须将它们转换为字符串。
function formatDate(userDate) {
userDate = new Date(userDate);
y = userDate.getFullYear();
m = userDate.getMonth() + 1;
d = userDate.getDate();
return y.toString() +
('0' + m.toString()).slice(-2) +
('0' + d.toString()).slice(-2);
}
此外，您还需要在月和日添加前导零。
示例：
console.log(formatDate('2/12/2015'));
将写入日志20150212
console.log(formatDate('1/1/2015'));
将写入日志20150101
答案 4 :(得分：0)
function formatDate(userDate) {
// format from M/D/YYYY to YYYYMMDD
var oldDate = String(userDate).split('/');
if (oldDate[0].length==1)oldDate[0]='0'+oldDate[0];
if (oldDate[1].length==1)oldDate[1]='0'+oldDate[1];
var newDate = [oldDate[2], oldDate[0], oldDate[1]];
return newDate.join('');
}
console.log(formatDate("12/31/2014"));
答案 5 :(得分：0)
function formatDate(userDate) {
userDate = new Date(userDate);
year= userDate.getFullYear().toString();
month = (userDate.getMonth() + 1).toString();
date = userDate.getDate().toString();
return year + (month=(month.length==1)?'0'+month:month) + (date=(date.length==1)?'0'+date:date);
}
console.log(formatDate("12/10/2017"));
更多细节：
（1）userDate = new Date（userDate）=＆gt;创建Date对象的实例。
（2）userDate.getFullYear（）从userDate获取年份。类似于月份的user.getMonth（）和日期的userDate.getDate（）...我已经将月份+1添加到月份，因为月份返回开始为0，即第6个月它返回5，因此添加了1。
（3）在退货声明年份+月份+日期是在有条件的情况下完成，以检查月份或日期是否为1或3之类的单个数字，然后在它之前添加0以使其成为01或03。
答案 6 :(得分：0)
function formatDate(userDate) {
var first = userDate.indexOf("/");
var last = userDate.lastIndexOf("/");
var months = userDate.substring(0, first);
var years = userDate.substring(last + 1);
var days = userDate.substring(last, 3);
return (years + months + days);
}
console.log(formatDate("12/31/2014"));
答案 7 :(得分：0)
function formatDate(userDate) {
let userdate = userDate.split('/');
let [month, day, year] = userdate;
if (day.length === 1) {
day = `0${day}`;
}
if (month.length === 1) {
month = `0${month}`;
}
return `${year}${month}${day}`;
}
console.log(formatDate("12/1/2014"));
答案 8 :(得分：0)
~\Miniconda3\lib\site-packages\matplotlib\axes\_axes.py in fill_between(self, x, y1, y2, where, interpolate, step, **kwargs)
5226
5227 # Handle united data, such as dates
-> 5228 self._process_unit_info(xdata=x, ydata=y1, kwargs=kwargs)
5229 self._process_unit_info(ydata=y2)
5230
~\Miniconda3\lib\site-packages\matplotlib\axes\_base.py in _process_unit_info(self, xdata, ydata, kwargs)
2123 return kwargs
2124
-> 2125 kwargs = _process_single_axis(xdata, self.xaxis, 'xunits', kwargs)
2126 kwargs = _process_single_axis(ydata, self.yaxis, 'yunits', kwargs)
2127 return kwargs
~\Miniconda3\lib\site-packages\matplotlib\axes\_base.py in _process_single_axis(data, axis, unit_name, kwargs)
2106 # We only need to update if there is nothing set yet.
2107 if not axis.have_units():
-> 2108 axis.update_units(data)
2109
2110 # Check for units in the kwargs, and if present update axis
~\Miniconda3\lib\site-packages\matplotlib\axis.py in update_units(self, data)
1496
1497 if neednew:
-> 1498 self._update_axisinfo()
1499 self.stale = True
1500 return True
~\Miniconda3\lib\site-packages\matplotlib\axis.py in _update_axisinfo(self)
1532 self.isDefault_label = True
1533
-> 1534 self.set_default_intervals()
1535
1536 def have_units(self):
~\Miniconda3\lib\site-packages\matplotlib\axis.py in set_default_intervals(self)
2170 if info.default_limits is not None:
2171 valmin, valmax = info.default_limits
-> 2172 xmin = self.converter.convert(valmin, self.units, self)
2173 xmax = self.converter.convert(valmax, self.units, self)
2174 if not dataMutated:
~\Miniconda3\lib\site-packages\pandas\plotting\_matplotlib\converter.py in convert(values, units, axis)
207 values = [PeriodConverter._convert_1d(v, units, axis) for v in values]
208 else:
--> 209 values = PeriodConverter._convert_1d(values, units, axis)
210 return values
211
~\Miniconda3\lib\site-packages\pandas\plotting\_matplotlib\converter.py in _convert_1d(values, units, axis)
213 def _convert_1d(values, units, axis):
214 if not hasattr(axis, "freq"):
--> 215 raise TypeError("Axis must have `freq` set to convert to Periods")
216 valid_types = (str, datetime, Period, pydt.date, pydt.time, np.datetime64)
217 if isinstance(values, valid_types) or is_integer(values) or is_float(values):
TypeError: Axis must have `freq` set to convert to Periods
答案 9 :(得分：0)
function formatDate(userDate) {
// format from M/D/YYYY to YYYYMMDD
let [month,date,year] = userDate.split("/");
month = (month.length === 1) ? "0"+month : month;
date = (date.length === 1) ? "0"+date : date;
return year+month+date
}
console.log(formatDate("1/31/2014"));