java.lang.NumberFormatException:对于输入字符串:"2019-11-27" [英] java.lang.NumberFormatException: For input string: “2019-11-27”

问题描述

在我的应用中，我从API服务器获取的日期为 String 值，我想将其解析为 int ，但是出现此错误:

In my app I am getting date from an API server as String value and I want to parse it to int but I get this error:

Caused by: java.lang.NumberFormatException: For input string: "2019-11-27"
at java.lang.Integer.parseInt(Integer.java: 521)
at java.lang.Integer.parseInt(Integer.java: 556)

我正在尝试解析为 int .我想将其传递给 BarEntry 类构造函数，它仅接受int值或将新的BarEntry浮动(int或float值，浮动).我需要它来显示图表.

I am trying to parse into an int. I want to pass it to BarEntry class constructor and it takes only int value or float new BarEntry(int or float values,float). I need it for showing Chart.

我的活动

try {
    Response response = client.newCall(request).execute();
    Log.d("Response", response.toString());
    JSONObject object = new JSONObject(Objects.requireNonNull(response.body()).string());
    JSONObject rates = object.getJSONObject("rates");
    Iterator < String > iterator = rates.keys();
    while (iterator.hasNext()) {
        String keyDate = iterator.next(); // this date which i want to parse to int value
        String cad = rates.getJSONObject(keyDate).getString("CAD");
        @SuppressLint("DefaultLocale") String value = String.format("%.4s", cad);
        float value1 = Float.parseFloat(value);
        int date = Integer.parseInt(keyDate);
        Log.d("TAG", date + "");
        //Log.d("TAG", value1 + "");

        //barEntries.add(new BarEntry(keyDate, value1));

    }
} catch (IOException | JSONException e) {
    e.printStackTrace();
    Log.d("ChartActivity", e.toString());
}

推荐答案

LocalDate和ThreeTenABP

    String dateString = "2019-11-27";
    LocalDate date = LocalDate.parse(dateString);
    int epochDay = (int) date.toEpochDay();

    System.out.println(epochDay);

此代码段输出:

18227

文档说明:

纪元日计数是第0天的简单递增计数是1970-01-01(ISO).

The Epoch Day count is a simple incrementing count of days where day 0 is 1970-01-01 (ISO).

所以我的建议是该数字适合输入 BarEntry 构造函数.

So my suggestion is that this number is fine for feeding into your BarEntry constructor.

toEpochDay()返回 long .如果您的构造函数不接受 long ，请转换为 int .在上面的代码中，我做了一个简单的转换.风险是，如果在遥远的将来或遥远的日期发生 int 溢出，我们将得到非常错误的结果.我宁愿进行范围检查来避免这种情况:

toEpochDay() returns a long. If your constructor doesn’t accept a long, convert to int. In the code above I did a simple cast. The risk is that we will get a very wrong result in case of int overflow for dates in the far future or the far past. I prefer to do a range check to avoid that:

    long epochDayLong = date.toEpochDay();
    if (epochDayLong < Integer.MIN_VALUE || epochDayLong > Integer.MAX_VALUE) {
        throw new IllegalStateException("Date " + date + " is out of range");
    }
    int epochDay = (int) epochDayLong;

结果与以前相同.此检查与我在评论中提到的 Math.toIntExact 方法所做的检查相同(可从Android API级别24获得).

The result is the same as before. This check is the same check that the Math.toIntExact method I mentioned in a comment does (available from Android API level 24).

我已将值 18227 转换为正常日期，并给出了日期是 1970/01/01 的年份，而在JSON中是 2019-11-27 的原因?和我该如何纠正?

I had converted this value 18227 to normal date and it gives date of this year 1970/01/01 and in JSON it's 2019-11-27 why? and how should i correct it?

让我猜，您实际上做了 new Date(18227).我的建议是，您完全避免使用 Date 类，并坚持使用Java.time(现代Java日期和时间API).您为何获得1970的原因是:18227是自该纪元以来的 days 天，而 Date 则是毫秒(自该纪元00:00 UTC开始)日).因此，您当天获得了00:00:18.227 UTC.上面的代码中已经有一个 LocalDate ，因此只需使用它即可.

Let me guess, you effectively did new Date(18227). My suggestion is that you avoid the Date class completely and stick to java.time, the modern Java date and time API. Why you got 1970 is: 18227 is a count of days since the epoch, and Date counts milliseconds (since 00:00 UTC on the epoch day). So you got 00:00:18.227 UTC on that day. We already have a LocalDate in the above code, so just use that.

    System.out.println(date);

2019-11-27

2019-11-27

如果您需要以相反的方式进行转换，那么当您知道如何进行转换时很容易:

Should you need to convert the opposite way, it’s easy when you know how:

    LocalDate convertedBack = LocalDate.ofEpochDay(epochDay);

结果是具有相同值的 LocalDate .

java.time在旧的和更新的Android设备上均可正常运行.它只需要至少 Java 6 .

java.time works nicely on both older and newer Android devices. It just requires at least Java 6.

• 在Java 8和更高版本以及更新的Android设备(API级别26起)中，内置了现代API.
• 在非Android Java 6和7中，获得ThreeTen反向端口，即现代类的反向端口(JSR 310的ThreeTen；请参见底部的链接).
• 在(较旧的)Android上，请使用Android版的ThreeTen Backport.称为ThreeTenABP.并确保使用子包从 org.threeten.bp 导入日期和时间类.

• In Java 8 and later and on newer Android devices (from API level 26) the modern API comes built-in.
• In non-Android Java 6 and 7 get the ThreeTen Backport, the backport of the modern classes (ThreeTen for JSR 310; see the links at the bottom).
• On (older) Android use the Android edition of ThreeTen Backport. It’s called ThreeTenABP. And make sure you import the date and time classes from org.threeten.bp with subpackages.

• LocalDate的文档.toEpochDay()
• Oracle教程:日期时间解释了如何使用java.time./li>
• Java规范请求(JSR)310 ，其中 java首先描述了.time .
• ThreeTen Backport项目，即 java.time 的反向端口Java 6和7(JSR-310的ThreeTen).
• ThreeTenABP ，ThreeTen Backport的Android版本
• 问题:如何在Android Project中使用ThreeTenABP ，非常详尽的解释.

• Documentation of LocalDate.toEpochDay()
• Oracle tutorial: Date Time explaining how to use java.time.
• Java Specification Request (JSR) 310, where java.time was first described.
• ThreeTen Backport project, the backport of java.time to Java 6 and 7 (ThreeTen for JSR-310).
• ThreeTenABP, Android edition of ThreeTen Backport
• Question: How to use ThreeTenABP in Android Project, with a very thorough explanation.

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