SimpleDateFormat parse(string str) doesn't thr

2019-01-09 12:41发布

Here is an example:

public MyDate() throws ParseException {
    SimpleDateFormat sdf = new SimpleDateFormat("yyyy/MM/d");
    sdf.setLenient(false);
    String t1 = "2011/12/12aaa";
    System.out.println(sdf.parse(t1));
}

2011/12/12aaa is not a valid date string. However the function prints "Mon Dec 12 00:00:00 PST 2011" and ParseException isn't thrown.

Can anyone tell me how to let SimpleDateFormat treat "2011/12/12aaa" as an invalid date string and throw an exception?

6条回答
Viruses.
2楼-- · 2019-01-09 13:18

Java 8 LocalDate may be used:

public static boolean isDate(String date) {
try {
        LocalDate.parse(date, DateTimeFormatter.ofPattern("yyyy/MM/dd"));
        return true;
    } catch (DateTimeParseException e) {
        return false;
    }
}

If input argument is "2011/12/12aaaaaaaaa", output is false;

If input argument is "2011/12/12", output is true

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家丑人穷心不美
3楼-- · 2019-01-09 13:24

The JavaDoc on parse(...) states the following:

parsing does not necessarily use all characters up to the end of the string

It seems like you can't make SimpleDateFormat throw an exception, but you can do the following:

SimpleDateFormat sdf = new SimpleDateFormat("yyyy/MM/d");
sdf.setLenient(false);
ParsePosition p = new ParsePosition( 0 );
String t1 = "2011/12/12aaa";    
System.out.println(sdf.parse(t1,p));

if(p.getIndex() < t1.length()) {
  throw new ParseException( t1, p.getIndex() );
}

Basically, you check whether the parse consumed the entire string and if not you have invalid input.

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疯言疯语
4楼-- · 2019-01-09 13:31

To chack whether a date is valid The following method returns if the date is in valid otherwise it will return false.

public boolean isValidDate(String date) {

        SimpleDateFormat sdf = new SimpleDateFormat("yyyy/MM/d");
        Date testDate = null;
        try {
            testDate = sdf.parse(date);
        }
        catch (ParseException e) {
            return false;
        }
        if (!sdf.format(testDate).equals(date)) {
            return false;
        }
        return true;

    }

Have a look on the following class which can check whether the date is valid or not

** Sample Example**

import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;

public class DateValidCheck {


    public static void main(String[] args) {

        if(new DateValidCheck().isValidDate("2011/12/12aaa")){
            System.out.println("...date is valid");
        }else{
            System.out.println("...date is invalid...");
        }

    }


    public boolean isValidDate(String date) {

        SimpleDateFormat sdf = new SimpleDateFormat("yyyy/MM/d");
        Date testDate = null;
        try {
            testDate = sdf.parse(date);
        }
        catch (ParseException e) {
            return false;
        }
        if (!sdf.format(testDate).equals(date)) {
            return false;
        }
        return true;

    }

}
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成全新的幸福
5楼-- · 2019-01-09 13:33

Take a look on the method documentation which says: ParseException if the beginning of the specified string cannot be parsed.

Method source code with javadoc:

/**
 * Parses text from the beginning of the given string to produce a date.
 * The method may not use the entire text of the given string.
 * <p>
 * See the {@link #parse(String, ParsePosition)} method for more information
 * on date parsing.
 *
 * @param source A <code>String</code> whose beginning should be parsed.
 * @return A <code>Date</code> parsed from the string.
 * @exception ParseException if the beginning of the specified string
 *            cannot be parsed.
 */
public Date parse(String source) throws ParseException
{
    ParsePosition pos = new ParsePosition(0);
    Date result = parse(source, pos);
    if (pos.index == 0)
        throw new ParseException("Unparseable date: \"" + source + "\"" ,
            pos.errorIndex);
    return result;
}
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贼婆χ
7楼-- · 2019-01-09 13:37

After it successfully parsed the entire pattern string SimpleDateFormat stops evaluating the data it was given to parse.

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