Failed to convert value of type java.lang.String to required type java.util.Date

Question

<form th:action="@{home}" method="get">
    <div class="form-group">
        <label>from date:</label> <input type="date" pattern="yyyy-MM-dd" name="d1" th:value="${d1}" />
        <label>to date:</label> <input type="date" pattern="yyyy-MM-dd"  name="d2" th:value="${d2}" />
        <button type="submit">Trouver</button>
    </div>
</form>

this is the controller code part:

@RequestParam(name = "d1", defaultValue = "1900-01-01") @DateTimeFormat(pattern = "yyyy-MM-dd") Date d1,
        @RequestParam(name = "d2", defaultValue = "2200-01-01") @DateTimeFormat(pattern = "yyyy-MM-dd") Date d2){

Im getting this message:

There was an unexpected error (type=Bad Request, status=400). Failed to convert value of type [java.lang.String] to required type [java.util.Date]; nested exception is org.springframework.core.convert.ConversionFailedException: Failed to convert from type [java.lang.String] to type [@org.springframework.web.bind.annotation.RequestParam @org.springframework.format.annotation.DateTimeFormat java.util.Date] for value 'Wed Jun 08 00:00:00 WET 2016'; nested exception is java.lang.IllegalArgumentException: Parse attempt failed for value [Wed Jun 08 00:00:00 WET 2016]

can you show your class declaration of type date please ? or try to add @DateTimeFormat(pattern = "yyyy-MM-dd") like this : @DateTimeFormat(pattern = "yyyy-MM-dd") private Date your_variable; — Abderrahmane BECHIKH, Dec 19 '19 at 10:14

score 12 · Answer 1 · answered Sep 20 '16 at 17:32

The pattern you give to your html input elements doesn't work as you expect. You are not setting a format for your date with this, and as you can see from your error message, the date that spring is trying to parse is

Wed Jun 08 00:00:00 WET 2016

not any date in the format you have set in both your html and in your controller (the html pattern doesn't modify the format that gets sent, it's there for validation purposes).

I've never worked with that, but you should either

just remove the complete pattern and format and see if that works (I guess it could)
set the correct date format in your controller pattern, according to the date format I posted above (and your error message).

Here:

@DateTimeFormat(pattern = "yyyy-MM-dd")

The pattern is not his issue. If you read the error message, it states that it cannot convert the type java.lang.String to java.util.Date. This will instead require the date being parsed into a date, by using SimpleDateFormat or something similar. — Mindsect Team, Sep 19 '17 at 10:44

score 1 · Answer 2 · answered Apr 18 '20 at 02:08

You have correctly set pattern in the Controller Class with the annotation @DateTimeFormat(pattern = "yyyy-MM-dd"). And please also make sure that you have: imported two required patterns in your Model/Entity Class as:

import javax.persistence.Temporal;
import javax.persistence.TemporalType;

@Temporal(TemporalType.DATE)
private Date date;

Hope it does work.cause It worked on mine.

Failed to convert value of type java.lang.String to required type java.util.Date

2 Answers2

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