Pass array as argument in Thymeleaf fragment
up vote
1
down vote
favorite
Is it possible to pass an array to a fragment like this:
<div th:replace="fragments :: test_fragment( {'one', 'two', 'three'} )"></div>
And iterate in the fragment like this:
<div th:fragment="test_fragment(numberArray)">
<span th:each="number : ${numberArray}" th:text="${number}"></span>
</div>
As a bonus, are multidimensional arrays also possible?
I am using Thymeleaf in a Spring Boot 2.0 project.
spring-boot thymeleaf
add a comment |
up vote
1
down vote
favorite
Is it possible to pass an array to a fragment like this:
<div th:replace="fragments :: test_fragment( {'one', 'two', 'three'} )"></div>
And iterate in the fragment like this:
<div th:fragment="test_fragment(numberArray)">
<span th:each="number : ${numberArray}" th:text="${number}"></span>
</div>
As a bonus, are multidimensional arrays also possible?
I am using Thymeleaf in a Spring Boot 2.0 project.
spring-boot thymeleaf
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is it possible to pass an array to a fragment like this:
<div th:replace="fragments :: test_fragment( {'one', 'two', 'three'} )"></div>
And iterate in the fragment like this:
<div th:fragment="test_fragment(numberArray)">
<span th:each="number : ${numberArray}" th:text="${number}"></span>
</div>
As a bonus, are multidimensional arrays also possible?
I am using Thymeleaf in a Spring Boot 2.0 project.
spring-boot thymeleaf
Is it possible to pass an array to a fragment like this:
<div th:replace="fragments :: test_fragment( {'one', 'two', 'three'} )"></div>
And iterate in the fragment like this:
<div th:fragment="test_fragment(numberArray)">
<span th:each="number : ${numberArray}" th:text="${number}"></span>
</div>
As a bonus, are multidimensional arrays also possible?
I am using Thymeleaf in a Spring Boot 2.0 project.
spring-boot thymeleaf
spring-boot thymeleaf
edited Nov 10 at 19:51
asked Nov 10 at 19:20
BigJ
74921432
74921432
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
I've found two ways: the fragment parameters and th:with.
Fragment parameter array:
<div th:replace="~{fragments :: test_fragment(arrayX = ${ {'a', 'b'} }) }"></div>
Frag parameter array multidimensional:
<div th:replace="~{fragments :: test_fragment(arrayX = ${ {{'a1','a2'},{'b1','b2'}} } ) }"></div>
th:with array:
<div th:insert="~{fragments :: test_fragment}" th:with="arrayX=${ {'a','b'} }"></div>
th:with array multidimensional:
<div th:insert="~{fragments :: test_fragment}" th:with="arrayX=${ {{'a1','a2'},{'b1','b2'}} }"></div>
Notice that I used th:insert when I used th:with. This is because th:replace would replace the div line and thus the th:with, which causes the arrayX to not be available.
add a comment |
up vote
1
down vote
Yes, it is possible. The following code should do the trick. The only difference, is the added ${}
, outside the array.
<div th:replace="fragments :: test_fragment(${ {'one', 'two', 'three'} })"></div>
Thanks! That did the trick. I also found another way, and a way for a multidimensional array so I'll use that as the answer.
– BigJ
Nov 12 at 17:49
Glad I could help! :)
– Alain Cruz
Nov 12 at 17:51
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I've found two ways: the fragment parameters and th:with.
Fragment parameter array:
<div th:replace="~{fragments :: test_fragment(arrayX = ${ {'a', 'b'} }) }"></div>
Frag parameter array multidimensional:
<div th:replace="~{fragments :: test_fragment(arrayX = ${ {{'a1','a2'},{'b1','b2'}} } ) }"></div>
th:with array:
<div th:insert="~{fragments :: test_fragment}" th:with="arrayX=${ {'a','b'} }"></div>
th:with array multidimensional:
<div th:insert="~{fragments :: test_fragment}" th:with="arrayX=${ {{'a1','a2'},{'b1','b2'}} }"></div>
Notice that I used th:insert when I used th:with. This is because th:replace would replace the div line and thus the th:with, which causes the arrayX to not be available.
add a comment |
up vote
1
down vote
accepted
I've found two ways: the fragment parameters and th:with.
Fragment parameter array:
<div th:replace="~{fragments :: test_fragment(arrayX = ${ {'a', 'b'} }) }"></div>
Frag parameter array multidimensional:
<div th:replace="~{fragments :: test_fragment(arrayX = ${ {{'a1','a2'},{'b1','b2'}} } ) }"></div>
th:with array:
<div th:insert="~{fragments :: test_fragment}" th:with="arrayX=${ {'a','b'} }"></div>
th:with array multidimensional:
<div th:insert="~{fragments :: test_fragment}" th:with="arrayX=${ {{'a1','a2'},{'b1','b2'}} }"></div>
Notice that I used th:insert when I used th:with. This is because th:replace would replace the div line and thus the th:with, which causes the arrayX to not be available.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I've found two ways: the fragment parameters and th:with.
Fragment parameter array:
<div th:replace="~{fragments :: test_fragment(arrayX = ${ {'a', 'b'} }) }"></div>
Frag parameter array multidimensional:
<div th:replace="~{fragments :: test_fragment(arrayX = ${ {{'a1','a2'},{'b1','b2'}} } ) }"></div>
th:with array:
<div th:insert="~{fragments :: test_fragment}" th:with="arrayX=${ {'a','b'} }"></div>
th:with array multidimensional:
<div th:insert="~{fragments :: test_fragment}" th:with="arrayX=${ {{'a1','a2'},{'b1','b2'}} }"></div>
Notice that I used th:insert when I used th:with. This is because th:replace would replace the div line and thus the th:with, which causes the arrayX to not be available.
I've found two ways: the fragment parameters and th:with.
Fragment parameter array:
<div th:replace="~{fragments :: test_fragment(arrayX = ${ {'a', 'b'} }) }"></div>
Frag parameter array multidimensional:
<div th:replace="~{fragments :: test_fragment(arrayX = ${ {{'a1','a2'},{'b1','b2'}} } ) }"></div>
th:with array:
<div th:insert="~{fragments :: test_fragment}" th:with="arrayX=${ {'a','b'} }"></div>
th:with array multidimensional:
<div th:insert="~{fragments :: test_fragment}" th:with="arrayX=${ {{'a1','a2'},{'b1','b2'}} }"></div>
Notice that I used th:insert when I used th:with. This is because th:replace would replace the div line and thus the th:with, which causes the arrayX to not be available.
answered Nov 12 at 17:46
BigJ
74921432
74921432
add a comment |
add a comment |
up vote
1
down vote
Yes, it is possible. The following code should do the trick. The only difference, is the added ${}
, outside the array.
<div th:replace="fragments :: test_fragment(${ {'one', 'two', 'three'} })"></div>
Thanks! That did the trick. I also found another way, and a way for a multidimensional array so I'll use that as the answer.
– BigJ
Nov 12 at 17:49
Glad I could help! :)
– Alain Cruz
Nov 12 at 17:51
add a comment |
up vote
1
down vote
Yes, it is possible. The following code should do the trick. The only difference, is the added ${}
, outside the array.
<div th:replace="fragments :: test_fragment(${ {'one', 'two', 'three'} })"></div>
Thanks! That did the trick. I also found another way, and a way for a multidimensional array so I'll use that as the answer.
– BigJ
Nov 12 at 17:49
Glad I could help! :)
– Alain Cruz
Nov 12 at 17:51
add a comment |
up vote
1
down vote
up vote
1
down vote
Yes, it is possible. The following code should do the trick. The only difference, is the added ${}
, outside the array.
<div th:replace="fragments :: test_fragment(${ {'one', 'two', 'three'} })"></div>
Yes, it is possible. The following code should do the trick. The only difference, is the added ${}
, outside the array.
<div th:replace="fragments :: test_fragment(${ {'one', 'two', 'three'} })"></div>
edited Nov 12 at 17:47
answered Nov 11 at 17:06
Alain Cruz
1,5231818
1,5231818
Thanks! That did the trick. I also found another way, and a way for a multidimensional array so I'll use that as the answer.
– BigJ
Nov 12 at 17:49
Glad I could help! :)
– Alain Cruz
Nov 12 at 17:51
add a comment |
Thanks! That did the trick. I also found another way, and a way for a multidimensional array so I'll use that as the answer.
– BigJ
Nov 12 at 17:49
Glad I could help! :)
– Alain Cruz
Nov 12 at 17:51
Thanks! That did the trick. I also found another way, and a way for a multidimensional array so I'll use that as the answer.
– BigJ
Nov 12 at 17:49
Thanks! That did the trick. I also found another way, and a way for a multidimensional array so I'll use that as the answer.
– BigJ
Nov 12 at 17:49
Glad I could help! :)
– Alain Cruz
Nov 12 at 17:51
Glad I could help! :)
– Alain Cruz
Nov 12 at 17:51
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53242576%2fpass-array-as-argument-in-thymeleaf-fragment%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown