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
asked Nov 10 at 19:20
BigJ
76921432
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
asked Nov 10 at 19:20
BigJ
76921432
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
asked Nov 10 at 19:20
BigJ
76921432
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
asked Nov 10 at 19:20
BigJ
76921432
asked Nov 10 at 19:20
BigJ
76921432
edited Nov 10 at 19:51
asked Nov 10 at 19:20
BigJ
76921432
asked Nov 10 at 19:20
BigJ
76921432
asked Nov 10 at 19:20
BigJ
76921432
76921432
add a comment |
add a comment |
2 Answers
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1
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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
76921432
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>
answered Nov 11 at 17:06
Alain Cruz
1,6841818
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 |
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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.
answered Nov 12 at 17:46
BigJ
76921432
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.
answered Nov 12 at 17:46
BigJ
76921432
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.
answered Nov 12 at 17:46
BigJ
76921432
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
76921432
answered Nov 12 at 17:46
BigJ
76921432
answered Nov 12 at 17:46
BigJ
76921432
answered Nov 12 at 17:46
BigJ
76921432
76921432
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>
answered Nov 11 at 17:06
Alain Cruz
1,6841818
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>
answered Nov 11 at 17:06
Alain Cruz
1,6841818
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>
answered Nov 11 at 17:06
Alain Cruz
1,6841818
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>
answered Nov 11 at 17:06
Alain Cruz
1,6841818
edited Nov 12 at 17:47
answered Nov 11 at 17:06
Alain Cruz
1,6841818
answered Nov 11 at 17:06
Alain Cruz
1,6841818
answered Nov 11 at 17:06
Alain Cruz
1,6841818
1,6841818
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 |
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