Pass array as argument in Thymeleaf fragment









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1
down vote

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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.










share|improve this question



























    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.










    share|improve this question

























      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.










      share|improve this question















      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 10 at 19:51

























      asked Nov 10 at 19:20









      BigJ

      74921432




      74921432






















          2 Answers
          2






          active

          oldest

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          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.






          share|improve this answer



























            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>





            share|improve this answer






















            • 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










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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.






            share|improve this answer
























              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.






              share|improve this answer






















                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.






                share|improve this answer












                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.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 12 at 17:46









                BigJ

                74921432




                74921432






















                    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>





                    share|improve this answer






















                    • 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














                    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>





                    share|improve this answer






















                    • 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












                    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>





                    share|improve this answer














                    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>






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Nov 12 at 17:47

























                    answered Nov 11 at 17:06









                    Alain Cruz

                    1,5211818




                    1,5211818











                    • 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










                    • 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

















                     

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