Pass array as argument in Thymeleaf fragment

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.

edited Nov 10 '18 at 19:51

asked Nov 10 '18 at 19:20

BigJ

81421532

add a comment |

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.

edited Nov 10 '18 at 19:51

asked Nov 10 '18 at 19:20

BigJ

81421532

add a comment |

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.

edited Nov 10 '18 at 19:51

asked Nov 10 '18 at 19:20

BigJ

81421532

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

edited Nov 10 '18 at 19:51

asked Nov 10 '18 at 19:20

BigJ

81421532

edited Nov 10 '18 at 19:51

asked Nov 10 '18 at 19:20

BigJ

81421532

edited Nov 10 '18 at 19:51

asked Nov 10 '18 at 19:20

BigJ

81421532

asked Nov 10 '18 at 19:20

BigJ

81421532

asked Nov 10 '18 at 19:20

BigJ

81421532

add a comment |

2 Answers
2

active

oldest

votes

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 '18 at 17:46

BigJ

81421532

add a comment |

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 '18 at 17:47

answered Nov 11 '18 at 17:06

Alain Cruz

1,7683818

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 '18 at 17:49

Glad I could help! :)

– Alain Cruz
Nov 12 '18 at 17:51

add a comment |

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2 Answers
2

active

oldest

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2 Answers
2

active

oldest

votes

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 '18 at 17:46

BigJ

81421532

add a comment |

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 '18 at 17:46

BigJ

81421532

add a comment |

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 '18 at 17:46

BigJ

81421532

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 '18 at 17:46

BigJ

81421532

answered Nov 12 '18 at 17:46

BigJ

81421532

answered Nov 12 '18 at 17:46

BigJ

81421532

answered Nov 12 '18 at 17:46

BigJ

81421532

add a comment |

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 '18 at 17:47

answered Nov 11 '18 at 17:06

Alain Cruz

1,7683818

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 '18 at 17:49

Glad I could help! :)

– Alain Cruz
Nov 12 '18 at 17:51

add a comment |

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 '18 at 17:47

answered Nov 11 '18 at 17:06

Alain Cruz

1,7683818

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 '18 at 17:49

Glad I could help! :)

– Alain Cruz
Nov 12 '18 at 17:51

add a comment |

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 '18 at 17:47

answered Nov 11 '18 at 17:06

Alain Cruz

1,7683818

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 '18 at 17:47

answered Nov 11 '18 at 17:06

Alain Cruz

1,7683818

edited Nov 12 '18 at 17:47

answered Nov 11 '18 at 17:06

Alain Cruz

1,7683818

answered Nov 11 '18 at 17:06

Alain Cruz

1,7683818

answered Nov 11 '18 at 17:06

Alain Cruz

1,7683818

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 '18 at 17:49

Glad I could help! :)

– Alain Cruz
Nov 12 '18 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 '18 at 17:49

Glad I could help! :)

– Alain Cruz
Nov 12 '18 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 '18 at 17:49

Glad I could help! :)

– Alain Cruz
Nov 12 '18 at 17:51

add a comment |

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