ned help on XSLT sequence in the loop

updated points

updated points

help please

you can put contains() in the tests, so you need two tests, preferably in when clauses
<xsl

        <xsl:choose>
            <xsl:when test="contains(translate(., 'LUNCH', 'lunch'), 'lunch') and contains(translate(., 'DINER', 'diner'),'dinner')">
                <xsl:text>Lunch and Dinner</xsl:text>
            </xsl:when>
            <xsl:when test="contains(translate(., 'LUNCH', 'lunch'), 'lunch') and contains(translate(., 'BREAKFST', 'breakfst'),'breakFast ')">
                <xsl:text>BreakFast and lunch</xsl:text>
            </xsl:when>
        </xsl:choose>

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The translate() function I used for making the serach for you keywords, case-insensitive

Note: sending 4 follow ups in a minute will not make you get a quicker answer

I think I did not ask the question properly.
meal can occurs maximun of 2 times. This will run the loop 2 times., If i need to write in c, or java
isLunch = false;
isBreakFast = false;
isDinner = false;
string bf= null;
string lunch = null;
string dinner= null;
for() // runs 2
{
   If (meal.equals("BreakFast")
   {
      isBreakFast = true;
     bf = meal;
   }
If (meal.equals("lunch")
   {
      IsLunch = true;
      lunch = meal;
   }
If (meal.equals("Dinner")
   {
     isDinner = true;
     dinner = meal;
   }


}/ end loop.
If(isDinner)
{
   println(lunch); //new line
  print(dinner); //new line
   
}
If(isBf)
{
 print(isBreakFast ); //new line
 println(lunch); //new line
   
}
I am loking some thing similar to this in XSLT.

Note: I am in hurry, Today is a ruch day. sorry for the updates
Thank you

You are in a hurry, OK, please ask the right question from the start then
Please explain what you want, do I really have to go out and interprete java code to get what you need?

Listen, this forum is about XSLT
- XSLT requires a completely different approach then procedural Javacode, so you procedural java code means nothing in this context
- input to XSLT is XML, I haven't( seen any XML so far

So please show the XML you start from or some examples of XML
based on that, show what you want as an output.
If you are really stressed and in a hurry, better learn to ask questions the right way

I am sorry.

I have one xml element, that can repeat 2 or more times. Possible values are Breakfast, lunch, or dinner.
if i loop through these by for each, the out put would be in the same order as input. But my requirement need a sequence.
input 1:

<ROOT>
   <Meal>lunch</Meal>
    <Meal>Breakfast</Meal>
</ROOT>
Out put should be always breakfast first if it is with lunch. ===>
breakfast
luncn
======
input 1:

<ROOT>
   <Meal>Dinner</Meal>
    <Meal>Lunch</Meal>
</ROOT>
Out put should be always Lunch first if it is with dinner. ===>

luncn
dinner

This would help

Thank you, this really helps.
You made a mistake by posting java code.
because you don't solve this in XSLT, using loops and variables
You can XPath directly in the XML structure

<xsl:template match="ROOT">
        <xsl:choose>
            <xsl:when test="contains(translate(Meal, 'LUNCH', 'lunch'), 'lunch') and contains(translate(Meal, 'DINER', 'diner'),'dinner')">
                <xsl:text>Lunch and Dinner</xsl:text>
            </xsl:when>
            <xsl:when test="contains(translate(Meal, 'LUNCH', 'lunch'), 'lunch') and contains(translate(Meal, 'BREAKFST', 'breakfst'),'breakFast ')">
                <xsl:text>BreakFast and lunch</xsl:text>
            </xsl:when>
        </xsl:choose>
</xsl:template>

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If you need them on two lines, you could do something like this
<xsl:text>Lunch&#10;Dinner</xsl:text>

and there is a bug, sorry
<xsl:when test="contains(translate(Meal, 'LUNCH', 'lunch'), 'lunch') and contains(translate(Meal, 'BREAKFST', 'breakfst'),'breakFast ')">
should be
<xsl:when test="contains(translate(Meal, 'LUNCH', 'lunch'), 'lunch') and contains(translate(Meal, 'BREAKFST', 'breakfst'),'breakfast ')">

I am geting this error
A sequence of more than one item is not allowed as the first argument of translate()
Thank you

yep, sorry, that is a mistake, try this

    <xsl:template match="ROOT">
        <xsl:choose>
            <xsl:when test="Meal[contains(translate(., 'LUNCH', 'lunch'), 'lunch')]  and Meal[contains(translate(., 'DINER', 'diner'),'dinner')]">
                <xsl:text>Lunch and Dinner</xsl:text>
            </xsl:when>
            <xsl:when test="Meal[contains(translate(., 'LUNCH', 'lunch'), 'lunch')] and Meal[contains(translate(., 'BREAKFST', 'breakfst'),'breakfast')]">
                <xsl:text>BreakFast and lunch</xsl:text>
            </xsl:when>
        </xsl:choose>
    </xsl:template>

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Than you. It works.  I am accepting solution. Can you explain me the what the does?
test="Meal[contains(lower-case(.), 'lunch')]  and Meal[contains(lower-case(.),'dinner')]">
Is it comparing every element in the NodeList

 Is there anyway we can do with out hard coding the values in the output.

Thank you

Thank you for your time.

welcome

Meal[contains(lower-case(.), 'lunch')]
is an XPath expression that selects the Meal child element that has a content that contains 'lunch' when lower-cased
In an test expression this will reult in false() if there isn't such a Meal child, or true() when there is

test="Meal[contains(lower-case(.), 'lunch')]  and Meal[contains(lower-case(.),'dinner')]">
checks whether there is a Meal child with 'lunch' AND one with 'dinner'. I made order not important that way

Which values would you like to not 'hardcode'?