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Showing field based on multiselect component label

Parthiban

Hi Team,

I want to show custom field based on Component selection. I have written code in Script Runner Behaviour , my code working perfectly when users select one component label. But its not behaving correctly when users select two or more component labels.

If users select two or more component labels , I have to show two or more custom fields.

Note : I have created 40 custom fields and 40 component labels for my client. When ever users select component I need to show/display respective custom fields.

My code is working when users select one component label.

@BaseScript FieldBehaviours fieldBehaviours
def log = Logger.getLogger("com.acme.CreateSubtask");
log.setLevel(Level.DEBUG);

def dashboardCost = getFieldById("customfield_12735");
def devCost = getFieldById("customfield_12736");
def sccdCost = getFieldById("customfield_12737");
def jiraCost = getFieldById("customfield_12738");

def components =getFieldById("components");
def comp= components.getValue()*.getName();

if (comp.toString().contains('Dashboard')){
dashboardCost.setHidden(false);
}
else if(comp.toString().contains('dev')){
devCost.setHidden(false);
}
else if (comp.toString().contains('SCCD')){
sccdCost.setHidden(false);
}
else if (comp.toString().contains('JIRA')){
jiraCost.setHidden(false);
}

Accepted answers

Joanna Choules

Hi,

What exactly is happening when you select more than one component label? Knowing this will make it easier for us to diagnose the problem and suggest a fix.

One initial suggestion I would make, however, is not to call toString() on comp. The comp variable is a list of strings so the contains() method will work fine with it, without needing to turn the whole thing into a string first.

All comments

Joanna Choules

Hi,

What exactly is happening when you select more than one component label? Knowing this will make it easier for us to diagnose the problem and suggest a fix.

Parthiban

Hi Jake,

Thanks for your suggestion , I removed toString() from my coding still its working fine when I select single component label.

If I select more than one component label , it always pick the first component label .

Because I have set the condition like wise, can you tell me how to modify my code.

Thanks,

Parthiban

Alexey Matveev

Hello,

I guess it does not work because you use else if. Only first condition fires. You need to rewrite it to

if (comp.toString().contains('Dashboard')){
dashboardCost.setHidden(false);
} else {

dashboardCost.setHidden(true);

}
if(comp.toString().contains('dev')){
devCost.setHidden(false);
} else {

devCost.setHidden(true);

}
if (comp.toString().contains('SCCD')){
sccdCost.setHidden(false);
} else {

sccdCost.setHidden(true);

}
if (comp.toString().contains('JIRA')){
jiraCost.setHidden(false);
} else {


jiraCost.setHidden(true);

}

Parthiban

Hi Alexey/Jake,

Thanks for writing me back.

Yup, I know my code was wrong . May I know how to achieve my requirement.

If I select more than one component label , how to get the list of Component and size of component selected.

Thanks,

Parthiban

Alexey Matveev

Your code is right. The problem is about if statements. I gave the the right code in my previous post. If you want to get the size of the list of the selected components you should do it like this

List<String> comp= components.getValue()*.getName();
def compSize = comp.size()

Joanna Choules

Hi Parthiban,

As Alexey has indicated, the reason that only the first label is getting picked is because your if-blocks are chained together with 'else'. As soon as one of the if-conditions succeeds, the code corresponding to that condition will run and all of the following if-blocks will be skipped over.

Alexey's code seems entirely correct to me (although as I said before, you can get rid of the toString() if you want) but I will suggest some slight alternatives for the sake of variety:

Because setHidden() takes a boolean, which is what your if-conditions are testing in the first place, you can in principle cut out the if-statements altogether and do something like this:
```
dashboardCost.setHidden(!comp.contains('Dashboard'))
```
Essentially we find out whether the components list includes 'Dashboard' and then set the hidden attribute of the dashboardCost field to the opposite of that (hence the use of '!').
You say in your original question that you have 40 each of custom fields and component labels, so I assume that the code you posted is just a small snippet of the whole thing. Given that you're doing essentially the same thing for each field/label pairing, there's going to be a lot of repetition in your code which will make it harder to modify later. You could eliminate a lot of this repetition by gathering all of the pairings together into a Map and looping over it:

def labelToFieldId =
  ['Dashboard': "customfield_12735",
   'dev': "customfield_12736",
   'SCCD': "customfield_12737",
   'JIRA': "customfield_12738",
   <etc...>
  ];

def components = getFieldById("components");
def comp = components.getValue()*.getName() ?: [];

labelToFieldId.each { label, fieldId ->
  def field = getFieldById(fieldId);
  field?.setHidden(!comp.contains(label));
}

Please note that I haven't tested the above code but it's just to give you the general idea of how you could do it.

Hope this is helpful.

Regards,

Parthiban

Hi Jake,

Thanks a lot.!! It worked exactly as per my need. I working on this for past one month on this requirement. Thanks for your great support.

Thanks,

Parthiban

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