I know it is possible to know what depth you are at with $category->depth, but is there a straght-forward way to check if it is indeed last level node?
Thanks
Within the category class there is a method called getLastPosition. It requires two parameters to be passed, but if you can get those it looks to me like it returns an integer of the last position. If depth also returns an integer then you should be able to compare the two to see if it comes out true.
You will need to pass the id of the parent category and the id of the shop to the getLastPosition method for it to work.
{if $category->getLastPosition($id_parent, $id_shop) == $category->depth}
{/if}
I think that should work, however I did not test it.
Related
I want to check if an entity is part of the users input.
Example:
entities['#PRODUKT_INTENT_STOP_LIST']?.contains($variables.tmpEntity)
As you can see by this example, the value of the entity#PRODUKT_INTENT_STOP_LIST
is a variable. I put this at a condition for a node, but this is not working.
If I use a hardcoded string instead of the variable it is working fine.
entities['#PRODUKT_INTENT_STOP_LIST']?.contains('Chart') works fine
but setting $variables.tmpEntity to 'Chart' a and then ask for
entities['#PRODUKT_INTENT_STOP_LIST']?.contains($variables.tmpEntity)
is not working.
Can someone tells me what's wrong here?
Still trying to understand what you are trying to do.But if you want to check whether an entity exist in your input or not you can do it by applying condition on size of that entity.
"context":{
"size":"<?#Entity.size()?>"
}
now if size is equals to 0 then entity does not exist.
I know this is a longer way but it also tells you how many times does that entity exist in your input.
Hi I used the wrong statement.
This statement should work:
entities[PRODUKT_INTENT_STOP_LIST]?.get($variables.countEntity).value==$variables.$variables.tmpEntity
$variables.countEntity : counter to iterate thru entity array #PRODUKT_INTENT_STOP_LIST to check if an entity value is equal $variables.tmpEntity
Regards
Following is a Node we having in DB
P:Person { name:"xxx", skill:"Java" }
and after awhile, we would like to change the Skill to skill array, is it possible?
P:Person { name:"xxx", skill:["Java", "Javascript"] }
Which Cypher query should I use?
If you have a single skill value in skill, then just do
MATCH (p:Person)
WHERE HAS (p.skill)
SET p.skill=[p.skill]
If there are multiple values you need to convert to an array such as P:Person { name:"xxx", skill:"Java","JavaScript" } then this should work:
MATCH (p:P)
SET p.skill= split(p.skill,",")
In fact, I think your real problem here is not how to get an array property in a node, but how to store it. Your data model is wrong in my opinion, storign data as array in neo4j is not common, since you have relations to store multiple skills (in your example).
How to create your data model
With your question, I can already see that you have one User, and one User can have 1..n skills.
I guess that one day (maybe tomorrow) you will need to know which users are able to use Java, C++, PHP, and every othre skills.
So, Here you can already see that every skill should have its own node.
What is the correct model in this case?
I think that, still with only what you said in question, you should have something like this:
(:Person{name:"Foo"})-[:KNOWS]->(:Skill{name:"Bar"})
using such a data model, you can get every Skill known by a Person using this query:
MATCH (:Person{name:"Foo"})-[:KNOWS]->(skill:Skill)
RETURN skill //or skill.name if you just want the name
and you can also get every Person who knows a Skill using this:
MATCH (:Skill{name:"Bar"})<-[:KNOWS]-(person)
RETURN person //Or person.name if you just want the name
Keep in mind
Storing array values in properties should be the last option when you are using neo4j.
If a property can be found in multiple nodes, having the same value, you can create a node to store it, then you will be able to link it the other nodes using relations, and finding every node having the property X = Y will be easier.
I would want to select the first instance of an element in a page where many number of such elements are present with 'ID' which will not be same always.
for example, visit, http://www.sbobet.com/euro which lists lot of sports and odds, where I want to click on the first odds.
and the html structure would be like this,
I want to click on this first span value and proceed with some test case.
Any help on how to achieve this ?
There could be two approaches two the problem:
1. If you are sure you will always need only the first instance:
driver.FindElementsByClassName("OddsR")[0];
If not, then you have collection of elemets and you can access an of those
2. Also, you can first identify any closest enclosing div and then you can use the same snippet as above:
driver.FindElementsByClassName("OddsR")[0];
This one is a better approach if page is a bit dynamic in nature
Use #class attribute. If OddsR class you are intrested in is the 1st one on the page then just use Driver.FindElement(By.ClassName("OddsR")). Webdriver will pick the 1st occurence (no matter if there are more)
Have checked your link and I agree with alecxe, you should probably start with div. But i would suggest a simpler selector :
css = "div.MarketBd span.OddsR"
The above selector will always point to the first span of "OddsR" class within div of "MarketBd" class.
Thanks for the response.
I am finally able to click on the element, by this XPATH,
"//span[#class='OddsR']"
This clicks on the first occurrence of 'OddsR' values, without giving any index.
could someone tell me in what way I can display the items in a view kanban with a specific color according to the state that is the record.
I'm trying something like this
<div t-attf-class="#{record.state=='scheduled' ? oe_kanban_color_#{kanban_getcolor(1)} : oe_kanban_color_#{kanban_getcolor(0)}">
but I looked ALL elements and not only those who are in the "scheduled".
Thanks :)
If you have copy/pasted exactly what you typed in the view definition, then your t-attf- class attribute is malformed, and all record will have the following class:
class="#{record.state=='scheduled' ? oe_kanban_color_1 : oe_kanban_color_0"
which, due to CSS class precedence, will cause them all to have the oe_kanban_color_1 style.
A few hints:
To avoid coloring some records, you can omit the oe_kanban_color_X entirely in some cases
You can use a t-att-class attribute to allow arbitrary Javascript expressions, depending on what you want to do. In contrast, t-attf-class only allows replacing placeholders.
When comparing field values with Javascript operators you normally want to use the value or raw_value of the field, rather that the Field object itself. value will only differ from raw_value when the value needs specific rendering, such as dates, numbers, etc.
The kanban_getcolor() function accepts any integer or string and returns one of the 10 default kanban color indexes.
Based on the above, the following might be closer to what you tried to do (note the t-att-class attribute:
<div t-att-class="record.state.value == 'scheduled' ?
'oe_kanban_color_1' :
'oe_kanban_color_0' ">
Alternatively, you could use t-attf-class and let kanban_getcolor() pick a color based on the state string:
<div t-attf-class="oe_kanban_color_#{kanban_getcolor(record.state.value)}">
That last example is similar to what is done in many default kanban views in the official OpenERP distribution.
I have a model:
class List:
data = ...
previous = models.ForeignKey('List', related_name='r1')
obj = models.ForeignKey('Obj', related_name='nodes')
This is one direction list containing reference to some obj of Obj class. I can reverse relation and get some list's all elements refering to obj by:
obj.nodes
But how Can I get the very last node? Without using raw sql, genering as little SQL queries by django as can.
obj.nodes is a RelatedManager, not a list. As with any manager, you can get the last queried element by
obj.nodes.all().reverse()[0]
This makes sense anyway only if there is any default order defined on the Node's Meta class, because otherwise the semantic of 'reverse' don't make any sense. If you don't have any specified order, set it explicitly:
obj.nodes.order_by('-pk')[0]
len(obj.nodes)-1
should give you the index of the last element (counting from 0) of your list
so something like
obj.nodes[len(obj.nodes)-1]
should give the last element of the list
i'm not sure it's good for your case, just give it a try :)
I see this question is quite old, but in newer versions of Django there are first() and last() methods on querysets now.
Well, you just can use [-1] index and it will return last element from the list. Maybe this question are close to yours:
Getting the last element of a list in Python
for further reading, Django does not support negative indexing and using something like
obj.nodes.all()[-1]
will raise an error.
in newer versions of Django you can use last() function on queryset to get the last item of your list.
obj.nodes.last()
another approach is to use len() function to get the index of last item of a list
obj.nodes[len(obj.nodes)-1]