How to change the RJP.MultiUrlPicker label Title on the umbraco backoffice code?
And where do I find the multi url picker code on the umbraco to make any changes?
The RJP.MultiUrlPicker package just uses the same dialog as the Umbraco Link Picker, which is used to insert links into rich text editors.
The view for this dialog can be found in the following location:
\umbraco\Views\common\dialogs\linkpicker.html
The label that you're looking to change (#content_nodeName) is actually being pulled from an XML language file, which you can modify by going to:
\umbraco\Config\Lang\en.xml
and changing the value of the key tag with alias nodeName.
If you do change the value of this property then you'll have to be careful when you upgrade your solution going forward, otherwise your changes may be overwritten.
Related
I'm writing my custom property type for module 2sxc. I mean I'd added a js file inside my app at path "/system/field-string-my-custom-field/index.js" and I'm writing my code there.
And I need to get id (or guid) and entity type of a current edited/created entity in modal window. I didn't find neither id, nor guid of that entity. The only place I've noticed in the html inside iframe that could tell type of entity is title on top of modal window, but that's also not good to take it from there, since it's actually a label, not the name of the entity, which might be different from the actual entity name. When I checked the code that is executed on submit button click - it has object formBuilderRef, which contains all the listed data I need (current entity's id, guid, type). But I didn't succeed in accessing it from my code for now.
Additional info:
I'm creating a custom property like this one. So on editing entity with my custom property type there's this modal window. Modal image is from internet but it should be enough to get an idea.
iframe's src property contains EntityId, but it looks like a bad place to take id from, since it contains an id of the entity used to open modal window of 2sxc module, but I could go to app settings and start editing totally different entity that is not on a current page at all. So I need an id of that currently edited entity and correct type, not a label.
So how do I get the data I need from there?
You can get them on
context.target.entity.id and context.target.entity.guid
see also: https://docs.2sxc.org/js-code/edit-form/formulas/context.html
while developing formulas, it's best to also watch the console, because on test-runs it will output all relevant objects data and context to the console.
In Sitecore MVC, FieldRenderer.Render(Item, "Field Name") automatically checks its field type. In another word, all XSLT extensions (sc:text, sc:image, sc:date, …) is affected by the RenderField pipeline.
But, I couldn't find sc:checkbox information and it doesn't show "CheckBox" in page edit mode. It just shows the value of the check box.
How can I make it show as like input type=checkbox in page editor?
Sorry for the link only answer but the general approach here is to use a Field Editor Button.
This will open a pop up window with the content editor style control for the field.
If using a custom experience button is not an option for you, you can output different markup (i.e. your input button) to the client when the page mode is in 'Edit' mode.
This will allow you to present the author with the controls you wish to display, but still output the normal output values for end users.
If you are able to use controller renderings, you can use the controller to load a different view to make this easier, but otherwise you can use if statements within your View to output different markup.
Let's say I defined a IBM WCM authoring template called "AT-Image"; it creates content item "CT-Image" that contains elements such as an Image of type "Image", Description of type "Short Text", Keywords of type "Text", and other fields that serve as meta for that image resource.
Now, I need to create a content item called "CT-Article" based off from its "AT-Article" authoring template. Let's say that "Article" item has 3 elements: Title, Summary, and Image. Title and Summary are of type "Short Text". But Image needs to be of type "CT-Image" and not Image or Image component.
Is this possible Out-of-the-Box or is this something I need to implement myself via WCM API?
From my point of view you should use a menu component in the following way:
Let's assume you have got your CT-Image and CT-Article in a common sitearea called "MyArticle" like this:
MyArticle
CT-Article
CT-Image
Furthermore you need a WCM Menu component which uses two selection criteria:
current sitearea as location
CT-Image as authoring template
So then rendering CT-Article with a presentation template PT-Article you can reference the menu component. Therefore, you reference the menu component in the presentation template markup. The menu component then will pick up the CT-Image and render it as it is located in the same sitearea as your CT-Article.
That's rather a content design approach then a technical approach. It's all there out-of-the-box and saves custom development effort.
If I understand correctly, basically you want to create pre-defined images and then pick one of them to display when you create an article. Correct?
Well, the easiest way would be to utilize what WCM already gives you: create an Image Component. You will be able to upload the image and also provide a description, but you won't be able to set keywords or metadata. The workaround for this would be to set the keywords/metadata as part of the article. If that is okay, this would probably be the best way to do it.
By doing that, all you would have to do is select the relevant Image Component from your Image Element in AT-Article.
If you really need the images to be their own content items (rather than components) so that they have their own keywords and other information, then the next "easiest" way is probably to use an HTML Element inside AT-Article. In this HTML Element you will have to use the "Insert Tag" option, and then choose an Element Tag and use "Selected" rather than "autofill" or "current" to navigate to the content item that you want (the one with the image).
The code will look like this:
[Element name="[path]" type="content" context="selected" key="[element]"]
Where [path] is the path to the content item and [element] is the element that you actually want to pull from that content item (such as image, description, keywords, etc.)
context="selected" is the key there. Then you just need to set up your Presentation Template to be able to use this information that you are pulling in from the HTML Element.
WCM 8.5 has the InContext tag.
The InContext tag renders a tag body within a specified context. The new context can be specified as a predefined context by using a UUID, or by path.
For eg. in this case you can use below where key="Image" in InContext tag is the Link element to the CT-Image content item :
[InContext uuid="[Element context='current' type='content' key='Image' format='id']"]
[Element context="current" type="auto" key="Image"]
[Element context="current" type="auto" key="Description"]
[/InContext]
I use the bartik theme in drupal7 as the default theme.And I completely override the page.tpl.php file.
design a two columns layout : the lefsidebar and the contentArea. Everything is ok except one problem.
the node's field label can not be set in the Manage display panel. And the node's body label was disappeared,and other's labels do exist but can not ,say, be set hidden ,above,inline.
I google the for solutions but bad luck no result.
Need your Help. Thanks.
I do not get the point in modifying the php file to do what you want. I think you just want to set a two-column layout, but you do not need to edit your code to do so. Have you tried the Display Suite module?
Adding a new Form in the project creates a default form with default properties. I have to change them manually, but in all my project I have the same properties (font, background color, etc.).
Is there a way I can change the default Ide template for Forms?
Not using inheritance as Adrian recommended is a mistake. It is the right way to do it because you can in one fell swoop change every form in your app by editing the base form properties.
But you want to change the template. That's easy to do as well. Start a new scratch project with one Form and change the properties you want to have customized. Click File + Export Template. Select Item template, Next. Tick the form, Next. Next. Give it a good name and description and click Finish. You can now select that template whenever you create a new form.
Create your own form control that is derived from the base one and has all the properties set just like you want them, then you can use this one all over your project. Also changing something will be very easy since you'll have to do it only in one place.