laytpl template syntax
{{ }} outputs a normal field, not escapes html official description, but null will be displayed when empty
laytpl template Solution to null display null
{{}} When giftName is empty in the template, a null string will be displayed on the page.
1 You can directly use a tri-item expression in the template to determine whether the string is empty.
as follows:
{{!?" ":}}
2 There is another method, using a || operator
{{||" "}} will be displayed as an empty string when empty
{{||"No giveaway"}} will be displayed when empty No giveaway Double quotes or single quotes wrapped strings are OK, but not without
{{|| }} Report an error directly without quotes Laytpl Error: SyntaxError: Unexpected token )
There is another exception that you can do without quotes
{{|| null }} and {{}} are no different
{{# JavaScript Expression }}
JS statement. Generally used for logical processing. Start with a separator plus a # sign. Note: If you want to output a function, the correct way to write it is: {{ fn() }}, instead of: {{# fn() }}
Try the if else syntax of the laytpl template
This is an example on the official website
{{# if(true){ }} start date:{{ fn() }} {{# } else { }} Deadlined {{# } }}
At first glance, I'll make it
{{# if(true){ }}
start date
{{# } else { }}
Deadlined
{{# } }}
Now it's clear at a glance
/doc/modules/