我想从sql视图层中获取样式(sld或css无关紧要,但更喜欢css),但是我无法实现。 这是我的层的SQL视图
select
value,
location,
'mark: symbol(square);mark-size: 6px;:mark {fill: #0099cc;}' STYLE
from TABLE t
这是我的风格:
/* @title point */
* {
[STYLE]
}
预览图层时,出现以下错误:
org.parboiled.errors.ParserRuntimeException: Error at line 3. Encountered "<EOF>" at line 1, column 5. Was expecting one of: "(" ... ":" ... <IDENTIFIER> ... "." ... "not" ... "in" ... "tequals" ... "before" ... "after" ... "during" ... "*" ... "/" ... "+" ... "-" ... "like" ... "ilike" ... "exists" ... "does-not-exist" ... "is" ... "between" ... "=" ... ">" ... "<" ... ">=" ... "<=" ... "<>" ... Parsing : STYLE.
有什么方法可以从数据库中获取样式信息?
答案 0 :(得分:0)
我认为不可能(正如您的问题所支持)。您当然可以从数据库中读取各个值,并在样式中允许使用表达式的任何地方使用它们。