构建“何处” Linq 查询中的子句

Question

在此查询中，我始终需要“正常”类型元素。
如果设置了 _includeX 标志，我也需要“工作空间”类型元素。
有没有一种方法可以将其写为一个查询？ 或者在提交查询之前基于_includeX构建where子句？

    if (_includeX) {
    query = from xElem in doc.Descendants(_xString)
        let typeAttributeValue = xElem.Attribute(_typeAttributeName).Value
        where typeAttributeValue == _sWorkspace ||
              typeAttributeValue == _sNormal
        select new xmlThing
        {
            _location = xElem.Attribute(_nameAttributeName).Value,
            _type = xElem.Attribute(_typeAttributeName).Value,
        }; 
}
else {
    query = from xElem in doc.Descendants(_xString)
        where xElem.Attribute(_typeAttributeName).Value == _sNormal
        select new xmlThing
        {
            _location = xElem.Attribute(_nameAttributeName).Value,
            _type = xElem.Attribute(_typeAttributeName).Value,
        }; 
}

Answer 1

您可以将其分解为一个单独的谓词：

Predicate<string> selector = x=> _includeX 
  ? x == _sWorkspace || x == _sNormal
  : x == _sNormal; 

query = from xElem in doc.Descendants(_xString)
      where selector(xElem.Attribute(_typeAttributeName).Value)
      select new xmlThing
      {
          _location = xElem.Attribute(_nameAttributeName).Value,
          _type = xElem.Attribute(_typeAttributeName).Value,
      };

或内联条件：

query = from xElem in doc.Descendants(_xString)
    let typeAttributeValue = xElem.Attribute(_typeAttributeName).Value
    where (typeAttributeValue == _sWorkspace && _includeX) ||
          typeAttributeValue == _sNormal
    select new xmlThing
    {
        _location = xElem.Attribute(_nameAttributeName).Value,
        _type = xElem.Attribute(_typeAttributeName).Value,
    }; 

或删除查询表达式用法并按以下方式执行：-

var all = doc.Descendants(_xString);
var query = all.Where( xElem=> {
      var typeAttributeValue = xElem.Attribute(_typeAttributeName).Value;
      return typeAttributeValue == _sWorkspace && includeX ) || typeAttributeValue == _sNormal;
})
.Select( xElem =>
    select new xmlThing
    {
        _location = xElem.Attribute(_nameAttributeName).Value,
        _type = xElem.Attribute(_typeAttributeName).Value,
    })

或将第一个和第三个组合起来并执行以下操作：

Predicate<string> selector = x=> _includeX 
  ? x == _sWorkspace || x == _sNormal
  : x == _sNormal; 

query = doc.Descendants(_xString)
      .Where(xElem => selector(xElem.Attribute(_typeAttributeName).Value))
      .Select(xElem => new xmlThing
      {
          _location = xElem.Attribute(_nameAttributeName).Value,
          _type = xElem.Attribute(_typeAttributeName).Value,
      };)

这完全取决于在您的上下文中最干净的工作方式。

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