Question

我的数据看起来像这样：
的节点

Name    Attribute    Date
14      A1           11-OCT-2015
14      A2           7-Nov-2015
12      B1           11-Nov-2015

载体

Node    V_NAME    color   Date
14      V1        blue   11-OCT-2015
14      V1        red    10-Nov-2015
14      V2        blue   7-Nov-2015
12      V3        black  11-Nov-2015
12      V4        black  11-Nov-2015

我希望得到如下结果

Node   Attribute    V_NAME   color 
14      A2           V1      red
14      A2           V2      blue
12      B1           V3      black
12      B1           V4      black

两个表中的日期列不相同

Answer 1

if (typeof Minicart != "undefined") {
    Minicart.prototype.updateItem = function(el) {
        var cart = this;
        var input = $j(this.selectors.quantityInputPrefix + $j(el).data('item-id'));
        var quantity = parseInt(input.val(), 10);
        cart.hideMessage();
        cart.showOverlay();
        $j.ajax({
            type: 'POST',
            dataType: 'json',
            url: input.data('link'),
            data: {qty: quantity, form_key: cart.formKey}
        }).done(function(result) {
            cart.hideOverlay();
            if (result.success) {
                cart.updateCartQty(result.qty);
                if (quantity !== 0) {
                    cart.updateContentOnUpdate(result);
                } else {
                    cart.updateContentOnRemove(result, input.closest('li'));
                }
            } else {
                cart.showMessage(result);
            }
        }).error(function() {
            cart.hideOverlay();
            cart.showError(cart.defaultErrorMessage);
        });
        return false;
    };
}

http://sqlfiddle.com/#!4/d3464/1

Answer 2

使用NOT EXISTS仅返回与其他日期不同的向量的向量。

select v.Node, n.Attribute, v.V_NAME, v.color
from nodes as n
  join vectors as v on n.name = v.node
where not exists (select 1 from vectors as v2
                  where v2.v_name = v.v_name
                    and v2.date > v.date)

替代答案，使用子查询来选择每个v_name的最新日期：

select v.Node, n.Attribute, v.V_NAME, v.color
from nodes as n
  join vectors as v on n.name = v.node
where (v.v_name, v.date) = (select v_name, max(date) from vectors group by v_name)

Answer 3

我使用两个cte来计算每个类别中最近的行。然后加入两个。

<强> SqlFiddleDemo

WITH n_node as (
    SELECT "Name", "Attribute",
           row_number() over (partition by "Name" order by "Date" DESC) rn
    FROM Nodes 
), 
n_vector as (
    SELECT "Node", "V_NAME", "color",
           row_number() over (partition by "Node", "V_NAME" order by "Date" DESC) rn
    FROM Vectors 
)
SELECT "Name", "Attribute", "V_NAME", "color"
FROM n_node
JOIN n_vector 
  ON n_node.rn = n_vector.rn
 AND n_node.rn = 1
 AND n_node."Name" = n_vector."Node"
ORDER BY "Name" DESC

<强>输出

| Name | Attribute | V_NAME | color |
|------|-----------|--------|-------|
|   14 |        A2 |     V1 |   red |
|   14 |        A2 |     V2 |  blue |
|   12 |        B1 |     V3 | black |
|   12 |        B1 |     V4 | black |

如何将一个表中的最新行连接到另一个表中的最新行（oracle）

3 个答案: