blocks|key|3987883|text|“清单中的所有项目都有一个id，该id存在于listcost中”。listCost需要与库存具有相同数量的项目(假设Id是唯一的)，可能会更多，才有机会返回true|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3987884|checkItemInventory.All(x+=>+listCost.Contains(x.Id))|code-block|syntax|javascript|3987885|“清单中至少有一个项目的id也在listCost中”。Listcost最少只能有一个id，以便有机会返回true。|3987886|checkItemInventory.Any(x+=>+listCost.Contains(x.Id))|3987887|正如您所看到的，这两个都不是您想要的，因为您似乎是在说，您想要检查listcost中的每一项是否也存在于库存中。这类似于最上面的代码，但反过来(“listCost中的所有项目都存在于inventory中”vs“所有inventory中的项目都存在于listcost中”|3987888|我想我会先从库存中做一个字典，除非它已经支持快速查找：|3987889|var+d+=+checkItemInventory.Select(x+=>+new+{+x.Id,+x.Id+}).ToDictionary();
var+boolResult+=+listCost.All(lc+=>+d.ContainsKey(lc));|3987890|如果库存很少，您可以使用此方法：|3987891|listCost.All(lc+=>+checkItemInventory.Any(cii+=>+cii.Id+==+lc));|3987892|请注意，在内部，它可能会执行以下操作：|3987893|bool+all+=+true;
foreach(lc+in+listCost){
++bool+found+=+false;
++foreach(cci+in+checkItemInventory)
++++if(lc+==+cci.Id){
++++++found+=+true;
++++++break;
++++}
++all+&=+found;
++if(!all)
++++return+false;
}
return+true;|3987894|这是大量的重复比较(对于listCost中的每一项，都会扫描整个库存)，可能会很慢|3987895|编辑|3987896|我要求澄清您如何存储您的库存和您的建设项目的成本。这是我的一个假设，以及基于它的解决方案是如何工作的：|3987897|假设你的清单中有物品的种类，并有一个计数来说明玩家携带了多少物品：|3987898|class+InventoryItem{
++int+ItemKindId+{+get;+set;}
++int+CountOf+{+get;+set;+}
}

player.Inventory.Add(new+InventoryItem()+{+
++ItemKindId+=+Constants.WOOD,+//1
++CountOf+=+10+//holding+10+items+of+wood
};
player.Inventory.Add(new+InventoryItem()+{+
++ItemKindId+=+Constants.STONE,+//2
++CountOf+=+5+//holding+5+items+of+stone
};|3987899|假设你有一个制作斧头的食谱，它需要1块木头和2块石头，但它按简单的顺序列出了它们：|3987900|int[]+axeRecipe+=+new+int[]+{+Constants.WOOD,+Constants.STONE,+Constants.STONE+};|3987901|对食谱进行分组可能是最简单的：|3987902|var+recipe+=+axeRecipe.GroupBy(item+=>+item)

/*
++now+we+have+a+grouping+of+the+recipe[item].Key+as+the+material+and+a
++recipe[item].Count()+of+how+much.+The+group+is+like+a+dictionary:

++recipe[Constants.WOOD]+=+new+List<int>{+Constants.WOOD+};
++recipe[Constants.STONE]+=+new+List<int>{+Constants.STONE,+Constants.STONE,+};
++A+group+item+has+a+Key+and+a+list+of+objects+that+have+that+key
++Because+my+recipe+was+simply+ints,+the+Key+is+the+same+number+as+all+the+
++items+in+the+list
*/

//for+all+items+in+the+recipe
grp.All(groupItem+=>
++//does+the+player+inventory+contain+any+item
++playerInventory.Any(inventoryItem+=>+
++++//where+the+material+kind+is+the+same+as+the+recipe+key+(material)
++++inventoryItem.ItemKindId+==+groupItem.Key+&&
++++//and+the+count+they+have+of+it,+is+enough+to+make+the+recipe
++++inventoryItem.CountOf+>=+groupItem.Count()
);|3987903|当然，如果您愿意，可以将其减少到一行：axeRecipe.GroupBy(...).All(...)|offset|length|style|CODE|3987904|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|J|V|0^^$0|@$1|2|3|4|5|6|7|1O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1S|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1T|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|1U|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|1V|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|1W|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|1X|8|@]|9|@]|A|$]]|$1|W|3|X|5|D|7|1Y|8|@]|9|@]|A|$E|F]]|$1|Y|3|Z|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|10|3|11|5|6|7|20|8|@]|9|@]|A|$]]|$1|12|3|13|5|6|7|21|8|@]|9|@]|A|$]]|$1|14|3|15|5|6|7|22|8|@]|9|@]|A|$]]|$1|16|3|17|5|D|7|23|8|@]|9|@]|A|$E|F]]|$1|18|3|19|5|6|7|24|8|@]|9|@]|A|$]]|$1|1A|3|1B|5|D|7|25|8|@]|9|@]|A|$E|F]]|$1|1C|3|1D|5|6|7|26|8|@]|9|@]|A|$]]|$1|1E|3|1F|5|D|7|27|8|@]|9|@]|A|$E|F]]|$1|1G|3|1H|5|6|7|28|8|@$1I|29|1J|2A|1K|1L]]|9|@]|A|$]]|$1|1M|3|-4|5|6|7|2B|8|@]|9|@]|A|$]]]|1N|$]]

&quot;All items in the inventory have an id that present in listcost&quot;. listCost needs to have the same number of items as inventory (assuming Id is unique) possibly more, to stand a chance of returning true
<pre><code>checkItemInventory.All(x =&gt; listCost.Contains(x.Id))
</code></pre>
&quot;At least one item in the inventory has an id that is also in listCost&quot;. Listcost could minimally have only one id in it, to stand a chance of returning true
<pre><code>checkItemInventory.Any(x =&gt; listCost.Contains(x.Id))
</code></pre>
As you can see, neither of these are what you want as you seem to be saying you want to check whether every item in listcost is also present in the inventory. This is like the top code, but the other way round (&quot;all items in listCost are present in inventory&quot; vs &quot;all items in inventory are present in listcost&quot;
I think I'd make a dictionary out of the inventory first, unless it's already something that supports a fast lookup:
<pre><code>var d = checkItemInventory.Select(x =&gt; new { x.Id, x.Id }).ToDictionary();
var boolResult = listCost.All(lc =&gt; d.ContainsKey(lc));
</code></pre>
If inventory is small, you could use this approach:
<pre><code>listCost.All(lc =&gt; checkItemInventory.Any(cii =&gt; cii.Id == lc));
</code></pre>
Just be mindful that internally it might do something like:
<pre><code>bool all = true;
foreach(lc in listCost){
 bool found = false;
 foreach(cci in checkItemInventory)
 if(lc == cci.Id){
 found = true;
 break;
 }
 all &amp;= found;
 if(!all)
 return false;
}
return true;
</code></pre>
Which is a lot of repeated comparisons (for every item in listCost, the whole inventory is scanned), could be slow
<h3>Edit</h3>
I asked for clarification of how you store your inventory and your costs of building items. Here's one assumption I made, and how a solutio based on it might work:
Assuming your inventory has the kind of item and a count saying how many of that item the player is carrying:
<pre><code>class InventoryItem{
 int ItemKindId { get; set;}
 int CountOf { get; set; }
}

player.Inventory.Add(new InventoryItem() { 
 ItemKindId = Constants.WOOD, //1
 CountOf = 10 //holding 10 items of wood
};
player.Inventory.Add(new InventoryItem() { 
 ItemKindId = Constants.STONE, //2
 CountOf = 5 //holding 5 items of stone
};
</code></pre>
Assuming you have a Recipe for making e.g. an axe, it needs 1 wood and 2 stone, but it lists them in simple order:
<pre><code>int[] axeRecipe = new int[] { Constants.WOOD, Constants.STONE, Constants.STONE };
</code></pre>
Might be easiest to group the recipe:
<pre><code>var recipe = axeRecipe.GroupBy(item =&gt; item)

/*
 now we have a grouping of the recipe[item].Key as the material and a
 recipe[item].Count() of how much. The group is like a dictionary:

 recipe[Constants.WOOD] = new List&lt;int&gt;{ Constants.WOOD };
 recipe[Constants.STONE] = new List&lt;int&gt;{ Constants.STONE, Constants.STONE, };
 A group item has a Key and a list of objects that have that key
 Because my recipe was simply ints, the Key is the same number as all the 
 items in the list
*/

//for all items in the recipe
grp.All(groupItem =&gt;
 //does the player inventory contain any item
 playerInventory.Any(inventoryItem =&gt; 
 //where the material kind is the same as the recipe key (material)
 inventoryItem.ItemKindId == groupItem.Key &amp;&amp;
 //and the count they have of it, is enough to make the recipe
 inventoryItem.CountOf &gt;= groupItem.Count()
);
</code></pre>
You can of course reduce this to a single line if you want: <code>axeRecipe.GroupBy(...).All(...)</code>

blocks|key|3984277|text|您可以将listCost映射到int列表，然后使用Except()和Any()检查是否包含所有项：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3984278|bool+containsAll+=+!listCost.Select(x+=>+x.Id).Except(checkItemInventory).Any();|code-block|syntax|javascript|3984279|entityMap^0|4|8|F|3|P|8|Y|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

You could map the <code>listCost</code> to a list of <code>int</code> and then use <code>Except()</code> and <code>Any()</code> to check whether all items are contained:

<pre><code>bool containsAll = !listCost.Select(x =&gt; x.Id).Except(checkItemInventory).Any();
</code></pre>

blocks|key|3984322|text|您可以使用Join创建基于方法的Linq查询，并使用结果检查列表的长度是否大于0。然后将其转换为布尔值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3984323|var+query+=+checkItemInventory.Join(listCost,
++++++++++++++++++++++++++++++++++++inventory+=>+inventory.Id,
++++++++++++++++++++++++++++++++++++cost+=>+cost.Id,
++++++++++++++++++++++++++++++++++++(inventory,+cost)+=>+new+{+id+=+inventory.Id+});

var+count+=+query.ToList().Count();
var+b+=+(count+>+0);|code-block|syntax|javascript|3984324|entityMap^0|5|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You could use a <code>Join</code> to create a method based Linq query and use the results to check if the length of the list is greater than 0. Then turn that into a boolean.

<pre><code>var query = checkItemInventory.Join(listCost,
 inventory =&gt; inventory.Id,
 cost =&gt; cost.Id,
 (inventory, cost) =&gt; new { id = inventory.Id });

var count = query.ToList().Count();
var b = (count &gt; 0);
</code></pre>

How to convert a query to bool?
I used the "ALL (x => x)" but did not give the answer I needed.

Code Line

<pre><code>checkItemInventory.Where(x =&gt; listCost.Contains(x.Id));
</code></pre>

In this case, the <code>listcost</code> would have <code>2</code> items, I needed to check if the <code>checkItemInventory</code> has these <code>2</code> items.

How can I convert this linq to bool

如何将查询转换为bool？我使用了"ALL (x => x)“，但是没有给出我需要的答案。代码行checkItemInventory.Where(x => listCost.Contains(x.Id));在本例中，listcost将包含2项，我需要检查checkItemInventory是否包含这些2项。

问如何将此linq转换为bool
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