blocks|key|2355333|text|int+counter+=+0;
if+(condition1)+counter%2B%2B;
if+(condition2)+counter%2B%2B;
if+(condition3)+counter%2B%2B;
if+(condition4)+counter%2B%2B;
if+(condition5)+counter%2B%2B;
if+(condition6)+counter%2B%2B;
if+(counter+>=+3)+{
++++//+something
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2355334|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>int counter = 0;
if (condition1) counter++;
if (condition2) counter++;
if (condition3) counter++;
if (condition4) counter++;
if (condition5) counter++;
if (condition6) counter++;
if (counter &gt;= 3) {
 // something
}
</code></pre>

blocks|key|5707478|text|你可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5707479|int+validated=0;

if(condition1){
++++validated%2B%2B;
}
if(condition2){
++++validated%2B%2B;
}
if(condition3){
++++validated%2B%2B;
}
if(condition4){
++++validated%2B%2B;
}
if(condition5){
++++validated%2B%2B;
}
if(condition6){
++++validated%2B%2B;
}
if(validated>=3){
++++++//do+your+stuffs
+}|code-block|syntax|javascript|5707480|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can do it as :

<pre><code>int validated=0;

if(condition1){
 validated++;
}
if(condition2){
 validated++;
}
if(condition3){
 validated++;
}
if(condition4){
 validated++;
}
if(condition5){
 validated++;
}
if(condition6){
 validated++;
}
if(validated&gt;=3){
 //do your stuffs
 }
</code></pre>

blocks|key|232067|text|您可以尝试计算正确的条件：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|232068|伪代码：|232069|int+counter+=+0;
if(A)+counter%2B%2B;
if(B)+counter%2B%2B;
if(C)+counter%2B%2B;
if(D)+counter%2B%2B;
if(E)+counter%2B%2B;
if(F)+counter%2B%2B;

if(counter+>=+3){
//do+stuff+here
}
counter+=+0;|code-block|syntax|javascript|232070|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You could try counting the correct conditions:

pseudo-code:

<pre><code>int counter = 0;
if(A) counter++;
if(B) counter++;
if(C) counter++;
if(D) counter++;
if(E) counter++;
if(F) counter++;

if(counter &gt;= 3){
//do stuff here
}
counter = 0;
</code></pre>

blocks|key|5707534|text|除了已经给出的答案之外，对于数量可变的条件，这里有一个更灵活的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5707535|int+conditions[6]+=+{condition1,+condition2,+condition3,+
+++++++++++++++++++++condition4,+condition5,+condition6};
int+counter+=+0;
for+(int+i+=+0;+i+<+sizeof(conditions)/sizeof(int);+i%2B%2B)+{
++++counter+%2B=+conditions[i];++//+Assuming+your+conditions+are+0+or+1.
}
if+(counter+>=+3)+{
++++//+do+something
}|code-block|syntax|javascript|5707536|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In addition to the answers already given, here is a more flexible solution for a variable number of conditions:

<pre><code>int conditions[6] = {condition1, condition2, condition3, 
 condition4, condition5, condition6};
int counter = 0;
for (int i = 0; i &lt; sizeof(conditions)/sizeof(int); i++) {
 counter += conditions[i]; // Assuming your conditions are 0 or 1.
}
if (counter &gt;= 3) {
 // do something
}
</code></pre>

blocks|key|232163|text|+@interface+Conditions
+@property+(nonatomic,+strong)+NSMutableArray+*conditions;

+-+(void)+addCondition:+(Condition*)+theCondition;
+-+(NSInteger)+count;
+-+(NSInteger)+satisfying:+(+void+(%5Eblock)(Condition*)++);
+@end|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|232164|将您的条件列表包装在一个对象中。当您需要知道是否满足以下三个条件时：|unstyled|232165|+if+([self.conditions+satisfying:+%5E(Condition+*c){+return+[c+isSatisfied];+})>3)+{...};|232166|如果这是一个一次性项目，那么这就是夸大其词--也许是荒谬的夸大其词。但如果长期维护是一个问题，这将使条件与实现的细节解耦。你可以避免长长的条件句列表。您可以轻松地添加或更改条件，如果它们的逻辑变得复杂，您就有一个很好的面向对象的接口来处理它。(条件可能是作为外观运行的协议，而不是对象)。|232167|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code> @interface Conditions
 @property (nonatomic, strong) NSMutableArray *conditions;

 - (void) addCondition: (Condition*) theCondition;
 - (NSInteger) count;
 - (NSInteger) satisfying: ( void (^block)(Condition*) );
 @end
</code></pre>

Wrap your list of conditions in an object. When you need to know if three conditions are satisfied:

<pre><code> if ([self.conditions satisfying: ^(Condition *c){ return [c isSatisfied]; })&gt;3) {...};
</code></pre>

This is overkill -- perhaps ludicrous overkill -- if this is a disposable project. But if long-term maintenance is an issue, this decouples the condition from details of implementation. You avoid the long list of conditionals. You can add or change conditions easily, and if their logical gets complicated, you have a nice object-oriented interface for coping with it. (Condition is probably an protocol operating as a Facade, not an object).

I have a small problem with the objective c and conditions. How can we show that on 6 conditions that I give at least 3 must be verified as correct?

Thank you for your answer!

Multiple conditions in objective C

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个关于目标c和条件的小问题。我们如何证明，在我给出的6个条件下，至少有3个条件必须被验证为正确？谢谢你的回答！

问objective C中的多个条件
EN

回答 5

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