blocks|key|424933|text|谓词的结构是A+&&+B+&&+(C+%7C%7C+D)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|424934|设置你的谓词|424935|NSPredicate+*aPredicate+=+[NSPredicate+predicateWithFormat:@"stateCode+==+%25d",+value];
NSPredicate+*bPredicate+=+[NSPredicate+predicateWithFormat:@"invoiceDate+>=++%25@",+givenDate];|code-block|syntax|javascript|424936|cPredicate和dPredicate也是如此。然后先用OR将c和d组合起来|424937|NSArray+*cdPredicateArray+=+@[cPredicate,+dPredicate];
NSPredicate+*cdPredicate+=+[NSCompoundPredicate+orPredicateWithSubpredicates:cdPredicateArray];|424938|然后所有的数据都带有和|424939|NSArray+*allPredicateArray+=+@[aPredicate,+bPredicate,+cdPredicate];
NSPredicate+*allPredicate+=+[NSCompoundPredicate+andPredicateWithSubpredicates:allPredicateArray];|424940|如果我误解了您问题，而您的结构是A+&&+B+&&+C+%7C%7C+D，那么您必须首先将A、B和C组合(与AND)，然后将结果与D(与OR)组合。|424941|entityMap^0|6|I|0|0|0|0|0|0|0|G|G|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|13|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|15|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|X|$]]

The structure of your predicate is <code>A &amp;&amp; B &amp;&amp; (C || D)</code>

Setup your predicates

<pre><code>NSPredicate *aPredicate = [NSPredicate predicateWithFormat:@"stateCode == %d", value];
NSPredicate *bPredicate = [NSPredicate predicateWithFormat:@"invoiceDate &gt;= %@", givenDate];
</code></pre>

Similar do the cPredicate and dPredicate. Then first combine c and d with OR

<pre><code>NSArray *cdPredicateArray = @[cPredicate, dPredicate];
NSPredicate *cdPredicate = [NSCompoundPredicate orPredicateWithSubpredicates:cdPredicateArray];
</code></pre>

and then all of them with AND

<pre><code>NSArray *allPredicateArray = @[aPredicate, bPredicate, cdPredicate];
NSPredicate *allPredicate = [NSCompoundPredicate andPredicateWithSubpredicates:allPredicateArray];
</code></pre>

If I misunderstood your question and your structure is <code>A &amp;&amp; B &amp;&amp; C || D</code> Then you have to combine A, B and C first (with AND) and then combine that result with D (with OR).

blocks|key|1484013|text|opportunitylines是一种多对多关系，因此一个DBOpportunity对象有多个DBOpportunityLines对象。假设最后一个条件|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1484014|1484015|+DBOpportunityLines.crmAccept+==+1或DBOpportunityLines.crmAccept+==+3|blockquote|1484016|1484017|1484018|对于任何相关对象，您都需要一个SUBQUERY：|1484019|NSDate+*givenDate+=+...;
NSPredicate+*predicate+=+[NSPredicate+predicateWithFormat:@"stateCode+==+1+AND+invoiceDate+>=+%25@+"
++++"AND+SUBQUERY(opportunitylines,+$x,+$x.crmAccept+==+1+OR+$x.crmAccept+==+3).@count+>+0",
++++givenDate];|code-block|syntax|javascript|1484020|备注:不幸的是，SUBQUERY在谓词中的用法很少有文档记录。在NSExpression类引用中有one+example。另请参见Quick+Explanation+of+SUBQUERY+in+NSPredicate+Expression。|1484021|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.apple.com/documentation/foundation/nsexpression/1411651-init|1|https://stackoverflow.com/questions/3810992/quick-explanation-of-subquery-in-nspredicate-expression^0|0|G|0|0|0|0|0|0|0|1D|B|0|1T|1J|1|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|17|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|18|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|19|8|@]|D|@]|E|$]]|$1|K|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|1B|8|@]|D|@]|E|$]]|$1|N|3|O|5|P|7|1C|8|@]|D|@]|E|$Q|R]]|$1|S|3|T|5|6|7|1D|8|@]|D|@$9|1E|A|1F|1|1G]|$9|1H|A|1I|1|1J]]|E|$]]|$1|U|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]|12|$5|X|Y|Z|E|$10|13]]]]

<code>opportunitylines</code> is a to-many relationship, so there are multiple DBOpportunityLines objects for one DBOpportunity object. Assuming that the last condition
<blockquote>
DBOpportunityLines.crmAccept == 1 OR DBOpportunityLines.crmAccept == 3
</blockquote>
should hold for any of the related objects, you need a SUBQUERY:
<pre><code>NSDate *givenDate = ...;
NSPredicate *predicate = [NSPredicate predicateWithFormat:@&quot;stateCode == 1 AND invoiceDate &gt;= %@ &quot;
 &quot;AND SUBQUERY(opportunitylines, $x, $x.crmAccept == 1 OR $x.crmAccept == 3).@count &gt; 0&quot;,
 givenDate];
</code></pre>
Remark: Unfortunately, the usage of SUBQUERY in predicates is poorly documented. There is <a href="https://developer.apple.com/documentation/foundation/nsexpression/1411651-init" rel="nofollow noreferrer">one example</a> in the NSExpression class reference. See also <a href="https://stackoverflow.com/questions/3810992/quick-explanation-of-subquery-in-nspredicate-expression">Quick Explanation of SUBQUERY in NSPredicate Expression</a>.

blocks|key|1484072|text|您还可以获取opportunityLines，然后获取父实体，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1484073|NSFetchRequest+*fetchRequest+=+[[NSFetchRequest+alloc]+init];
NSEntityDescription+*entity+=+[NSEntityDescription+entityForName:@"opportunityLines"+inManagedObjectContext:context];
[fetchRequest+setEntity:entity];

NSDate+*yourDate+=+[NSDate+date];
NSPredicate+*predicate+=+[NSPredicate+predicateWithFormat:@"(crmAccept==1+OR+crmaccept==3)+AND+opportunity.stateCode==1+AND+opportunity.invoiceDate>=%25@",+yourDate];
[fetchRequest+setPredicate:predicate];

NSError+*error;

//So+here+you+have+your+array+of+opportunitylines
NSArray+*opportunityLines+=+[context+executeFetchRequest:fetchRequest+error:&error];

//And+this+is+how+you+get+your+opportunity+objects
NSArray+*opportunities+=+[opportunityLines+valueForKeyPath:@"@distinctUnionOfObjects.opportunity"];|code-block|syntax|javascript|1484074|entityMap^0|6|G|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 also fetch your <code>opportunityLines</code> and then get the parent entities like this:

<pre><code>NSFetchRequest *fetchRequest = [[NSFetchRequest alloc] init];
NSEntityDescription *entity = [NSEntityDescription entityForName:@"opportunityLines" inManagedObjectContext:context];
[fetchRequest setEntity:entity];

NSDate *yourDate = [NSDate date];
NSPredicate *predicate = [NSPredicate predicateWithFormat:@"(crmAccept==1 OR crmaccept==3) AND opportunity.stateCode==1 AND opportunity.invoiceDate&gt;=%@", yourDate];
[fetchRequest setPredicate:predicate];

NSError *error;

//So here you have your array of opportunitylines
NSArray *opportunityLines = [context executeFetchRequest:fetchRequest error:&amp;error];

//And this is how you get your opportunity objects
NSArray *opportunities = [opportunityLines valueForKeyPath:@"@distinctUnionOfObjects.opportunity"];
</code></pre>

I have following CoreData objects Model 

<img src="https://i.stack.imgur.com/lPXP2.png" alt="enter image description here">

Now I am having issue in making a predicate with following conditions.

Fetch all those <code>DBOpportunity</code> WHERE

DBOpportunity.stateCode == 1

AND

DBOpportunity.invoiceDate >= GIVEN_DATE

AND

DBOpportunityLines.crmAccept == 1 OR DBOpportunityLines.crmAccept == 3

I have tried lots of examples and programming guide by the <a href="http://developer.apple.com/library/ios/#documentation/Cocoa/Conceptual/Predicates/Articles/pUsing.html#//apple_ref/doc/uid/TP40001794-SW3" rel="nofollow noreferrer">apple</a> but can't able to achieve this.

CoreData NSPredicate with relationship

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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我有以下CoreData对象模型​​现在，我在使用以下条件进行谓词时遇到了问题。获取所有位于以下位置的DBOpportunityDBOpportunity.stateCode == 1和DBOpportunity.invoiceDate >= GIVEN_DATE和DBOpportunityLines.crmAccep...

问带关系的CoreData NSPredicate
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