Question

有没有办法直接从预先格式化的字符串创建一个nspredicate而不调用predicateWithFormat？最后的字符串看起来像：

(inpatient=1) AND (dischargedate!=<null>) AND ((attending=SMITH) OR (admitting=SMITH) OR (consulting contains[cd] SMITH) OR (attending=JONES) OR (admitting=JONES) OR (consulting contains[cd] JONES))

    NSMutableString *preds = [[NSMutableString alloc] initWithString:@""];

    NSArray *provs = [self.providerCode componentsSeparatedByString:@"|"];
    for (NSString *prov in provs) {
        [preds appendFormat:@" (attending=%@) OR (admitting=%@) OR (consulting contains[cd] %@) ", prov, prov, prov];
    }
    NSString *final = [NSString stringWithFormat:@"(inpatient=%@) AND (dischargedate!=%@) AND (%@)", [NSNumber numberWithBool: self.inpatients], [NSNull null], preds];

    [fetchRequest setPredicate:[NSPredicate predicateWithFormat:final]];

Answer 1

是的，你可以，但你需要稍微修改格式字符串。

而不是：

[preds appendFormat:@" (attending = %@)", prov];

你需要这样做：

[preds appendFormat:@" (attending = '%@')", prov];

请注意%@修饰符周围使用单引号。这就是谓词知道它是一个恒定值的方式。

但是，即使你走这条路线，你仍然会使用你想要避免的predicateWithFormat:。您也可能在格式字符串中使用NSNull时遇到问题。

我建议做更像这样的事情：

NSArray *provs = [self.providerCode componentsSeparatedByString:@"|"];

NSMutableArray *providerPredicates = [NSMutableArray array];
NSPredicate *template = [NSPredicate predicateWithFormat:@"attending = $prov OR admitting = $prov OR consulting CONTAINS[cd] $prov"];
for (NSString *prov in provs) {
  NSDictionary *substitutions = [NSDictionary dictionaryWithObject:prov forKey:@"prov"];
  NSPredicate *p = [template predicateWithSubstitutionVariables:substitutions];
  [providerPredicates addObject:p];
}

NSPredicate *final = [NSPredicate predicateWithFormat:@"inpatient = 1 AND dischargedate != nil"];

if ([providerPredicates count] > 0) {
  NSPredicate *providers = nil;
  if ([providerPredicates count] > 1) {
    providers = [NSCompoundPredicate orPredicateWithSubpredicates:providerPredicates];
  } else {
    providers = [providerPredicates objectAtIndex:0];
  }

  final = [NSCompoundPredicate andPredicateWithSubpredicates:[NSArray arrayWithObjects:final, providers, nil]];
}

这是使用几个不同的整洁的东西：

谓词变量。您解析格式字符串@"attending = $prov OR admitting = $prov OR consulting CONTAINS[cd] $prov"一次，然后每次有不同的提供商时，只需用新值代替$prov
构造复合谓词。您可以在NSCompoundPredicate上使用某些类方法将多个谓词转换为单个分组OR或AND谓词。

使用预先构建的字符串创建谓词

1 个答案: