NSPredicate - 无法解析格式字符串 [英] NSPredicate - Unable to parse the format string

问题描述

我想写一个 NSPredicate 来获取my_column值为 193e00a75148b4006a451452c618ccec 的行得到以下崩溃。

由于未捕获异常而终止应用程序
'NSInvalidArgumentException'，原因：'无法解析格式
stringmy_column = 193e00a75148b4006a451452c618ccec'

我的谓词语句

  fetchRequest.predicate = [NSPredicate predicateWithFormat：[NSString stringWithFormat：@％@ == \％@ \，attributeName，itemValue]]; 
  

也尝试过

  fetchRequest.predicate = [NSPredicate predicateWithFormat：[NSString stringWithFormat：@％@ ==％@，attributeName，itemValue]]; 
  

此

  fetchRequest.predicate = [NSPredicate predicateWithFormat：[NSString stringWithFormat：@％@ =％@，attributeName，itemValue]]; 
  

和此

 code> fetchRequest.predicate = [NSPredicate predicateWithFormat：[NSString stringWithFormat：@％@ = \％@ \，attributeName，itemValue]]; 
  

请帮助。

我发现这个，当我尝试与Martin R的回答

  fetchRequest.predicate = [NSPredicate predicateWithFormat：@ ％@ ==％@，attributeName，itemValue]; 
  
 
 
 
  attributeName我传递一个''，所以我摘掉了attributeName并硬编码，然后
解决方案
将字符串用单引号括起来：
  [NSPredicate predicateWithFormat：@my_column ='193e00a75148b4006a451452c618ccec'] 
  
 
 b $ b 
或更好，使用参数替换：
  [NSPredicate predicateWithFormat：@ my_column =％@，@193e00a75148b4006a451452c618ccec] 
  
第二种方法避免了如果搜索字符串包含特殊字符
，例如'或。
 
 
 
 备注：在建立谓词时不要使用 stringWithFormat 。 stringWithFormat 和
  predicateWithFormat 处理％K 和％@  ，所以结合这些
两个方法是非常容易出错（和不必要的）。
 
I am trying to write a NSPredicate to fetch rows with my_column value with this string "193e00a75148b4006a451452c618ccec" and I get the below crash.

  
Terminating app due to uncaught exception
  'NSInvalidArgumentException', reason: 'Unable to parse the format
  string "my_column=193e00a75148b4006a451452c618ccec"'
My predicate statement
fetchRequest.predicate=[NSPredicate predicateWithFormat:[NSString stringWithFormat:@"%@==\"%@\"",attributeName,itemValue]];
also tried this
fetchRequest.predicate=[NSPredicate predicateWithFormat:[NSString stringWithFormat:@"%@ == %@",attributeName,itemValue]];
this
fetchRequest.predicate=[NSPredicate predicateWithFormat:[NSString stringWithFormat:@"%@ = %@",attributeName,itemValue]];
and this
fetchRequest.predicate=[NSPredicate predicateWithFormat:[NSString stringWithFormat:@"%@=\"%@\"",attributeName,itemValue]];
Please help.

I found out this, when I was trying with Martin R's answer
fetchRequest.predicate=[NSPredicate predicateWithFormat:@"%@==%@",attributeName,itemValue];
attributeName I pass comes with a ' ' so I took off attributeName and hardcoded it, then it works fine.
解决方案
Enclose the string in single quotes:
[NSPredicate predicateWithFormat:@"my_column = '193e00a75148b4006a451452c618ccec'"]
or better, use argument substitution:
[NSPredicate predicateWithFormat:@"my_column = %@", @"193e00a75148b4006a451452c618ccec"]
The second method avoids problems if the search string contains special characters
such as ' or ".

Remark: Never use stringWithFormat when building predicates. stringWithFormat and
predicateWithFormat handle the %K and %@ format differently, so combining these
two methods is very error-prone (and unnecessary).

                        
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