ios NSPredicate - 无法解析格式字符串
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NSPredicate - Unable to parse the format string
提问by Satheeshwaran
I am trying to write a NSPredicateto fetch rows with my_column value with this string "193e00a75148b4006a451452c618ccec" and I get the below crash.
我正在尝试NSPredicate使用此字符串“ 193e00a75148b4006a451452c618ccec”编写带有 my_column 值的行来获取行，并且出现以下崩溃。
Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: 'Unable to parse the format string "my_column=193e00a75148b4006a451452c618ccec"'
由于未捕获的异常“NSInvalidArgumentException”而终止应用程序，原因：“无法解析格式字符串“my_column=193e00a75148b4006a451452c618ccec””
My predicate statement
我的谓词陈述
fetchRequest.predicate=[NSPredicate predicateWithFormat:[NSString stringWithFormat:@"%@==\"%@\"",attributeName,itemValue]];
also tried this
也试过这个
fetchRequest.predicate=[NSPredicate predicateWithFormat:[NSString stringWithFormat:@"%@ == %@",attributeName,itemValue]];
this
这个
fetchRequest.predicate=[NSPredicate predicateWithFormat:[NSString stringWithFormat:@"%@ = %@",attributeName,itemValue]];
and this
和这个
fetchRequest.predicate=[NSPredicate predicateWithFormat:[NSString stringWithFormat:@"%@=\"%@\"",attributeName,itemValue]];
Please help.
请帮忙。
I found out this, when I was trying with Martin R's answer
当我尝试使用 Martin R 的答案时，我发现了这一点
fetchRequest.predicate=[NSPredicate predicateWithFormat:@"%@==%@",attributeName,itemValue];
attributeName I pass comes with a ' ' so I took off attributeName and hardcoded it, then it works fine.
我传递的 attributeName 带有一个“ ”，所以我去掉了 attributeName 并对其进行了硬编码，然后它就可以正常工作了。
回答by Martin R
Enclose the string in single quotes:
将字符串括在单引号中：
[NSPredicate predicateWithFormat:@"my_column = '193e00a75148b4006a451452c618ccec'"]
or better, use argument substitution:
或者更好，使用参数替换：
[NSPredicate predicateWithFormat:@"my_column = %@", @"193e00a75148b4006a451452c618ccec"]
The second method avoids problems if the search string contains special characters
such as 'or ".
如果搜索字符串包含特殊字符（例如'或 ），则第二种方法可避免出现问题"。
Remark:Never use stringWithFormatwhen building predicates. stringWithFormatand
predicateWithFormathandle the %Kand %@format differently, so combining these
two methods is very error-prone (and unnecessary).
备注：stringWithFormat在构建谓词时切勿使用。stringWithFormat并且
predicateWithFormat以不同的方式处理%K和%@格式，因此将这两种方法结合起来非常容易出错（并且没有必要）。
回答by Thilina Chamin Hewagama
I think you have missed '' when checking for equality,
我认为您在检查相等性时错过了 ''，
now try,
现在试试，
[NSPredicate predicateWithFormat:@"my_column='193e00a75148b4006a451452c618ccec'"];
回答by Anupdas
Try
尝试
//Case Insensitive
fetchRequest.predicate=[NSPredicate predicateWithFormat:@"%K = [cd] %@",attributeName,itemValue];
//Case sensitive
fetchRequest.predicate=[NSPredicate predicateWithFormat:@"%K = %@",attributeName,itemValue];
