如何比较两个Java字符串而不必担心Parse查询中的任何一个的大小写？

Question

I'm making a program where a user searches for items based on item name or description. I'm using query.whereContains(key, value) since I want it to check if either contains the keyword they're looking for. My question is how can I compare the keyword(the value) to the item name or description but regardless of case. I know Java's equalsignoreCase() exists, but that returns a boolean which doesn't match the whereContains() method's paramters.

Example of what I mean:

If the user searches for "eXAmplE", which will be the keyword, and a description of an item in the cloud is "this is an ExAMplE" exists, the query should include that. How would I go about implementing this?

Answer 1

by normalizing the case of all Strings involved before comparing them.

something like this:

String original = "SoMeCrazyCASE";
String lowerCased = original.toLowerCase(Locale.ROOT);
//now work with the lowercased one

note - its important to lower case in the root locale to avoid issues with the current language on the machine youre running on

Answer 2

original.equalsIgnoreCase(other)

这可能比执行s.toLowerCase（）更快，因为不需要创建额外的对象。

Answer 3

Try the following:

if(myDescription.toLowerCase().contains((keyword.toLowerCase())){
    //do stuff here
}

More complex queries may need you to implement the search using patterns.

Hope it helps.

Answer 4

Try this

 String str1 = "hello";
    String str2 = "HELLO";
    if(str1.equalsIgnoreCase(str2)){
        //YOUR LOGIc
    }

Answer 5

Try this.

String str1="HAI";
String str2="hai";
if(str1.equalsIgnoreCase(str2))
{
    //Use your code
}