How to use `string.startsWith()` method ignoring the case?

Question

I want to use string.startsWith() method but ignoring the case.

Suppose I have String "Session" and I use startsWith on "sEsSi" then it should return true.

How can I achieve this?

There was no reason to close this question. Anyway, the best solution is to use `string.regionMatches(true, 0, prefix, 0, prefix.length());` which doesn't incur the cost of "normalizing" both strings. — isapir, Dec 06 '16 at 19:11
@AlexanderAbakumov Fine by me. The correct answer is posted in my comment above. It is much more efficient than changing the case of strings. It'd be good to upgrade it to an answer rather than a comment. — isapir, Oct 09 '18 at 21:39

score 119 · Accepted Answer · edited Oct 07 '16 at 14:46

119

Use toUpperCase() or toLowerCase() to standardise your string before testing it.

edited Oct 07 '16 at 14:46

Buhake Sindi

87,898
29
167
228

answered Oct 03 '13 at 08:19

Nemesis

2,750
2
22
20

15

The big problem here is the performance. For small Strings no problem, but if you have a big one... I mean... Do you really will do a toUpperCase in a 1mb String just to compare 4-10 inicial characters? – Dyorgio Apr 07 '16 at 16:52
7

Hum even if i can't see why i would like to compare the starting part of a 1mb String then if only 4 - 10 initial characters interest me i use substring and then normalize and done... – Nemesis Apr 09 '16 at 20:31
4

This answer (and all other answers found herein) are wrong. If you look at the implementation of `String.equalsIgnoreCase()` you will discover that you need to compare **both** lowercase and uppercase versions of the Strings before you can conclusively return `false`. See http://stackoverflow.com/a/38947571/14731 for an alternative answer. – Gili Aug 14 '16 at 23:37
Hum if you have uped your string and then test agains an upped prefix i can't see when this will fail. Ii seem's to me that's what 'standardise a string' means... I used it many times with no error in real life projects. – Nemesis Sep 12 '16 at 08:19
16

This answer is not true because toLowerCase() uses Locale.getDefault() internally to change the chars and that might lead to false negatives. Let's say you want to compare "İstanbul" with "istanbul". In Turkish "İ" s lower case equivalent is "ı". So when I try to do that `"İstanbul".toLowerCase().startsWith("istanbul".toLowerCase())` (in a Locale.Turkish environment), it will give me false. – Mustafa Berkay Mutlu Jan 29 '18 at 11:24
3

@Nemesis, This answer is wrong, having it as the accepted answer could be misleading for a lot of people. Would you please edit your answer and link to Rohit Jain answer. – Moha the almighty camel Jul 28 '19 at 16:42
This is a wrong answer since the original value should not be changed for a conditional function. – A.G. Sep 01 '23 at 00:15

score 102 · Answer 2 · edited Jul 06 '22 at 07:53

102

~~One option is to convert both of them to either lowercase or uppercase:~~

"Session".toLowerCase().startsWith("sEsSi".toLowerCase());

This is wrong. See: https://stackoverflow.com/a/15518878/14731

Another option is to use String#regionMatches() method, which takes a boolean argument stating whether to do case-sensitive matching or not. You can use it like this:

String haystack = "Session";
String needle = "sEsSi";
System.out.println(haystack.regionMatches(true, 0, needle, 0, needle.length()));  // true

It checks whether the region of needle from index 0 till length 5 is present in haystack starting from index 0 till length 5 or not. The first argument is true, means it will do case-insensitive matching.

And if only you are a big fan of Regex, you can do something like this:

System.out.println(haystack.matches("(?i)" + Pattern.quote(needle) + ".*"));

(?i) embedded flag is for ignore case matching.

edited Jul 06 '22 at 07:53

Ingo Karkat

167,457
16
250
324

answered Oct 03 '13 at 08:18

Rohit Jain

209,639
45
409
525

1

Actually, suppose there is a table which is having more than 1 lakh entries , and I use a for loop on particular column to parse all those entries and in that, I try to use this lowercase option above to search an entry starting with particular pattern. So, this is causing a performance issue. If there is no match found at all, then internally it will parse through all the entries, which will take lot of time, as simultaneously its getting converted to lowercase as well. So is there any other option, to resolve performance issue? – Sheetal Bhatewara Oct 08 '13 at 12:47
@Rohit Jain, actually this: `System.out.println("Session".toLowerCase().startsWith("sEsSi".toLowerCase()));` is not a propper solution. You can try this: `System.out.println("SessIon".toLowerCase().startsWith("sEsSi".toLowerCase()));` with vm arguments `-Duser.country=TR -Duser.language=tr` In this example the `I` is uppercase `i` – DPM Mar 02 '18 at 10:06

score 20 · Answer 3 · answered Jul 23 '20 at 09:35

20

I know I'm late, but what about using StringUtils.startsWithIgnoreCase() from Apache Commons Lang 3 ?

Example :

StringUtils.startsWithIgnoreCase(string, "start");

Just add the following dependency to your pom.xml file (taking the hypothesis that you use Maven) :

<dependency>
    <groupId>org.apache.commons</groupId>
    <artifactId>commons-lang3</artifactId>
    <version>3.11</version>
</dependency>

answered Jul 23 '20 at 09:35

Sébastien Vandamme

575
7
24

Perfect -- thank you. This is the answer I was looking for. – Mass Dot Net Dec 13 '21 at 17:14

score 1 · Answer 4 · answered Oct 03 '13 at 08:18

1

myString.toLowerCase().startsWith(starting.toLowerCase());

answered Oct 03 '13 at 08:18

agad

2,192
1
20
32

score 1 · Answer 5 · answered Oct 03 '13 at 08:18

1

try this,

String session = "Session";
if(session.toLowerCase().startsWith("sEsSi".toLowerCase()))

answered Oct 03 '13 at 08:18

newuser

8,338
2
25
33

A.G. · Answer 6 · 2023-09-01T02:17:37.587

0

You can use String.search() function with regex that will give you the index of the first substring match of the search string. Then you can test the returned index to see if it is 0. 'i' at the end makes the search case insensitive.

const session = 'Session';
const startsWith = session.search(/sEsSi/i) === 0;

edited Sep 01 '23 at 02:17

answered Sep 01 '23 at 00:54

A.G.

304
2
11

score -1 · Answer 7 · answered Oct 03 '13 at 08:19

-1

You can use someString.toUpperCase().startsWith(someOtherString.toUpperCase())

answered Oct 03 '13 at 08:19

Prasad Kharkar

13,410
5
37
56

score -1 · Answer 8 · edited Aug 02 '23 at 06:24

-1

myString.StartsWith(anotherString, StringComparison.OrdinalIgnoreCase)

edited Aug 02 '23 at 06:24

Procrastinator

2,526
30
27
36

answered Jul 27 '23 at 23:21

TTTT

9
1

The question is about Java, not .NET/C#. – General Grievance Aug 02 '23 at 15:41

score -3 · Answer 9 · answered Oct 03 '13 at 08:18

-3

You can always do

"Session".toLowerCase().startsWith("sEsSi".toLowerCase());

answered Oct 03 '13 at 08:18

RamonBoza

8,898
6
36
48

score -3 · Answer 10 · answered Jul 28 '21 at 05:05

-3

StartsWith(String value, bool ignoreCase, CultureInfo? culture) e.g:

string test = "Session";
bool result = test.StartsWith("sEsSi", true, null);
Console.WriteLine(result);

point: in VS by right-clicking on StartsWith and then "pick definition" can see all overloading method

answered Jul 28 '21 at 05:05

pedram

705
7
6

2

The question is about java. – Kenster Sep 17 '21 at 15:12

How to use `string.startsWith()` method ignoring the case?

10 Answers10

Linked

Related