java - How can I compare the string with the keys in hash map -

-

i having difficulties compare string (str) keys in hashmap. can see below code, have multiple keys. comparing str keys couldn't values specific key. please note map contains coin.

it works if directly input string eg entry.getkey().getkeyone().tostring().contains("coin")

string str = "coin"; for(map.entry<pair, string> entry : map.entryset()) {     if(entry.getkey().getkeyone().tostring().contains(str)||         entry.getkey().getkeytwo().tostring().contains(str)) {             entry.getvalue();     } }  public class pair {     private string keyone;     private string keytwo;      pair(string one,string two) {         this.keyone=one;         this.keytwo=two;     }      public string getkeyone() {         return keyone;     }      public string getkeytwo() {        return keytwo;     } }

advice: long using own class pair in map must override equals() , should override hashcode().

you example string literal works, because "coin" in string pool , references same string.

runs here:

public class pair {      private string keyone;     private string keytwo;      pair(string one, string two) {         this.keyone = one;         this.keytwo = two;      }      public string getkeyone() {         return keyone;     }      public string getkeytwo() {         return keytwo;      }      public static void main(string[] args) {         map<pair, string> map = new hashmap();         map.put(new pair("coin", "5"), "u");         map.put(new pair("bill", "100"), "h");         map.put(new pair("10", "5coin"), "t");          string str = new string("coin");         (map.entry<pair, string> entry : map.entryset()) {             if (entry.getkey().getkeyone().tostring().contains(str)                     || entry.getkey().getkeytwo().tostring().contains(str)) {                 system.err.println(entry.getvalue());             }         }     } }

Comments

Post a Comment

Popular posts from this blog

asp.net - How to correctly use QUERY_STRING in ISAPI rewrite? -

-
i'm having difficult time , hope guys can help. using helicon isapi rewrite version 3.1.0.104 on our iis server. edit http.conf file , have tried , still fail. this trying do: redirect url: https://www.domain.com/switch-by-version?version=2.8.5.2594 to: https://test.domain.com/load/load.aspx?tver=2.8.5.2594 the version number @ end source url change, , need target url have same version number @ end of url example above shows. i tried following not work: rewritecond %{query_string} ^version=(\d\d?)\.(\d\d?)\.(\d\d?)\.(.*)$ [nc] rewriterule ^/switch-by-version(.*)$ https://test.domain.com/load/load.aspx?tver=%1 [r=307,nc,l] any appreciated! thanks. if it's running in equivalent context of .htaccess , rule shouldn't start / . here's simpler version: rewritecond %{query_string} ^version=(\d+\.\d+\.\d+(?:\..+)?)$ [nc] rewriterule ^switch-by-version/?$ https://test.domain.com/load/load.aspx?tver=%1 [r=307,nc,l]
Read more

jsf - "PropertyNotWritableException: Illegal Syntax for Set Operation" error when setting value in bean -

-
this question has answer here: how send form input values , invoke method in jsf bean 1 answer i error when trying send values jsf page bean. i want set values fields in bean, create airplane instance , add database. error telling me use illegal syntax setter part of bean. why error? didnt find on site helped me problem. here jsf part: <h4>add plane</h4> <c:if test="#{not empty listairplanebb.error}"> <div class="alert alert-danger"> <span class="glyphicon glyphicon-remove"></span> <h:outputtext value="#{listairplanebb.error}" /> </div> </c:if> <h:form> <table class="table"> <thead> ...
Read more

arrays - Algorithm to find ideal starting spot in a circle -

-
so problem have been debating such: your given array of integers representing circle. then, have pick spot in circle start. place start, compare value @ array number of steps took there , if steps less or equal number, include in final set. find place start have elements in set. ex, a=[0, 1, 2] if start @ index=0, then: a[0]=0 < =0 0 included a[1]=1 < =1 1 included a[2]=2 < =2 2 included final set: {0,1,2} if start @ index=1, then: a[1]=1 > 0 1 not included a[2]=2 > 1 2 not included here loop around a[0]=0 > 2 0 included final set: {0} if start @ index=2, then: a[2]=2 > 0 2 not included, here loop around a[0]=0 < = 1 0 included a[1]=1 < = 2 1 included final set: {0,1} so in trivial case, starting position index=0 best position results in final set elements. brute force method of finding obvious, trying find more efficient method. attempts far have been examine trying find max intersect of viable starting ranges calculated each element in arr...
Read more