Java 如何解析并获取出现在值端的键数?计数应该是7
这里的名称是键,Java 如何解析并获取出现在值端的键数?计数应该是7,java,Java,这里的名称是键,{后面的右边是一个值。如何解析和获取值端的“键”计数数 {name={doc_count_error_upper_bound=0.0, sum_other_doc_count=0.0, buckets=[{key=x, doc_count=165324.0}, {key=y, doc_count=100875.0}, {key=z, doc_count=99981.0}, {key=A, doc_count=76631.0}, {key=B, doc_count=68793.0}
{
后面的右边是一个值。如何解析和获取值端的“键”计数数
{name={doc_count_error_upper_bound=0.0, sum_other_doc_count=0.0,
buckets=[{key=x, doc_count=165324.0}, {key=y, doc_count=100875.0},
{key=z, doc_count=99981.0}, {key=A, doc_count=76631.0}, {key=B,
doc_count=68793.0}, {key=C, doc_count=50717.0}, {key=D,
doc_count=50034.0}]}
此代码适用于您:
SEVERE: Servlet.service() for servlet [spring-web] in context with path [/Reports] threw exception [Request processing failed; nested
exception is java.lang.ClassCastException:
com.google.gson.internal.LinkedTreeMap cannot be cast to
java.lang.String] with root cause
java.lang.ClassCastException: com.google.gson.internal.LinkedTreeMap cannot be cast to
java.lang.String
publicstaticvoidmain(字符串[]args){
字符串s=“{name={doc\u count\u error\u upper\u bound=0.0,sum\u other\u doc\u count=0.0,bucket=[{key=x,doc\u count=165324.0},{key=y,doc\u count=100875.0},{key=z,doc\u count=99981.0},{key=A,doc count 76631.0},{key=B,doc count 68793.0},{key=C,doc key=507.0},{;
int i=0;
整数计数=0;
而(i
O/p:7String value=pair.getValue();String arrr[]=value.split(“,”);System.out.println(value);int senderValue=0;for(int i=0;i
键的次数
吗?另外,请编辑问题并在那里添加代码。是的,只想在值侧添加键计数,因此计数应该是7?
SEVERE: Servlet.service() for servlet [spring-web] in context with path [/Reports] threw exception [Request processing failed; nested
exception is java.lang.ClassCastException:
com.google.gson.internal.LinkedTreeMap cannot be cast to
java.lang.String] with root cause
java.lang.ClassCastException: com.google.gson.internal.LinkedTreeMap cannot be cast to
java.lang.String
public static void main(String[] args) {
String s = "{name={doc_count_error_upper_bound=0.0, sum_other_doc_count=0.0, buckets=[{key=x, doc_count=165324.0}, {key=y, doc_count=100875.0}, {key=z, doc_count=99981.0}, {key=A, doc_count=76631.0}, {key=B, doc_count=68793.0}, {key=C, doc_count=50717.0}, {key=D, doc_count=50034.0}]}";
int i = 0;
int count = 0;
while (i < s.length()) {
i = s.indexOf("key", i);
if (i != -1) {
count++;
i++;
}
else {
break;
}
}
System.out.println(count);
}