若串S1=“ABCDEFG”,S2=“9898”,执行s1.substring(0,s2.length()),其结果为()
A: "ABCD"
B: "BCDE"
C: "ABC"
D: "DEFG"
A: "ABCD"
B: "BCDE"
C: "ABC"
D: "DEFG"
举一反三
- 若串S1=‘ABCDEFG’, S2=‘9898’ ,S3=‘###’,S4=‘012345’,执行concat(replace(S1,substr(S1,4,3),S3),substr(S4,index(S2,‘8’),length(S2)))其结果为
- 若串 S1= ‘ ABCDEFG ’ , S2= ‘ 9898 ’ ,S3= ‘...,length(S2))) 其结果为()
- 若串S1=‘ABCDEFG’,S2=‘9898’,S3=‘...,length(S2)))其结果为()
- 若串S1=‘ABCDEFG’, S2=‘9898’ ,S3=‘###’,S4=‘...8’),length(S2)))其结果为
- 设字符串S1=″ABCDEFG″,S2=″PQRST″,则运算 S=CONCAT(SUBSTR(S1,2,LENGTH(S2)),SUBSTR(S1,LENGTH(S2),2)) 后S的结果为() A: ″BCQR″ B: ″BCDEF″ C: ″BCDEFG″ D: ″BCDEFEF″