j=0
for i=1 to length(字符串)
  if substr(字符串,i,1)=指定的字符
    j=j+1
  endif基本上就是这种格式

解决方案 »

  1.   

    09:46:37 SQL> select length('abcabcaaadeca')-length(
    09:46:58   2  replace('abcabcaaadeca','a','')) from dual;LENGTH('ABCABCAAADECA')-LENGTH(REPLACE('ABCABCAAADECA','A',''))
    ---------------------------------------------------------------
                                                                  6已用时间:  00: 00: 00.16
    09:47:09 SQL>