update hr_user_qy0331_ok l
set l.login_name_new = substr(l.login_name,
1,
instr(l.login_name, '@') - 1) ||
to_char(to_date(l.从身份证号码截取出生日期, 'yyyy/mm/dd'), 'mmdd') ||
substr(l.login_name, instr(l.login_name, '@')) where l.从身份证号码截取出生日期 is not null
and l.同名否 = 'same' and l.login_name is not null
亲,沙发正空着,还不快来抢?
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