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全体に10桁のUNIX日付スタンプが散らばっている大きなテキスト文字列があります。これらの10桁の数字をテキスト文字列で検索し、通常の日付形式に置き換えるクエリを作成しようとしています。私は次のようにconvertステートメントを持っています：

TRUNC(DATE '1970-01-01' + [timestamp]/86400)、値を入力すると完全に機能します。

例：デュアルからTRUNC（DATE '1970-01-01' + 1022089483/86400）を選択します。=2002年5月22日

しかし、私は見つけて交換するための適切な方法を見つけるのに苦労しています。また、正規表現は使用できません。だから、これが私の理論上のSQLです：

replace([column],'[sql to find 10-digit number]'
          ,TRUNC(DATE '1970-01-01' + [10-digit number]/86400))

サンプルテキストは次のとおりです。

1022089483 blah blah blah blah blah 1022094450 blah blah blah blah blah blah 1022095218 blah blah blah blah

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1 に答える 1

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楽しみのために、これを難しい方法で試してみようと思いました。を使用して 10 桁の数字の先頭を検索する 1 つの関数とinstr、それを繰り返し呼び出して、見つかったものを書式設定された日付に置き換える関数。これが賢明なアプローチであるか、または何らかの形で効率的であるかはまったくわかりません...

create or replace function epoch_offset(p_value in varchar2, p_start number)
return number is
    l_value varchar2(4000);
    offset number := 0;
    prev_offset number := 0;
    digit_count number := 0;
    epoch_start number := 0;
    pos number := p_start;
begin
    -- replace all digits with a single one, to make searching with instr
    -- simpler
    l_value := translate(p_value, '1234567890', '9999999999');

    while true loop
        -- find the next digit, starting as pos; first time through, pos
        -- will be the p_start we were given, then it tracks where we have
        -- got to
        offset := instr(l_value, '9', pos);

        if offset = 0 then
            -- we didn't find a digit, check if we already had a 10-digit
            -- number and have just reached the end
            if digit_count = 10 then -- and pos > length(p_value) then
                -- original value ends with a timestamp; so we have a 10-digit
                -- number
                exit;
            else
                -- no more digits, and last set we saw was short than 10; so
                -- l_value does not contain any 10-digit numbers (at least,
                -- after p_start)
                epoch_start := 0;
                exit;
            end if;
        end if;

        if prev_offset > 0 and offset != prev_offset + 1 then
            -- we've found a digit, but there's a gap since the last one
            if digit_count = 10 then
                -- the gap denotes the end of a 10-digit number, which is
                -- what we're looking for
                exit;
            end if;

            -- we've potentially started a new 10-digit number, so reset
            epoch_start := offset;
            digit_count := 0;
            prev_offset := 0;
        else
            -- we've found a sequential digit
            prev_offset := offset;
        end if;

        -- mark where we are
        if digit_count = 0 then
            -- start of a potential digit-sequence, make a note
            epoch_start := offset;
        end if;
        digit_count := digit_count + 1;
        pos := offset + 1;
    end loop;

    return epoch_start;
end epoch_offset;
/

create or replace function epoch_replace(p_value in varchar2,
    p_start in number default 1)
return varchar2 as
    l_pos number;
    l_time number;
    l_value varchar2(4000);
begin
    -- for this iteration, find the start of a 10-digit number, starting
    -- from p_start (1 on first iteration, by default)
    l_pos := epoch_offset(p_value, p_start);
    if l_pos > 0 then
        -- found a 10-digit number; call this recursively before modifying -
        -- this means we'll replace numbers with dates working from the end,
        -- so the positions don't need to be adjusted for the difference
        -- between the number and date lengths
        l_value := epoch_replace(p_value, l_pos + 10);
        -- get the 10-digit number...
        l_time := to_number(substr(l_value, l_pos, 10));
        -- ... and convert it to a date, with the rest of the original value
        -- around it
        return substr(l_value, 1, l_pos - 1)
            || to_char(trunc(DATE '1970-01-01' + l_time/86400), 'DD-mon-RR')
            || substr(l_value, l_pos + 10);
    else
        -- didn't find a 10-digit number, so return what we started with
        return p_value;
    end if;
end epoch_replace;
/

それがつまずく他のエッジケースがあるかもしれませんが、いくつかの明白なものでそれを試みました:

with tmp_tab as (
    select '1022089483 blah blah blah blah blah 1022094450 blah blah blah blah blah blah 1022095218 blah blah blah blah' as value from dual
    union all
    select 'blah 1022089483 blah 1022094450' from dual
    union all
    select 'blah 1022089483 98765 1022094450 1234' from dual
    union all
    select 'blah 1022089483 98765 1022094450 1022095218 1234 123456789 12345678901 123' from dual
)
select epoch_replace(value) from tmp_tab;

EPOCH_REPLACE(VALUE)
------------------------------------------------------------------------------------------------------------------------
22-may-02 blah blah blah blah blah 22-may-02 blah blah blah blah blah blah 22-may-02 blah blah blah blah
blah 22-may-02 blah 22-may-02
blah 22-may-02 98765 22-may-02 1234
blah 22-may-02 98765 22-may-02 22-may-02 1234 123456789 12345678901 123
于 2012-08-10T17:53:12.583 に答える