Question

function createWorker(main){
    var blob = new Blob(
        ["(" + main.toString() + ")(self)"],
        {type: "text/javascript"}
    );
    return new Worker(window.URL.createObjectURL(blob));
}

// Worker
var worker = createWorker(function(self){
    setInterval(function(){
        self.postMessage(Date.now());
    }, 250);
});
worker.onmessage = function(e) {
    console.log("Worker: " + e.data);
}

这样可以正常使用

t2 - 3百万 t4 - 3百万总共我有600万条记录，由于某些原因我的临时空间被填满所以我想要的是在插入每10000条记录后提交。

怎么办？

Answer 1

我会使用BULK COLLECT和FORALL。

请参阅Incremental Commit Processing with FORALL和Bulk Processing with BULK COLLECT and FORALL了解一些可以适应您案例的例子。

Answer 2

希望这段代码有所帮助。

--Test data creation
CREATE TABLE TEST_SO_BULK
AS
SELECT LEVEL COL1,'AVRAJIT'||LEVEL COL2 FROM DUAL
CONNECT BY LEVEL < 100000;


--Create another table to insert

CREATE TABLE TEST_SO1
AS
SELECT * FROM TEST_SO_BULK
WHERE 1=2;


--Bulk collect with limit clause
set serveroutput on;
DECLARE
TYPE lv
IS
  TABLE OF TEST_SO_BULK%ROWTYPE;
  lv_tab lv;
  CURSOR lvsql IS
  SELECT * FROM TEST_SO_BULK;
BEGIN
  OPEN lvsql;
  LOOP
  FETCH lvsql BULK COLLECT INTO lv_tab LIMIT 10000;
  dbms_output.put_line(lv_tab.COUNT);
  FORALL i IN lv_tab.FIRST..lv_tab.LAST
  INSERT INTO TEST_SO1 VALUES
  (
  lv_tab(i).col1,
  lv_tab(i).col2
  );
  EXIT WHEN lvsql%NOTFOUND;
  END LOOP; 
END;

--Check data count
select count(1) from test_so1;

过程：10000条记录后提交

2 个答案: