Question

我想结合2个表，但得到错误

var firstTabPath, secondTabPath, thirdTabPath,
        firstTabTitle, secondTabTitle, thirdTabTitle,
        firstTabContent, secondTabContent, thirdTabContent;

const getTabsTitlesInOrder = () => {
    for(var i = 0; i <tabsJson.tabs.length; i ++){
        switch(tabsJson.tabs[i].order){
            case 0:
                firstTabTitle = tabsJson.tabs[i].title;
                getFirstTabContent(firstTabTitle);
                break;
            case 1:
                secondTabTitle = tabsJson.tabs[i].title;
                getSecondTabContent(secondTabTitle);
                break;
            case 2:
                thirdTabTitle = tabsJson.tabs[i].title;
                getThirdTabContent(thirdTabTitle);
                break;
        }
    }
};

const getFirstTabContent = (firstTabTitle) => {
    switch (firstTabTitle) {
        case "Dummy Table":
            firstTabContent = <DummyTable/>;
            firstTabPath = "/dummyTable";
            break;
        case "Dummy Chart":
            firstTabContent = <DummyChart/>;
            firstTabPath = "/dummyChart";
            break;
        case "Dummy List":
            firstTabContent = <DummyList/>;
            firstTabPath = "/dummyList";
            break;
    }
};

const getSecondTabContent = (secondTabTitle) => {
    switch (secondTabTitle) {
        case "Dummy Table":
            secondTabContent = <DummyTable/>;
            secondTabPath = "/dummyTable";
            break;
        case "Dummy Chart":
            secondTabContent = <DummyChart/>;
            secondTabPath = "/dummyChart";
            break;
        case "Dummy List":
            secondTabContent = <DummyList/>;
            secondTabPath = "/dummyList";
            break;
    }
};

const getThirdTabContent = (thirdTabTitle) => {
    switch (thirdTabTitle) {
        case "Dummy Table":
            thirdTabContent = <DummyTable/>;
            thirdTabPath = "/dummyTable";
            break;
        case "Dummy Chart":
            thirdTabContent = <DummyChart/>;
            thirdTabPath = "/dummyChart";
            break;
        case "Dummy List":
            thirdTabContent = <DummyList/>;
            thirdTabPath = "/dummyList";
            break;
    }
};

class TabContent extends React.Component{
    render() {
        return(
            <Switch>
                <Route exact path={firstTabPath} component={firstTabContent}/>
                <Route path={secondTabPath} component={secondTabContent}/>
                <Route path={thirdTabPath} component={thirdTabContent}/>
            </Switch>
        );
    }
};

class TabButtons extends React.Component{
    render() { 
        getTabsTitlesInOrder();
        return(
            <div>
                <div className="tabTitleWrapper">
                    <Link className="tabLink" to={firstTabPath}>
                        {firstTabTitle}
                    </Link>
                </div>
                <div className="tabTitleWrapper">
                    <Link className="tabLink" to={secondTabPath}>
                        {secondTabTitle}
                    </Link>
                </div>
                <div className="tabTitleWrapper">
                    <Link className="tabLink" to={thirdTabPath}>
                        {thirdTabTitle}
                    </Link>
                </div>
            </div>
        );
    }
};

class App extends React.Component{
    render() { 
        return(
            <div>
                <TabButtons />
                <TabContent />
            </div>
        );
    }
};

ReactDOM.render((
    <BrowserRouter>
        <App />
    </BrowserRouter>),
    document.getElementById('tab')
);

但是得到错误：

proc sql;
select * from Table1
outer union corr
select * from table2;

如果我理解这是正确的并且基于UNION ALL two SELECTs with different column types - expected behaviour?，则第一列是不同的数据类型，不能继续使用union（这是真的）

ERROR: The type of column EntryId from the left hand side of the OUTER UNION set operation is
       different from EntryId on the right hand side

但是，我想要使用的列具有相同的格式

RecordID num label='RecordID' format=20. informat=20.
and
RecordID num label='RecordID' format=11. informat=11.

并且在每个表中它们是第3列和第4列。

有没有办法使用该列作为引用将这些表联合起来，而不是原始的？

我试着无济于事：

Pseu char(64) label='Pseu' format=$64. informat=$64.
Pseu char(64) label='Pseu' format=$64. informat=$64.

它来自http://support.sas.com/documentation/cdl/en/proc/61895/HTML/default/viewer.htm#a002473694.htm上给出的OUTER UNION CORRESPONDING示例，并且基于我想要的内容：

proc sql;
select * from Table1
outer union corr
select * from table2
on Table1.Pseu=Table2.Pseu;

ERROR: Found "on" when expecting ;

Answer 1

或许这样的事情：

proc sql;
select * from Table1
outer union corr
select p, r as r2, z from table2

列r。

的列别名

使用常规UNION：

select p, r, y, null, null from Table1
outer union corr
select p, null, null, r as r2, z from table2

Answer 2

我的搜索和jarlh提供的答案是正确的。

由于要合并的数据集中的列的大小和数量，问题出现了。我必须确保在联合数据集中没有重复列名称（在我的总共600列中，有些名称相似），所以我不得不重命名列

如何基于sas中的特定列结合2个表

2 个答案: