如何在HIve中将table1的一列追加到table2？

编程问答 2022-06-19

问题描述

enter image description here 

Animated.View

输出

import React from "react";
import { StyleSheet,View } from "react-native";
import { PanGestureHandler,State } from "react-native-gesture-handler";
import Animated,{  add,clockRunning,cond,debug,divide,eq,floor,not,set,useCode,} from "react-native-reanimated";
import {
    snapPoint,timing,useClock,usePanGestureHandler,useValue,} from "react-native-redash";
import { height,width } from "../../../../../assets/styles/main";
import SingleImage from "./SingleImage";

   const MultipleImages = ({ data,content,app }) => {

   const styles = StyleSheet.create({
   container: {
            flex: 1,backgroundColor: "black",},pictures: {
            width: width * data.photo.length,height,flexDirection: "row",picture: {
            width,overflow: "hidden",image: {
            ...StyleSheet.absoluteFillObject,width: undefined,height: undefined,});

const clock = useClock();
const index = useValue(0);
const offsetX = useValue(0);
const translateX = useValue(0);
const {
    gestureHandler,state,veLocity,translation,} = usePanGestureHandler();
const snapPoints = data.photo.map((_,i) => i * -width);
const to = snapPoint(translateX,veLocity.x,snapPoints);

useCode(
    () => [
        cond(eq(state,State.ACTIVE),[
            set(translateX,add(offsetX,translation.x)),]),cond(eq(state,State.END),timing({ clock,from: translateX,to })),set(offsetX,translateX),cond(not(clockRunning(clock)),[
                set(index,floor(divide(translateX,-width))),debug("index",index),],[]
);
return (
    <View style={styles.container}>
        <PanGestureHandler {...gestureHandler}>
            <Animated.View style={{ width: width,height: width }}>
                <Animated.View
                    style={[styles.pictures,{ transform: [{ translateX }] }]}
                >
                    {data.photo.map((item) => (
                        <SingleImage
                            app={app}
                            postId={data._id}
                            photo={item}
                            content={content}
                        />
                    ))}
                </Animated.View>
            </Animated.View>
        </PanGestureHandler>
    </View>
);
};

export default MultipleImages;

解决方法

假设第x列和第a列的值将始终增加，并且那些相应的记录需要并排显示-您可以使用下面的逻辑。但是问题仍然不十分清楚（因为理想情况下，您需要一个排序列或一个ID列才能将它们放在一起。在您的情况下，这些列可能是x＆a，但从问题中尚不清楚）。但是，以下查询适用于您的示例。

Select Tb1.a,Tb1.b,Tb2.x
from
 (Select a,b,row_number() over(order by a) as rn1 from Table2) Tb1
 join (Select x,row_number() over(order by x) as rn2 from Table1) Tb2
 on Tb1.rn1 = Tb2.rn2 ;
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