【发布时间】:2019-03-17 22:57:03
【问题描述】:
- 我使用了第一个索引过滤的数组值结果,并根据不同的条件呈现它们
- 我很惊讶如何使用映射过滤数组值的所有索引值并将其呈现在 UI 上
- 你们能帮我解决您的建议问题吗
- 完整代码可在 stackblitz 及其链接中找到:https://stackblitz.com/edit/react-geum6v?file=index.js
- 下面提供了一个用户过滤值和尝试的多个索引数组值的代码 sn-p:
- 能否告诉我在为 map 方法执行多个索引数组值时我在哪里出错了
index.js(JS部分)
sportsZipSearch = () => {
const { zip1, dob1, age1, state1, check, count } = this.state;
const newArray = students.filter((el) => { return ((el.zip === zip1) && (parseInt(el.Age) <= parseInt(age1[Object.keys(age1)[0]])) || (el.sports_state != (state1)) && check ) });
const newArray1 = newArray[0].Sports_games.map((el1) => { return Object.keys(el1) });
//const sorted = newArray1[Object.keys(newArray1)].sort();
console.log(age1);
console.log(zip1);
console.log("here--", this.state.zip1, this.state.age1);
console.log(newArray);
console.log(newArray1);
//console.log(sorted);
console.log(newArray.Subjects);
console.log(count);
this.setState({ result: newArray, zipValue: true })
}
index.js (html 部分)
Show list Sports: {result.length && result[0].Sports_games.map(g => { var visibility = (check && g[Object.keys(g)[0]]) ? 'block' : 'none'; return <div style={{display: visibility}}>{Object.keys(g)[0]} {g[Object.keys(g)[0]]}</div> } )}<br/><br/>
Show list Sports: {result.map(result_new => result_new.length && result_new.Sports_games.map(g => { var visibility = (check && g[Object.keys(g)]) ? 'block' : 'none'; console.log(visibility);return <div style={{display: visibility}}>{Object.keys(g)} {g[Object.keys(g)]}</div> } ))}<br/><br/>
【问题讨论】:
标签: javascript html reactjs