Question

如何将列表字符串转换为列表布尔值？

List<String> listString = ["true", "false, "true"]

List<bool> listBool = []

我试过了

listBool = listString.map((f)=>(f.contains("true")? listBool.add(true): listBool.add(false)));

我的预期结果是

listBool = [true, false, true]

Answer 1

您可以尝试使用 forEach 方法，如下所示：

    List<String> listString = ["true", "false", "true"];
    List<bool> listBool = <bool>[];
    listString.forEach((item) => item == "true" ? listBool.add(true) : listBool.add(false));

Answer 2

您可以使用List<String> listString = ["true", "false", "true"]; List<bool> listBool = listString.map((e) => e == "true").toList();

const express = require("express");
const bodyParser = require("body-parser");
const cors = require("cors");
const Pool = require("pg").Pool;
const pool = new Pool({
    user: "mgr@stanpgtest",
    host: "stanpgtest.postgres.database.azure.com",
    database: "my db name",
    password: "my pass",
    port: 5432
});

const app = express();
app.use(cors());
app.use(bodyParser.json());

app.listen(8000, () => {
    console.log(`Server is running, listening to port 8000`);
});

app.post("/api/v1/nodo2059e7", (req, res) => {
    const { label, status, priority } = req.body;
    pool.query(
        "select now()",
        (error, results) => {
            if (error) {
              throw error;
            }
            res.send(results);
        }
    );
});

将列表布尔值转换为列表字符串颤振

2 个答案: