Question

我试图根据服务器的内容以不同的方式处理服务器响应，但是我什至无法运行与响应数据无关的东西。

这是我在jQuery中的请求：

function Register() { var form = $('#signup-form') $.post("http://localhost:3000/register", form.serialize(), (data) => console.log("1")) .done((data) => console.log("2")) .fail((data) => console.log("3")) .always((data) => console.log("4")); }

在Express中：

rota.post('/register', (req, res) => {
const name = req.body.name;
const age = req.body.age;
const pass = sha256(req.body.pass);
const request = new sql.Request();
request.query(`SELECT CAST(CASE WHEN EXISTS(SELECT * FROM Usuario WHERE username = '${name}') THEN 1 ELSE 0 END AS BIT) AS re`).then(
    result => {
        if (result.recordset[0].re == true) {
            console.log("usuário já registrado");
            res.send("1");
        } else {
            request.query(`INSERT INTO Usuario(username, userage, userpass) VALUES('${name}', '${age}', '${pass}')`);
        }
    }).catch(result => {})
})

我在做什么错了？

Answer 1

重新发送状态以完成请求，否则该请求将保持打开状态。在 catch块中插入 400 状态响应，以使前端知道请求已失败并进入fail()函数：

request.query(`SELECT CAST(CASE WHEN EXISTS(SELECT * FROM Usuario WHERE username = '${name}') THEN 1 ELSE 0 END AS BIT) AS re`).then(
    result => {
        if (result.recordset[0].re == true) {
            console.log("usuário já registrado");
            res.send("1");
        } else {
            request.query(`INSERT INTO Usuario(username, userage, userpass) VALUES('${name}', '${age}', '${pass}')`);
            res.status(200).send();  // here add a status of 200 to signal success
        }
    }).catch(result => {
       res.status(400).send();  // send also a 400 status response to signal failure
    })
})

无法获得express.js响应

1 个答案: