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Remove times between 2 other ones
app.post('/checkTimes', (req, res) => {
    let date = req.body.date;
    let times = [
        "16:00",
        "16:30",
        "17:00",
        "17:30",
        "18:00",
        "18:30",
        "19:00",
        "19:30",
        "20:00",
        "20:30",
        "21:00",
    ];

    connection.query("SELECT time FROM appointments WHERE date = ?", [date], (err, result) => {
        if(err) throw err;

        let timesToReturn = [];
        // Get all times that are not between start time and end time
        for(let i = 0; i < times.length; i++) {
            let found = false;
            for(let j = 0; j < result.length; j++) {
                if(times[i] == result[j].time) {
                    found = true;
                    break;
                }
            }
            if(!found) timesToReturn.push(times[i]);
        }

        res.status(200).send(timesToReturn);
    });
});

So, I got this. It only checks start time right now, but I want it to remove all times that are between start and end. Not sure how to do that if anyone could help me i'd be much appreciated for example if I got an appointment between 16:00 and 17:30 I want it to remove 16:00, 16:30 and 17:00

about 4 years ago · Juan Pablo Isaza
1 Respuestas
Responde la pregunta

0

app.post('/checkTimes', (req, res) => {
    let date = req.body.date;
    let times = [
        "16:00",
        "16:30",
        "17:00",
        "17:30",
        "18:00",
        "18:30",
        "19:00",
        "19:30",
        "20:00",
        "20:30",
        "21:00",
    ];

    connection.query("SELECT time, endTime FROM appointments WHERE date = ?", [date], (err, [result]) => {
        if(err) throw err;

        if(result) {
            let timesToReturn = [...times.slice(0, times.indexOf(result.time)), ...times.slice(times.indexOf(result.endTime))];
            res.status(200).send(timesToReturn);
        } else {
            res.status(200).send(times);
        }
    });
});
about 4 years ago · Juan Pablo Isaza Denunciar
Responde la pregunta
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