当我想获取数据时出现此错误：TypeError：无法读取未定义的属性“数据”

Question

I want to retrieve API data and display its data on my HTML page all in a table.

I am getting this error while trying to display my data on my HTML page

When I try to use the .map

When i want to fetch data i have this error : TypeError: Cannot read property 'data' of undefined When i want to fetch data i have this error : TypeError: Cannot read property 'data' of undefined

// Convert csv to Json
function csvJSON(csv) {
    const lines = csv.split('\n')
    const result = []
    const headers = lines[0].split(',')

    for (let i = 1; i < lines.length; i++) {        
        if (!lines[i])
            continue
        const obj = {}
        const currentline = lines[i].split(',')

        for (let j = 0; j < headers.length; j++) {
            obj[headers[j]] = currentline[j]
        }
        result.push(obj)
    }
    return result
}
    url = "https://api7.esv2.com/v2/Api/Bounces?apiKey=xxxxxxxxxxxxxxxx&startDate=2020-10-25&endDate=2020-10-25&bounceType=1"
// Fetch Data
function fetchData() {
   fetch(url)
        .then(response => {
            if (!response.ok)  {
                throw Error("ERROR");
            }
            $(document).ready(function() {
                $.ajax({url}).then(function(data) {
                return console.log(csvJSON(data));
        })
    })
})
        .then(data => {
            console.log(data);
            const html = data.data.map(user => {
                return `
                
                <div class="user">
                <p>Date:${user.Date}</p>
                <p>Email:${user.Email} </p>
                <p>BounceType:${user.BounceType}</p>
                </div>
                `;
            }).join("");
            console.log(html);
            document.querySelector('#app')
            .insertAdjacentHTML("afterbegin", html) 
        })
            .catch(error => {
                console.log(error);
        });
    }

    
    fetchData();
    ```

Answer 1

You are returning console.log in the function, try returning without console.log

$(document).ready(function() {
    $.ajax({ url }).then(function(data) {
      return csvJSON(data); // changed
    });
  });