如何從嵌套的json數據獲取csv

Question

嗨，我正在嘗試從嵌套的json數組中獲取csv文件，我也正在獲取csv，但問題是Seller_name和drop_seller名稱顯示為[object，object]，因為嵌套的json數據附加了我的csv文件的圖像文件 這是我的數組

 {
"id":"1925",
"booking_id":"1442931120",
"appointment_time":"2015-09-29 16:00:00",
"booking_address":"Sector 15 gurgaon",
"booking_type":"later",
"category_id":"9",
"category_name":"Wash & Iron",
"stage":"accepted",
"lat":"74.83908",
"invoice_price":null,
"lon":"87.7333",
"user_name":" ",
"user_mobile_number":"8800292916",
"user_id":"1782",
"booking_time":"2015-09-22 19:09:55",
"price":null,
"seller_id":"52",
"payment_stage":"individual",
"discount":null,
"visting_charge":"0",
"rating":null,
"promo_code":"",
"drop_address":"",
"drop_time":"2015-10-01 20:18:28",
"comments":"",
"platform":"Web",
"order_return":"0",
"cloths_count":null,
"dc_cloths_count":null,
"cloths_weight":null,
"categories":null,
"service_type":"regular",
"category_type":"0",
"seller":{
"name":"Dinesh",
"mobile_number":"9818782536",
"distance":"10",
"lat":"28.4356578",
"lon":"77.1110703",
"id":"52",
"picture":"resources\/seller\/images\/20150709004010_559e2c4248c73.jpg",
"provider_picture":"resources\/seller\/provider-pictures\/logo_559e2c4248d4a.png"
},
"dropSeller":{
"name":"Dinesh",
"mobile_number":"9818782536",
"distance":"10",
"lat":"28.4356578",
"lon":"77.1110703",
"id":"52",
"picture":"resources\/seller\/images\/20150709004010_559e2c4248c73.jpg",
"provider_picture":"resources\/seller\/provider-pictures\/logo_559e2c4248d4a.png"
}
}

這是我的javascript文件

 function JSONToCSVConvertor(JSONData, ReportTitle, ShowLabel) {
    //If JSONData is not an object then JSON.parse will parse the JSON string in an Object
    var arrData = typeof JSONData != 'object' ? JSON.parse(JSONData) : JSONData;

    var CSV = '';    
    //Set Report title in first row or line

    CSV += ReportTitle + '\r\n\n';

    //This condition will generate the Label/Header
    if (ShowLabel) {
        var row = "";

        //This loop will extract the label from 1st index of on array
        for (var index in arrData[0]) {

            //Now convert each value to string and comma-seprated
            row += index + ',';
        }

        row = row.slice(0, -1);

        //append Label row with line break
        CSV += row + '\r\n';
    }

    //1st loop is to extract each row
    for (var i = 0; i < arrData.length; i++) {
        var row = "";

        //2nd loop will extract each column and convert it in string comma-seprated
        for (var index in arrData[i]) {
            row += '"' + arrData[i][index] + '",';
        }

        row.slice(0, row.length - 1);

        //add a line break after each row
        CSV += row + '\r\n';
    }

    if (CSV == '') {        
        alert("Invalid data");
        return;
    }   

    //Generate a file name
    var fileName = "MyReport_";
    //this will remove the blank-spaces from the title and replace it with an underscore
    fileName += ReportTitle.replace(/ /g,"_");   

    //Initialize file format you want csv or xls
    var uri = 'data:text/csv;charset=utf-8,' + escape(CSV);

    // Now the little tricky part.
    // you can use either>> window.open(uri);
    // but this will not work in some browsers
    // or you will not get the correct file extension    

    //this trick will generate a temp <a /> tag
    var link = document.createElement("a");    
    link.href = uri;

    //set the visibility hidden so it will not effect on your web-layout
    link.style = "visibility:hidden";
    link.download = fileName + ".csv";

    //this part will append the anchor tag and remove it after automatic click
    document.body.appendChild(link);
    link.click();
    document.body.removeChild(link);
}

而且我像我之前顯示的那樣傳遞了一個json對象，我也得到了csv文件，但是問題出在Seller_name和drop_seller名稱上，顯示為[object，object]剩余數據正確，我想是因為這些值在嵌套json中如何獲取這些名字請幫忙

Answer 1

如果我理解正確，那么您只希望顯示賣方名稱，而不是顯示其所有屬性？

如果是這種情況，只需添加一個對象檢查，然后顯示其名稱即可。

for (var index in arrData[i]) {
    var data = arrData[i][index];
    if(typeof data === 'object'){
        data = data.name;
    }
    row += '"' + data + '",';
}

Answer 2

我嘗試了上述解決方案，發現如果字段具有null值，則它不能正常工作，因為null的類型會返回一個對象。 如果仍在尋找解決方案，請嘗試以下解決方案。 希望這可以幫助！

 for (var index in arrData[i]) {
        var data = arrData[i][index];
        if(typeof data === 'object' && data != null){
            data = data.name;
        }
        row += '"' + data + '",';
    }