使用JavaScript將多個JSON對象添加或推送到API JSON數組

Question

我有一個CSV文件，其中包含已轉換為JSON的數據。 現在，我正在嘗試弄清楚如何將多個JSON對象發送到API JSON數組。 我目前正在使用ractive作為我的框架選擇。 通過單擊“發送JSON數據”按鈕來添加JSON對象。

CSV到JSON的轉換

<html>

<body>

<?php 

    function csvToJson($fname) {
        if (!($fp = fopen($fname, 'r') )) {
            die("Can't open file");
        }

        $key = fgetcsv($fp, "1024", ",");

        $json = array();
            while ($row = fgetcsv($fp, "1024", ",")) {
                $json[] = array_combine($key, $row);
        }

        fclose($fp);

        foreach ( $json as $k=>$v ) {


            $json[$k]['dateRequested']     = $json[$k]['DATE']; 
            $json[$k]['assignedAgent']     = $json[$k]['AGENT'];
            $json[$k]['finalCompanyName']  = $json[$k]['COMPANY NAME'];
            $json[$k]['unitNumber']        = $json[$k]['UNIT'];
            $json[$k]['floorNumber']       = $json[$k]['FLOOR_LEVEL'];
            $json[$k]['buildingName']      = $json[$k]['BUILDING NAME'];
            $json[$k]['streetNumber']      = $json[$k]['NUMBER'];
            $json[$k]['streetName']        = $json[$k]['STREET'];
            $json[$k]['subdivisionName']   = $json[$k]['SUBDIVISION_COMPOUND_VILLAGE'];
            $json[$k]['barangayName']      = $json[$k]['BARANGAY'];
            $json[$k]['districtNum']       = $json[$k]['DISTRICT'];
            $json[$k]['cityMunicipality']  = $json[$k]['CITY_MUNICIPALITY'];
            $json[$k]['provinceName']      = $json[$k]['PROVINCE'];
            $json[$k]['region']            = $json[$k]['REGION'];
            $json[$k]['area']              = $json[$k]['AREA'];
            $json[$k]['longitude']         = $json[$k]['LONGITUDE'];
            $json[$k]['latitude']          = $json[$k]['LATITUDE'];
            $json[$k]['secRanking']        = $json[$k]['Sec Ranking'];
            $json[$k]['affiliate']         = $json[$k]['Affiliates'];
            $json[$k]['profiling']         = $json[$k]['Profiling'];
            $json[$k]['contactNumber']     = $json[$k]['CONTACT NUMBER'];
            $json[$k]['industry']          = $json[$k]['INDUSTRY']; 
            $json[$k]['amAssigned']        = $json[$k]['AM ASSIGNED'];
            $json[$k]['source']            = $json[$k]['SOURCE'];
            $json[$k]['gbu']               = $json[$k]['GBU'];
            $json[$k]['retention']         = $json[$k]['RETENTION'];
            $json[$k]['sec']               = $json[$k]['SEC'];
            $json[$k]['tier']              = $json[$k]['TIER'];
            $json[$k]['acqui']             = $json[$k]['ACQUI'];
            $json[$k]['finalReco']         = $json[$k]['FINAL RECO'];
            $json[$k]['remarksFindings']   = $json[$k]['FINAL REMARKS'];

            unset($json[$k]['DATE']);
            unset($json[$k]['AGENT']);
            unset($json[$k]['COMPANY NAME']);
            unset($json[$k]['UNIT']);
            unset($json[$k]['FLOOR_LEVEL']);
            unset($json[$k]['BUILDING NAME']);
            unset($json[$k]['NUMBER']);
            unset($json[$k]['STREET']);
            unset($json[$k]['SUBDIVISION_COMPOUND_VILLAGE']);
            unset($json[$k]['BARANGAY']);
            unset($json[$k]['DISTRICT']);
            unset($json[$k]['CITY_MUNICIPALITY']);
            unset($json[$k]['PROVINCE']);
            unset($json[$k]['REGION']);
            unset($json[$k]['AREA']);
            unset($json[$k]['LONGITUDE']);
            unset($json[$k]['LATITUDE']);
            unset($json[$k]['Sec Ranking']);
            unset($json[$k]['Affiliates']);
            unset($json[$k]['Profiling']);
            unset($json[$k]['CONTACT NUMBER']);
            unset($json[$k]['INDUSTRY']);
            unset($json[$k]['AM ASSIGNED']);
            unset($json[$k]['SOURCE']);
            unset($json[$k]['GBU']);
            unset($json[$k]['RETENTION']);
            unset($json[$k]['SEC']);
            unset($json[$k]['TIER']);
            unset($json[$k]['ACQUI']);
            unset($json[$k]['FINAL RECO']);
            unset($json[$k]['FINAL REMARKS']);

        }

        return json_encode($json, JSON_PRETTY_PRINT | JSON_UNESCAPED_SLASHES);

    }

 ?>



<?php
$json_data = csvToJson("lms.csv");
?>

Java腳本

<div id="container">

</div>


<script src="https://cdnjs.cloudflare.com/ajax/libs/ractive/0.9.0-build-48/ractive.js"></script>


<script id="template" type="text/ractive">
     <a class="button cta" on-click="addData">Send JSON Data</a>
</script>

<script type="text/javascript">

var app = new Ractive({
    el       : '#container',
    template : '#template',
});

var proxy       = 'http://192.168.1.126/lms-dev-noel/proxy.php';
var endpoint    = 'account/leads/';
var rt          = 'POST';
var url         = proxy+'?endpoint='+endpoint+'&rt='+rt;

var lms_json    = <?php echo json_encode($json_data); ?>;

var jobjects = JSON.parse(lms_json);
console.log(jobjects)




app.on('addData', function(event) {
    $.ajax({
        type     : 'POST',
        url      : url,
        data     : jobjects,
        dataType : 'json',
        success  : function() {
            console.log(jobjects);
        },
        error    : function(error) {
            console.log('Error')
        }

    })

})


</script>

</body>

</html>

Answer 1

JSON只是一種表示法，它不會改變任何東西。 解析JSON字符串后，您應該能夠像往常一樣將對象/值推送到所需的鍵上。

array[key].push(object);

例如：

jobjects['myKey1']['myKey2'].push({"key1":"value1", "key2":"value2"});

參考：向JSON對象添加新的數組元素

使用JavaScript將多個JSON對象添加或推送到API JSON數組

問題描述

1 個解決方案

解決方案1
0 2017-01-16 10:05:39

使用JavaScript將多個JSON對象添加或推送到API JSON數組

問題描述

1 個解決方案

解決方案1 0 2017-01-16 10:05:39

解決方案1
0 2017-01-16 10:05:39