Jquery 将JSON对象与不使用数据表的数组一起使用

Jquery 将JSON对象与不使用数据表的数组一起使用,jquery,arrays,json,object,datatables,Jquery,Arrays,Json,Object,Datatables,我正在努力将JSON数据用于数据表。很抱歉设置了格式,但这是我在开发人员工具中得到的: Object pharmacy:Array[3] 0:"Walmart" 1:"Safeway" 2:"Kroger Pharmacy" length:3 __proto__:Array[0] price: Array[3] 0:58.14 1:65.45 2:66.76 length:3 __proto__:Array[0] ... 我用这个来理解: $.ajax({ url: 'test.php

我正在努力将JSON数据用于数据表。很抱歉设置了格式,但这是我在开发人员工具中得到的:

Object
pharmacy:Array[3]
0:"Walmart"
1:"Safeway"
2:"Kroger Pharmacy"
length:3
__proto__:Array[0]
price: Array[3]
0:58.14
1:65.45
2:66.76
length:3
__proto__:Array[0]
...
我用这个来理解:

$.ajax({
    url: 'test.php',
    type: 'POST',
    dataType: 'JSON',
    data: {
      drug: picked_drug,
        },
    success: function(data) {
        var all_data = JSON.parse(data);
        final_data = all_data.data.price_detail;
        console.log(final_data); //this outputs object above.
}
});
然后,我以如下方式渲染表:

var drugtable = $("#drug_datatable").DataTable({
            "data": final_data,              
            "paging": true,
            "dom": '<"top">Bt<"bottom"><"clear">',
            "pageLength": 25,
            "order": [
                [0, "desc"]
            ],
            "columns": [
            {
                "data": "pharmacy",
        "searchable": false,
        "width": "20%",
        "className": "lang_body_2",
        "title": "Attribute"
    }, 
    {
                "data": "price",
        "searchable": false,
        "width": "20%",
        "className": "lang_body_2",
        "title": "Attribute"
    }, 
              ],
        });
var drugtable=$(“#druge_datatable”).datatable({
“数据”:最终数据,
“分页”:正确,
“dom”:“Bt”,
“页面长度”:25,
“命令”:[
[0,“描述”]
],
“栏目”:[
{
“数据”:“药房”,
“可搜索”:错误,
“宽度”:“20%”,
“类名”:“lang_body_2”,
“标题”:“属性”
}, 
{
“数据”:“价格”,
“可搜索”:错误,
“宽度”:“20%”,
“类名”:“lang_body_2”,
“标题”:“属性”
}, 
],
});
问题是我的表返回“表中没有可用数据”。我肯定我的对象的格式有问题,但我不知道要更改什么


谢谢

您的数据结构正确吗? 您的数据看起来像[“Safeway”,“Kroger Pharmacy”],价格:[58.14,65.45,66.76]}

你可能想要[{药房:“沃尔玛”,价格:58.14},{药房:“安全之路”,价格:65.45},{药房:“克罗格药房”,价格:66.76}]

这将使数据变得平坦

    // simulated the return from ajax as i understand your data
    function stuff () {
        this.pharmacy = ["Walmart","Safeway", "Kroger Pharmacy"];
        this.price = [58.14, 65.45, 66.76]
    }
    var list = new stuff();

    // now flatening for datatables
    var keys = Object.keys(list);
    var listLen = list[keys[0]].length;
    var final_data = [];
    // create the list for data table
    for (var i = 0; i < listLen; ++i) {

        var item = {};
        for (var j = 0; j < keys.length; ++j) {
            item[keys[j]] = list[keys[j]][i];
        }
        final_data[final_data.length] = item;
    }

我无法实际测试这段代码,因为我无法从源代码中获取数据,但根据我们的讨论,这是我的最佳选择

<script type="text/javascript">

    // this flattens the data passed back by ajax
    function flattenData(list) {
        // now flatening for datatables
        var keys = Object.keys(list);
        var listLen = list[keys[0]].length;
        var fd= [];
        // create the list for data table
        for (var i = 0; i < listLen; ++i) {

            var item = {};
            for (var j = 0; j < keys.length; ++j) {
                item[keys[j]] = list[keys[j]][i];
            }
            fd[final_data.length] = item;
        }
        return fd;
    }


    // function is called from the success callback function to create the DataTable
    function createTable(tableData) {

        var drugtable = $("#drug_datatable").DataTable({
            "data": tableData,
            columns: [{ data: "pharmacy" }, { data: "price" }],
        });
    }

    $(function ($) {

        $.ajax({
            url: 'test.php',
            type: 'POST',
            dataType: 'JSON',
            data: {
                drug: picked_drug,
            },
            success: function (data) {
                var all_data = JSON.parse(data);
                var final_data = flattenData(all_data.data.price_detail);
                createTable(final_data)
                console.log(final_data); //this outputs object above.
            }
        });

    })
</script>

//这将使ajax传回的数据平坦化
功能数据(列表){
//现在,数据表正在变平
var keys=Object.keys(列表);
var listLen=列表[键[0]]。长度;
var-fd=[];
//创建数据表的列表
对于(变量i=0;i
是的……这正是我想要的……我正努力用一个
语句来实现它。想法?所以你不能在数据源处更改数据,以使其平坦化?数据来自一个API,所以我只能尝试这样的方法:
$.each(final_data,function(index,val){item={}item[index]=val;})我有一个想法,但是重新创建itI需要几分钟的时间。我还没有测试datatable部分,还没有安装在我的开发环境中。我确实做了一些关于阵列同步的假设
<script type="text/javascript">

    // this flattens the data passed back by ajax
    function flattenData(list) {
        // now flatening for datatables
        var keys = Object.keys(list);
        var listLen = list[keys[0]].length;
        var fd= [];
        // create the list for data table
        for (var i = 0; i < listLen; ++i) {

            var item = {};
            for (var j = 0; j < keys.length; ++j) {
                item[keys[j]] = list[keys[j]][i];
            }
            fd[final_data.length] = item;
        }
        return fd;
    }


    // function is called from the success callback function to create the DataTable
    function createTable(tableData) {

        var drugtable = $("#drug_datatable").DataTable({
            "data": tableData,
            columns: [{ data: "pharmacy" }, { data: "price" }],
        });
    }

    $(function ($) {

        $.ajax({
            url: 'test.php',
            type: 'POST',
            dataType: 'JSON',
            data: {
                drug: picked_drug,
            },
            success: function (data) {
                var all_data = JSON.parse(data);
                var final_data = flattenData(all_data.data.price_detail);
                createTable(final_data)
                console.log(final_data); //this outputs object above.
            }
        });

    })
</script>