将JQuery AJAX中的多维数组发送到PHP

Wasif Khalil 发表于 Dev

瓦西夫·哈利勒（Wasif Khalil）

我想从JQuery AJAX向PHP发送多维数组，但是这样在PHP中接收

Array
(
[recordid] => 38
[locations] => [object Object],[object Object]
)

我一定是在做一些愚蠢的错误。这是代码。它从表中获取记录并发送到PHP

$(document).on('click','.savenow',function(){
    recordid = $(this).data('id');

    locations = [];

    $('.selectrec').each(function () {
        parent = $(this).parent().parent();

        name    = parent.find('td').eq(5);
        address = parent.find('td').eq(6);
        lat     = parent.find('td').eq(1);
        lng     = parent.find('td').eq(2);

        row = [name,address,lat,lng];

        locations.push(row);
    });

    locations = locations.toString();
    $.ajax({
        type: "POST",
        url:'/record/saveSearchedLocations',
        data: { recordid: recordid,locations:locations },
        dataType: 'json',
        success: function (data) {
            console.log(data);
        },
        error:function(data){
          alert("something went wrong, please try again.");
        }
    });

});

这是我正在接收数据的PHP函数：

function saveSearchedLocations(){
    print_r($_POST);
}

理查德·伯纳兹

使用JSON.stringify()而不是toString()像这样：

将您的AJAX调用更改为：

$(document).on('click','.savenow',function(){
    recordid = $(this).data('id');

    locations = [];

    $('.selectrec').each(function () {
        parent = $(this).parent().parent();

        name    = parent.find('td').eq(5);
        address = parent.find('td').eq(6);
        lat     = parent.find('td').eq(1);
        lng     = parent.find('td').eq(2);

        row = [name,address,lat,lng];

        locations.push(row);
    });

    ajaxData = { recordid : recordid,locations : locations }
    $.ajax({
        type: "POST",
        url:'/record/saveSearchedLocations',
        data: JSON.stringify(ajaxData),
        dataType: 'json',
        success: function (data) {
            console.log(data);
        },
        error:function(data){
          alert("something went wrong, please try again.");
        }
    });

});