javascript - 我可以使用预设变量放置 json 调用字符串吗？

Question

这个问题已经有答案了:

Accessing an object property with a dynamically-computed name (19 个回答)

已关闭 10 年前。

<?php
date_default_timezone_set('America/Los_Angeles');
$time = date('Gi', time());
$day = date('l', time());
?>

<script type="text/javascript">
$(function() {
  $(".show_hide").click( function()
       {
        var locTime = <?php echo json_encode($time) ?>;
        var locDate = <?php echo json_encode($day) ?>;

$.getJSON( "url_to_json", function(data) {
    for (var i = 0; i < data.location[locDate].length; i++) { 
    console.log("sucess1");
    xr_mon= data.location.locDate[i];
    console.log("sucess2");
     if (locTime >= xr_mon.kai && locTime < xr_mon.guan ){
        console.log("cafe is open!");
        $('.xr').show(); 
        break;
        }


    else {
        console.log("cafe is closed");
        $('.xr').hide();
    }
    }
});
       }
  );
});

JSON

{ "location": 
{
    "Monday": [
    {"kai": 700, "guan": 1400},
    {"kai": 1700, "guan": 2100}
    ],

    "Tuesday": [
    {"kai": 700, "guan": 1400},
    {"kai": 1700, "guan": 2100}
    ],

    "Wednesday": [
    {"kai": 700, "guan": 1400},
    {"kai": 1700, "guan": 2100},
    {"kai": 2200, "guan": 2400},
    {"kai": 0, "guan": 200}
    ],

    "Thursday": [
    {"kai": 700, "guan": 1400},
    {"kai": 1700, "guan": 2100},
    {"kai": 2200, "guan": 2400},
    {"kai": 0, "guan": 200}
    ],

    "Friday": [
    {"kai": 700, "guan": 1400},
    {"kai": 1700, "guan": 2100},
    {"kai": 2200, "guan": 2400},
    {"kai": 0, "guan": 200}
    ],

    "Saturday": [
    {"kai": 1000, "guan": 1500},
    {"kai": 1700, "guan": 2100},
    {"kai": 2200, "guan": 2400},
    {"kai": 0, "guan": 200}
    ],

    "Sunday": [
    {"kai": 1000, "guan": 1500},
    {"kai": 1700, "guan": 2100},
    {"kai": 2200, "guan": 2400},
    {"kai": 0, "guan": 200}
    ]
}
}

其中 kai = 开放时间和 guan = 关门时间

locDate 是通过 php 服务器端日期提取的，我想替换用我的 JSON 调用字符串生成的日期，该字符串包含工作日所在的部分。

从这里

  for (var i = 0; i < data.location.'locDate'.length; i++)

到此

  for (var i = 0; i < data.location.Thursday.length; i++)

我该如何以正确的方式做到这一点。有人建议使用 [locDate]，但它不会替换 php 日期数据中的变量。

Answer 1

使用括号表示法。

for (var i = 0; i < data.location[locDate].length; i++)

另一个例子:

thing = [1, 2, 3, 4, 5]
//The following lines do the same thing:
thing.pop()
thing['pop']()