Question

我有以下对象：

{
    builders: [
        {
            name: "Branches > Build",
            slaves: [
                "slave-01.local"
            ]
        },
        {
            name: "Branches > Build1",
            slaves: [
                "slave-02.local"
            ]
        }
    ]
}

如果他们的父tr id与json的name属性匹配，我想更新html表td。

<table>
  <tr id="Branches > Build" class="odd">
    <td>
      value of slaves of Branches > Build should be here
    </td>
  <tr id="Branches > Build1" class="even">
    value of slaves of Branches > Build1 should be here
    <td>
    </td>
<table>

我如何实现这一目标？我正在使用jquery。

顺便说一下。我无法获得没有空格的id，因此必须使用它。

Answer 1

如果您无法将ID更改为有效ID，则可以使用属性选择器[id=""]

var obj = "the jsonfile";
var objBuilder = obj.builders;
for (var key in obj) {
    var item = $('[id="' + obj[key].name + '"]')
    if (item.length) {
        item.find('td').text(obj[key].slaves);
    }
}

DEMO修复了html标记以及

Answer 2

你能试试吗，

    var json = {
        builders: [
            {
                name: "Branches > Build",
                slaves: [
                    "slave-01.local"
                ]
            },
            {
                name: "Branches > Build1",
                slaves: [
                    "slave-02.local"
                ]
            }
        ]
    }

    $("table tr").each(function(index, obj){

        console.log(index+obj.id + json.builders[0].name);

        if(obj.id == json.builders[0].name){

            $(this).find('td').html('test');
        }

    });

Answer 3

请参阅basic types

ID and NAME tokens must begin with a letter ([A-Za-z]) and may be followed by any number
 of letters, digits ([0-9]), hyphens ("-"), underscores ("_"), colons (":"), and periods   (".").

IDREF and IDREFS are references to ID tokens defined by other attributes. IDREF is a single token and IDREFS is a space-separated list of tokens.

NUMBER tokens must contain at least one digit ([0-9]).

所以你可能必须找到一种更好的方法来解决它并使用正确的ID。我建议你去使用"Aungular"这样的javascript框架，这样可以让你的工作变得更简单，你可以用它做更多的事情。

如何将对象值与JQuery ID进行比较？

3 个答案: