问题在于您正在使用的这段代码:
html_content.position[0][0].toLowerCase() = "navegantes";
html_content.position
实际上是一个数组数组的数组(尽管在您的示例数据中,每个内部数组的长度仅为 1 ......这总是真的吗?),所以您需要一个括号运算符来测试html_content.position[i][0][0]
这是一个包含您的 JSON 数据的示例,查看每个团队,并找到相应的position
:(请参阅下面的工作 JSfiddle 演示)
解决方案代码:
var matches = [];
for(var teamName in json.teams)
{
for(var i = 0, len = json.html_content.position.length; i < len; i++)
{
if(json.html_content.position[i][0][0].toLowerCase() === teamName.toLowerCase())
{
// found a match. do something with it..
matches[matches.length] = {Name: teamName, Value: json.teams[teamName]};
break;
}
}
}
JSON:(供参考,因为以前仅在第 3 方网站上发布)
var json = {
"response":true,
"teams":{
"navegantes":"img\/magallanes.jpg",
"tigres":"img\/tigres.jpg",
"caribes":"img\/caribes.jpg",
"leones":"img\/leones.jpg",
"aguilas":"img\/aguilas.jpg",
"tiburones":"img\/tiburones.jpg",
"bravos":"img\/bravos.jpg",
"cardenales":"img\/cardenales.jpg",
"maga":"img\/magallanes.jpg",
"tigr":"img\/tigres.jpg",
"cari":"img\/caribes.jpg",
"leon":"img\/leones.jpg",
"agui":"img\/aguilas.jpg",
"tibu":"img\/tiburones.jpg",
"brav":"img\/bravos.jpg",
"card":"img\/cardenales.jpg"
},
"html_content":{
"position":[
[
[
"Navegantes",
"14",
"10",
"4",
"0"
]
],
[
[
"Tigres",
"14",
"10",
"4",
"0"
]
],
[
[
"Caribes",
"14",
"9",
"5",
"1"
]
],
[
[
"Leones",
"14",
"9",
"5",
"1"
]
],
[
[
"Tiburones",
"13",
"5",
"8",
"4.5"
]
],
[
[
"Aguilas",
"14",
"5",
"9",
"5"
]
],
[
[
"Bravos",
"14",
"4",
"10",
"6"
]
],
[
[
"Cardenales",
"13",
"3",
"10",
"6.5"
]
]
],
"current":[
[
"MAGA",
"CARI",
"7:00 pm",
"PUERTO LA CRUZ"
],
[
"AGUI",
"LEON",
"4:00 pm",
"CARACAS"
],
[
"BRAV",
"TIGR",
"5:30 pm",
"MARACAY"
],
[
"TIBU",
"CARD",
"5:30 pm",
"BARQUISIMETO"
]
],
"next":[
[
"MAGA",
"CARI",
"6:00 pm",
"PUERTO LA CRUZ"
],
[
"AGUI",
"LEON",
"1:00 pm",
"CARACAS"
],
[
"TIBU",
"TIGR",
"5:30 pm",
"MARACAY"
],
[
"BRAV",
"CARD",
"2:00 pm",
"BARQUISIMETO"
]
],
"previous":[
]
}
};
在此处查看示例小提琴:http: //jsfiddle.net/VqHpJ/
示例小提琴创建了一个数组,matches
该数组是代表团队的对象集合Name
和Value
相关图像的集合,并将它们吐出到列表中。(当然,您并没有说找到匹配项后要做什么,因此您可以根据需要进行调整)。
这是匹配两个数组(一个嵌套循环)中的元素的相当典型的模式,在我的示例中,它假设应该只找到一个匹配项,在这种情况下,它将break
跳出嵌套循环并开始寻找下一个团队。因此,在最坏的情况下,性能为 O(n^2)。
这有点棘手(但并不困难),因为该teams
对象并没有真正的团队数组,它具有每个团队名称的特定属性。因此,解决方案是teams
使用for(var teamName in json.teams)
. 如果您是生成 JSON 的函数的作者,您可能需要考虑修改它以生成一组团队,例如:
var json = {
"response":true,
"teams":
[
{ "Name": "navegantes", "ImageUrl": "img\/magallanes.jpg" },
{ "Name": "tigres", "ImageUrl": "img\/tigres.jpg"},
{ "Name": "caribes", "ImageUrl": "img\/caribes.jpg"},
...
]
...
}