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所以我有一个 php 页面，它以正确的格式返回一个 JSON 对象。喜欢:
[
{
"name":"Users",
"parent":"null",
"children":[ {
"name": "adsd", "parent": "Users", "children": []
}
,
{
"name": "ca", "parent": "Users", "children": []
}
,
{
"name":"ChanakaP",
"parent":"Users",
"children":[ {
"name": "Carlos Puyol rejects Barcelona sporting director role", "parent": "ChanakaP"
}
,
{
"name": "\r\nDiego Costa could leave Atletico Madrid", "parent": "ChanakaP"
}
,
{
"name": "FIFA insist Lionel Messi's award for best men's player was not rigged", "parent": "ChanakaP"
}
,
{
"name": "\r\nReal madrid interested in Van De Beek", "parent": "ChanakaP"
}
,
{
"name": "Hazard scores debut goal", "parent": "ChanakaP"
}
]
}
,
{
"name": "dsd", "parent": "Users", "children": []
}
,
{
"name": "ggggggggggggggggg", "parent": "Users", "children": []
}
,
{
"name":"james123",
"parent":"Users",
"children":[ {
"name": "first post", "parent": "james123"
}
]
}
,
{
"name": "new", "parent": "Users", "children": []
}
,
{
"name": "new123", "parent": "Users", "children": []
}
,
{
"name": "test", "parent": "Users", "children": []
}
]
}
]
我需要获取此 JSON 对象并将其以以下形式传输到我的脚本中的变量
var obj = k;
其中 k 是从此 URL 返回的 JSON 对象。我导入了 jQuery 并尝试了 $.getJSON 方法但没有成功。因此，如果您有一个带有 JSON 对象的 URL“ex.com”，我将如何将它原封不动地保存到一个变量中。因此，如果我将 url 的输出直接复制到一个变量中，例如:
var obj2 = copiedJSON;
我的函数按预期工作，但是当我像这样使用 $.getJSON 时:
$.getJSON('http://localhost/CS3744-N/GoalLiga/viz', function(json){
obj2 = json;
});
我的函数返回一个错误。 JSON 函数是否更改返回的输出。