I'm trying to load some data via jQuery.getJSON() but it does not work:
我正在尝试通过jQuery.getJSON()加载一些数据,但它不起作用:
here is my JSON:
这是我的JSON:
{didwork=true,userid=123}
or it is
或者是
{didwork=false,userid=0}
here is my Javascript:
这是我的Javascript:
$.ajax({
data["username"] = "u"
data["password"] = "p";
url: https://www.myurl.com/json.php,
dataType: 'json',
data: data,
success: function(json){
//fill it into div
}
});
2 个解决方案
#1
7
your json string is wrong. it has to be
你的json字符串是错误的。它一定要是
{"didwork":true,"userid":123}
or
{"didwork":false,"userid":0}
never use =
and always use "
从不使用=并始终使用“
#2
4
Your javascript is wrong..
你的javascript错了..
you need to move the data
initialization outside the ajax call..
plus the url needs to be quoted.. (between '
)
你需要在ajax调用之外移动数据初始化..加上url需要引用..(在'之间)
var data = {};
data["username"] = "u";
data["password"] = "p";
this could also be represented with
这也可以用。来表示
var data = {'username': 'u', 'password': 'p'};
and the call
和电话
$.ajax({
url: 'https://www.myurl.com/json.php',
dataType: 'json',
data: data,
success: function(json){
//fill it into div
}
});
Your json is wrong
你的json错了
should be {"didwork":true,"userid":123}
应该是{“didwork”:true,“userid”:123}
If the url is to a different site then the one making the call it will fail due to same origin policy
如果网址是到另一个网站,那么进行呼叫的网站将因相同的原始策略而失败
#1
7
your json string is wrong. it has to be
你的json字符串是错误的。它一定要是
{"didwork":true,"userid":123}
or
{"didwork":false,"userid":0}
never use =
and always use "
从不使用=并始终使用“
#2
4
Your javascript is wrong..
你的javascript错了..
you need to move the data
initialization outside the ajax call..
plus the url needs to be quoted.. (between '
)
你需要在ajax调用之外移动数据初始化..加上url需要引用..(在'之间)
var data = {};
data["username"] = "u";
data["password"] = "p";
this could also be represented with
这也可以用。来表示
var data = {'username': 'u', 'password': 'p'};
and the call
和电话
$.ajax({
url: 'https://www.myurl.com/json.php',
dataType: 'json',
data: data,
success: function(json){
//fill it into div
}
});
Your json is wrong
你的json错了
should be {"didwork":true,"userid":123}
应该是{“didwork”:true,“userid”:123}
If the url is to a different site then the one making the call it will fail due to same origin policy
如果网址是到另一个网站,那么进行呼叫的网站将因相同的原始策略而失败