I get id and category name from mysql database.
我从mysql数据库中获取id和类别名称。
When I am alerting a result, I get:
当我提醒结果时,我得到:
[{"id":"197","category":"Damskie"},"id":"198","category":"M\u0119skie"}]
(Is this object?)
(这个对象?)
-
How can I print a result like this:
如何打印如下结果:
Damskie
Damskie
M\u0119skie
中号\ u0119skie
-
M\u0119ski - has bad encoding. It should be Męskie. How can I change this?
M \ u0119ski - 编码错误。它应该是Męskie。我怎么能改变这个?
4 个解决方案
#1
24
var arrofobject = [{"id":"197","category":"Damskie"},{"id":"198","category":"M\u0119skie"}];
$.each(arrofobject, function(index, val) {
console.log(val.category);
});
#2
6
What you have from the server is a string like below:
你从服务器得到的是一个如下字符串:
var data = '[{"id":"197","category":"Damskie"},{"id":"198","category":"M\u0119skie"}]';
Then you can use JSON.parse
function to change it to an object. Then you access the category like below:
然后,您可以使用JSON.parse函数将其更改为对象。然后您访问如下所示的类别:
var dataObj = JSON.parse(data);
console.log(dataObj[0].category); //will return Damskie
console.log(dataObj[1].category); //will return Męskie
#3
2
Your result is currently in string format, you need to parse it as json.
您的结果当前是字符串格式,您需要将其解析为json。
var obj = $.parseJSON(result);
alert(obj[0].category);
Additionally, if you set the dataType of the ajax call you are making to json
, you can skip the $.parseJSON()
step.
另外,如果你设置了json调用的ajax调用的dataType,你可以跳过$ .parseJSON()步骤。
#4
0
var arrofobject = [{"id":"197","category":"Damskie"},{"id":"198","category":"M\u0119skie"}];
var data = arrofobject.map(arrofobject => arrofobject);
console.log(data)
for more details please look at jQuery.map()
有关更多详细信息,请查看jQuery.map()
#1
24
var arrofobject = [{"id":"197","category":"Damskie"},{"id":"198","category":"M\u0119skie"}];
$.each(arrofobject, function(index, val) {
console.log(val.category);
});
#2
6
What you have from the server is a string like below:
你从服务器得到的是一个如下字符串:
var data = '[{"id":"197","category":"Damskie"},{"id":"198","category":"M\u0119skie"}]';
Then you can use JSON.parse
function to change it to an object. Then you access the category like below:
然后,您可以使用JSON.parse函数将其更改为对象。然后您访问如下所示的类别:
var dataObj = JSON.parse(data);
console.log(dataObj[0].category); //will return Damskie
console.log(dataObj[1].category); //will return Męskie
#3
2
Your result is currently in string format, you need to parse it as json.
您的结果当前是字符串格式,您需要将其解析为json。
var obj = $.parseJSON(result);
alert(obj[0].category);
Additionally, if you set the dataType of the ajax call you are making to json
, you can skip the $.parseJSON()
step.
另外,如果你设置了json调用的ajax调用的dataType,你可以跳过$ .parseJSON()步骤。
#4
0
var arrofobject = [{"id":"197","category":"Damskie"},{"id":"198","category":"M\u0119skie"}];
var data = arrofobject.map(arrofobject => arrofobject);
console.log(data)
for more details please look at jQuery.map()
有关更多详细信息,请查看jQuery.map()