[英]JSON Object w/variables in it
我需要創建一個JSON對象“bio”,其中包含很多變量:
var bio = {
"name": "my name",
"role": "Web Developer",
"contacts":
{
"mobile": "my phone",
"email": "my@email.address",
"github": "https://github.com/",
"twitter": "https://twitter.com/",
"location": "Los Angeles, CA"
},
"welcomeMessage": "Welcome to my online resume.",
"skills": ["HTML5","CSS3","JavaScript","Bootstrap", "Angular", "CoffeeScript", "W3"],
"biopic": "http://placehold.it/150x150",
"display": displayFunc(){
}
};
當我嘗試運行時:
var formattedName = HTMLheaderName.replace('%data%', bio.name);
var formattedRole = HTMLheaderRole.replace('%data%', bio.role);
$('#header').prepend(formattedRole);
$('#header').prepend(formattedName);
什么都沒發生。 我認為錯誤是在“contacts”變量中的某個地方,因為如果我注釋掉聯系人及其下面的所有內容,則會顯示名稱和角色。 但是,如果我評論出welcomeMessage以及下面的所有內容,我仍然一無所獲。
編輯:對於這個類,聯系人變量必須是:
contacts : an object with
mobile: string
email: string
github: string
twitter: string (optional)
location: string
如果你在displayFunc(){}
之前添加function
,那么它應該工作(如提到的@BubbleHacker)
在這里演示: https : //jsfiddle.net/mrLh3pz7/
var bio = {
"name": "my name",
"role": "Web Developer",
"contacts": [
{
"mobile": "my phone",
"email": "my@email.address",
"github": "https://github.com/",
"twitter": "https://twitter.com/",
"location": "Los Angeles, CA"
}
],
"welcomeMessage": "Welcome to my online resume.",
"skills": ["HTML5","CSS3","JavaScript","Bootstrap", "Angular", "CoffeeScript", "W3"],
"biopic": "http://placehold.it/150x150",
"display": function displayFunc(){}
};
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