我正在使用JQuery dForm插件来创建表单。它适用于以下代码:
$("#myform").dform({
"action" : "index.html",
"method" : "get",
"html" :
[
{
"type" : "p",
"html" : "You must login"
},
{
"name" : "username",
"id" : "txt-username",
"caption" : "Username",
"type" : "text",
"placeholder" : "E.g. user@example.com"
},
{
"name" : "password",
"caption" : "Password",
"type" : "password"
},
{
"type" : "submit",
"value" : "Login"
}
]
});
但是如果我将表单定义移动到JSON文件中,如:
"action" : "index.html",
"method" : "get",
"html" :
[
{
"type" : "p",
"html" : "You must login"
},
{
"name" : "username",
"id" : "txt-username",
"caption" : "Username",
"type" : "text",
"placeholder" : "E.g. user@example.com"
},
{
"name" : "password",
"caption" : "Password",
"type" : "password"
},
{
"type" : "submit",
"value" : "Login"
}
]
尝试使用以下命令从json加载表单定义:
$("#myform").dform('test.json', function(data) {
this //-> Generated $('#myform')
data //-> data from path/to/form.json
});
表单无法加载。我是否需要在JSon文件或代码中添加其他内容?
非常感谢任何帮助。
谢谢, 卡洛斯
答案 0 :(得分:0)
这是JSon中的一个问题。基本上JSon DForm定义可以是:
{"id":"form","action":"","html" :[{"type" : "p","html" : "You must login"},{"name" : "username","id" : "txt-username","caption" : "Username","type" : "text","placeholder" : "E.g. user@example.com","onkeypress":"myKeypress"},{"name" : "password","caption" : "Password","type" : "password","onkeypress":"myKeypress"}]}
卡洛斯。