【发布时间】:2014-01-27 21:23:52
【问题描述】:
来自他们的官方文档:
Custom ajax validation made simple.
parsley-remote-method="POST" to change default GET call.
parsley-remote-datatype="jsonp" if you make cross domain ajax call and expect jsonp
Parsley will accept these valid returns with a 200 response code: 1, true, { "success": "..." } and assume false otherwise
You can show frontend server-side specific error messages by returning { "error": "your custom message" } or { "message": "your custom message" }
但是,如果我想在 post 请求中发送变量,我该怎么办?
【问题讨论】:
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我认为,如果您在输入中放置名称属性,则输入具有相同名称的变量,输入值将与发布请求一起发送。
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如果我想发送多个值怎么办?
标签: javascript php ajax