使用Play Framework以JSON形式提交表单 [英] Submit Form as JSON with Play Framework

问题描述

我试图将表单作为JSON对象提交，因为我想创建一个带有play的REST API.

I am trying to submit a form as a JSON object because I want to create a REST API with play.

我遇到的问题是Play告诉我这不是有效的JSON.

The issue that I have is that Play tells me that is not a valid JSON.

我的FORM代码:

@(form : Form[Product]) @main("Create Form"){
@helper.form(routes.Products.createProduct, 'enctype -> "application/json"){
    @helper.inputText(form("name"))
    <button>Commit</button>
} }

控制器代码:

// Read JSON an tell if it has a name Path
@BodyParser.Of(BodyParser.TolerantJson.class)
public static Result createProduct() {
    JsonNode json = request().body().asJson();
    String name = json.findPath("name").textValue();
    if (name == null) {
        return badRequest("Not JSON");
    } else {
        return ok(name);
    }
}

什么是最好的方法?阅读有关使用Ajax提交的内容，但是由于我是play的新手，所以我不知道如何使用Play的表单语法来做到这一点.

Whats the best way to do this? a read about submitting with Ajax but because I am new with play I don´t figure it out the way to do this with Play´s form syntax.

推荐答案

您可以轻松地使用jQuery(确保您的头部包含jQuery)和基于

You can do it easily with jQuery (so make sure you have jQuery included in your head) and formAsJson() function based on serializeObject function.

@helper.form(routes.Products.createProduct(), 'id -> "myform") {
    @helper.inputText(jsonForm("name"))
    <button>Commit</button>
}

<script>
    $.fn.formAsJson = function(){
        var o = {};
        var a = this.serializeArray();
        $.each(a, function () {
            if (o[this.name] !== undefined) {
                if (!o[this.name].push) {
                    o[this.name] = [o[this.name]];
                }
                o[this.name].push(this.value || '');
            } else {
                o[this.name] = this.value || '';
            }
        });
        return JSON.stringify(o)
    };

    var $myform = $("#myform");
    $myform.on('submit', function () {
        $.ajax({
            url: $myform.attr('action'),
            type: $myform.attr('method'),
            contentType: "application/json",
            data: $myform.formAsJson(),
            success:function(){
                alert("Great! Everything's OK!");
            },
            error: function(){
                alert("Booo, something wrong :(");
            }

        });
        return false;
    });
</script>

和您的createProduct()操作可能类似于:

and your createProduct() action could look just like:

public static Result createProduct() {
    JsonNode json = request().body().asJson();
    String name = json.findPath("name").textValue();
    if (json==null || name==null || ("").equals(name.trim())){
        return badRequest();
    }

    return ok();
}

这篇关于使用Play Framework以JSON形式提交表单的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！