在PHP中解析转换为GSON字符串的Java ArrayList

我使用MultipartEntityBuilder将包含JSON对象的Java ArrayList发送到PHP网页。我正在使用GSON将ArrayList转换为字符串，以便将数据发送到PHP网页。这一切都很好。这是我的Java代码：

for(List<JSONObject> rows: listOfValidationReports){
            HttpPost httppost = new HttpPost(url);
            Gson gson = new Gson(); 
            String rowsString = gson.toJson(rows);
            httpContext.setAttribute(HttpClientContext.COOKIE_STORE, cookieStore);
            MultipartEntityBuilder multipartEntity = MultipartEntityBuilder
                    .create();
            multipartEntity.setMode(HttpMultipartMode.BROWSER_COMPATIBLE);
            multipartEntity.addTextBody("validationReportRows", rowsString);
            httppost.setEntity(multipartEntity.build()); // Set parameter
            HttpResponse response = client.execute(httppost, httpContext);
}

我的问题是，当我将这些数据发送到PHP时，我不确定如何解析它并将其放入可以使用的PHP结构中。有很多关于如何发送数据的答案，但找不到任何关于如何在PHP中实际解析数据的答案

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我是否需要构建某种正则表达式匹配器来从数组中提取每个JSON对象，或者是否有方法将字符串转换回PHP ArrayList等价物，在该等价物中我可以循环遍历列表中的每个JSON对象。谢谢

不要自己滚动，使用内置功能。它将递归您发送的内容，并创建适当的对象、数组、关联数组等…可能的副本

[{"map":{"severity":"WARNING","offset":11902,"column":9,"length":22,"resourceUri":"test-fm.txt","lineNumber":411,"message":"Enum name should be less than 20 characters"}},{"map":{"severity":"WARNING","offset":12329,"column":9,"length":21,"resourceUri":"test-fm.txt","lineNumber":431,"message":"Enum name should be less than 20 characters"}},{"map":{"severity":"WARNING","offset":12544,"column":9,"length":39,"resourceUri":"test-fm.txt","lineNumber":441,"message":"Enum name should be less than 20 characters"}},{"map":{"severity":"WARNING","offset":12672,"column":9,"length":21,"resourceUri":"test-fm.txt","lineNumber":446,"message":"Enum name should be less than 20 characters"}},{"map":{"severity":"WARNING","offset":393,"column":3,"length":134,"resourceUri":"test-types-test@2016-06-01.txt","lineNumber":19,"message":"y extesion, version is missing"}}]