json verilerini php'ye nasıl geçireceğim?

Question

Dört metin kutum var ve tek bir düğmeye tıkladığınızda başka bir metin kutusu seti ekleyecektir. Sorunum, json verilerini işlem formuna nasıl aktarabilirim. i herhangi bir yardım muchly takdir

input type = "submit butonu bir id = "test" eklendi

<form action="<?php echo base_url().'resume/update'?>" method="post" id="send_form"> 


<!-- Experience Start --> 
       <div class="row"> 
        <div class="col-sm-12"> 
         <p>&nbsp;</p> 
         <h2>Experience</h2> 
        </div> 
       </div> 
       <div class="row experience"> 
        <div class="col-sm-6"> 
         <div class="form-group"> 
          <label for="resume-employer">Employer</label> 
          <input type="text" class="form-control" name="resumeEmployer" id="resume-employer" value="<?php echo set_value('resumeEmployer'); ?>" id="resume-employer" placeholder="Company name"> 
         </div> 
        </div> 
        <div class="col-sm-6"> 
         <div class="form-group"> 
          <label for="resume-experience-dates">Start/End Date</label> 
          <input type="text" class="form-control" name="resumeExperienceDates" name="<?php echo set_value('resumeExperienceDates'); ?>" id="resume-experience-dates" placeholder="e.g. April 2010 - June 2013"> 
         </div> 
        </div> 
       </div> 
       <div class="row"> 
        <div class="col-sm-6"> 
         <div class="form-group"> 
          <label for="resume-job-title">Job Title</label> 
          <input type="text" class="form-control" name="resumeJobTitle" id="resume-job-title" value="<?php echo set_value('resumeJobTitle'); ?>" placeholder="e.g. Web Designer"> 
         </div> 
        </div> 
        <div class="col-sm-6"> 
         <div class="form-group" id="resume-responsibilities-group"> 
          <label for="resume-responsibilities">Responsibilities (Optional)</label> 
          <input type="text" class="form-control" name="resumeResponsibilities" id="resume-responsibilities" value="<?php echo set_value('resumeResponsibilities');?>" placeholder="e.g. Developing new websites"> 
         </div> 
        </div> 
       </div> 
       <div class="row"> 
        <div class="col-sm-12"> 
         <hr class="dashed"> 
        </div> 
       </div> 
       <div class="row"> 
        <div class="col-sm-12"> 
         <p><a id="add-experience">+ Add Experience</a></p> 
         <hr> 
        </div> 
       </div> 
       <!-- Experience Start --> 
<div class="row text-center"> 
         <div class="col-sm-12"> 
          <p>&nbsp;</p> 
          <input type="submit" id="test" class="btn btn-primary btn-lg" value="Submit" /> 
         </div> 
        </div> 


    </form> 

$("#test").click(function(){ 


       var array = $('.experience').map(function() { 
        var obj = {}; 
        $(this).next().addBack().find('input:text').each(function() { 
         obj[this.id] = this.value; 
        }); 
        return obj; 
       }).get(); 

       $('#json').text(JSON.stringify(array, null, 2)); 


      }); 

ve benim form işlemi TIA: İşte benim kodudur.

Answer 1

Kullanım json_encode ve ref linkte json_decode işlevi örneği,

json_encode($arr); 
var_dump(json_decode($json)); 
http://php.net/manual/en/function.json-encode.php 
var_dump(json_decode($json, true));