Question

我有四个文本框，当你点击添加一个按钮时，它会添加另一组文本框。我的问题是如何将json数据传递给动作表单。这是我的代码：

 $("#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));


            });

和我的表单行动

    <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>

我添加了一个id =＆＃34; test＆＃34;在输入类型=＆＃34;提交按钮

非常感谢任何帮助。 TIA

Answer 1

在参考链接中使用json_encode和json_decode功能示例，

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

我将如何将json数据传递给php

1 个答案: