php - 为什么我的ajax无法获取输入表单中当前输入的字符串？

Question

我有一个简单的表单，然后我将表中的一些数据放置到每个输入表单值属性中。现在我的问题是，每当我在输入表单中输入新内容以更新数据时，它都无法获取当前输入的字符串，我不知道如何解决这个问题，因为我认为它正在获取回显的值在此更新页面上当前输入的字符串的，

这是我的前端

    <fieldset id="personaldetails">
    <legend>Personal Details</legend>
    Resume Title: <input type="text" name="resumetitle" id="resumetitle" value="<?php echo $v['ResumeTitle']; ?>" size="50" maxlength="50" /><br />
    Name: <input type="text" name="cvname" id="cvname" size="30" maxlength="30" value="<?php echo $v['Name']; ?>" /><br />
    DOB: <input type="text" id="datepicker" name="dob" value="<?php $date = new DateTime($v['DOB']); echo $date->format('m/d/Y'); ?>" /><br />
    Gender:  <input type="radio" name="gender" id="gender-male" value="1" <?php if($v['Gender'] == 1){ echo "checked"; } ?>/> <b>Male</b> |
     <input type="radio" name="gender" id="gender-female" value="0" <?php if($v['Gender'] == 0){ echo "checked"; } ?>/> <b>Female</b><br /><br />
    <input type="hidden" name="cvid" id="cvid" value="<?php echo $v['ResumeID']; ?>" />
    <button name="pdetails" id="pdetails">Update</button>
    </fieldset><br /><br />

//这是我的js

$(document).ready(function(){
 var resumetitle = $('#resumetitle').val();
 var cvid = $('input[type="hidden"]').val();
 var name = $('#cvname').val();
 var dob = $('#datepicker').val();
 var gender = $('input[name="gender"]:checked').val();


$('button#pdetails').click(function(){
   $.ajax({
      type: "POST",
      url: "classes/ajax.resumeupdate.php",
      data: "resumeid="+cvid+"&resumetitle="+resumetitle+"&name="+name+"&dob="+dob+"&gender="+gender,
      success: function(){
        //window.location = "resumeview.php?cvid="+cvid;
      },
   });
});


});

//这是我的 php 代码

require 'class.resume.php';

$db = new Resume();

if(isset($_POST['resumetitle']) || isset($_POST['name']) || isset($_POST['dob']) ||
   isset($_POST['gender']) || isset($_POST['cvid'])){
    $result =  $db->updatepdetails($_POST['resumetitle'],$_POST['name'],$_POST['dob'],$_POST['gender'],$_POST['cvid']);
    if($result){
      echo "success!";
    } else {
      echo "failed! ".$db->error;
    }
   }

Answer 1

您仅读取文档中准备好的值，将该代码移至点击事件中:

$(document).ready(function(){
   $('button#pdetails').click(function(){
     var resumetitle = $('#resumetitle').val();
     var cvid = $('input[type="hidden"]').val();
     var name = $('#cvname').val();
     var dob = $('#datepicker').val();
     var gender = $('input[name="gender"]:checked').val();
     $.ajax({
       type: "POST",
       url: "classes/ajax.resumeupdate.php",
       data: "resumeid="+cvid+"&resumetitle="+resumetitle+"&name="+name+"&dob="+dob+"&gender="+gender,
       success: function(){
           //window.location = "resumeview.php?cvid="+cvid;
       },
     });
   });
});