Question

我通过javascript动态地在表单上生成附加字段，并且每一行包含三个字段，因此我希望能够使用ajax将显示在php中的所有字段中的所有值发布到

am能够动态生成其他字段并发布到php，但只发送了一个列。

+这里是表格

<table id="nameBoxWrap">
    <tr id="nameBox">
    <input type="text" id="phonenumber" name="phonenumber[]"  placeholder="phone number">
    <input type='text' id='firstname' name='firstname[]' placeholder='firstname'>
    <input type='text' id='lastname' name='lastname[]'  placeholder='lastname'>
    <input type="button" value="Load More" onclick="addNameSection()" >
    <input type="button" value="Submit" id="submitdata1"  style="margin-left:50px"><br>
    <input type="hidden" id="addSectionCount" value="1" name="addSectionCount">

    </tr>
</table>

+此处代码会动态生成其他字段

  <script type="text/javascript">
    function addNameSection(){
    var addSectionCount=$("#addSectionCount").val();
    addSectionCount++;
    $("#addSectionCount").val(addSectionCount);
    $("#nameBoxWrap").append('<tr id="nameBox'+addSectionCount+'"><td></td>'
        +'<td><input type="text" placeholder="phone number" id="phonenumber'+addSectionCount+'" name="phonenumber[]"></td>'
        +'<td><input type="text" id="firstname'+addSectionCount+'" name="firstname[]"  placeholder="firstname"></td>'
        +'<td><input type="text" id="lastname'+addSectionCount+'" name="lastname[]"  placeholder="lastname"><td>'
        +'<td><input type="button" value="REMOVE" onclick=removeNameSection("'+addSectionCount+'")></td></tr>');
}
</script>

+脚本，用于将表单数据提交到数据库

<script>


    $(function(){
        $('#submitdata1').on('click',function(e){
            e.preventDefault();  // do not allow the default form action
            var phoneS = $('input[name="phonenumber[]"]').serializeArray();
           var firstn = $('input[name="firstname[]"]').serializeArray();
          var lastname = $('input[name="lastname[]"]').serializeArray();

              $.ajax({
                method: "POST",
                url: "senddata.php",
                data:lastname 
                     });

        });
    });

    </script>

+这是我的PHP脚本

    foreach($_POST['lastname'] as $key=>$value){
            $phone = $_POST['phonenumber'][$key];
          $lastname = $value;
          $firstname = $_POST['firstname'][$key];
          $query =mysqli_query($db,"insert into itequipment(phone_number,firstname,lastname) values ('$phone','$firstname','$lastname')");  

    }

+我想在数据库表的一行中存储名字，姓氏，电话号码，但只能发送和存储姓氏。

Answer 1

尝试一下：

$('#submitdata1').on('click',function(e){
  e.preventDefault();  // do not allow the default form action
  
  var list = {};
  var i = 0;
  
   $( "tr" ).each(function() {
     var inputs = $(this).find('input');
     list[i] = {};

      inputs.each(function() {
        $input = $(this);
        list[i][$input.attr('name')] = $input.val();
      });

      i++;
  });

  var json_data = JSON.stringify(list);
  var json = JSON.parse(json_data);  
  
  $.ajax({
      url : 'senddata.php.php',
      type : 'POST',
      dataType : 'json',
      data : json
  });
   
});

<table>
  <tr class="line"> <!-- Line 1 -->
    <td>
      <input type="text" name='phonenumber' value="phone_1">
    </td>
    <td>
    <input type="text" name='firstname' value="firstname_1">
    </td>
    <td>
      <input type="text" name='lastname' value="lastname_1">
    </td>
  </tr>
  <tr class="line"> <!-- Line 2 -->
    <td>
      <input type="text" name='phonenumber' value="phone_2">
    </td>
    <td>
    <input type="text" name='firstname' value="firstname_2">
    </td>
    <td>
      <input type="text" name='lastname' value="lastname_2">
    </td>
  </tr>
</table>

console.log(json)的结果：

{0: {…}, 1: {…}}

0:
firstname: "firstname_1"
lastname: "lastname_1"
phonenumber: "phonenumber_1"

1:
firstname: "firstname_2"
lastname: "lastname_2"
phonenumber: "phonenumber_2"

通过由javascript动态创建的ajax获取多个php值

1 个答案: