jQuery POST 表單數據

Question

當我單擊提交時，我希望將所有表單數據發布到 process.php。 然后在 process.php 上，我想回顯 POST 數據，最后將結果 div 中的所有內容替換為 process.php 中所做的內容。

<script type="text/javascript">
    $(document).ready(function(){
        $("#myform").submit( function () {    
            $.ajax({   
                type: "POST",
                dataType: "html", 
                cache: false,  
                url: "process.php",   
                success: function(data){
                    $("#results").html(data);                       
                }   
            });   
            return false;   
        });

        //$("#myform").submit( function () {
            //$('#results').html("yay");                    
        //}  
            // });  
        //} );          
    });
</script>

<form name="myform" id="myform" action="" method="POST">  
<!-- The Name form field -->
    <label for="name" id="name_label">zoom</label>  
    <input type="text" name="zoom" id="zoom" size="30" value=""/>  
    <br>
</select>


<!-- The Submit button -->
    <input type="submit" name="submit" value="Submit"> 
</form>

<!-- FORM END ----------------------------------------  -->


<!-- RESULTS START ---------------------------------------- -->
    <div id="results">nooooooo<?PHP $_SESSION[''] ?><div>
    <!-- <input type="image" name="mapcoords" border="0" src="mapgen.php"> ---- -->
<!-- RESULTS END ---------------------------------------- -->

Answer 1

您可以調用$.post傳遞表單數據序列化。 像這樣：

<script type="text/javascript">
        $(document).ready(function(){
            $("#myform").submit( function () {    
              $.post(
               'process.php',
                $(this).serialize(),
                function(data){
                  $("#results").html(data)
                }
              );
              return false;   
            });   
        });
</script>

Answer 2

$("#myform").submit( function () {    
    $.ajax({   
        type: "POST",
        data : $(this).serialize(),
        url: "process.php",   
        success: function(data){
            $("#results").html(data);                       
        }   
    });   
    return false;   
});

測試一下

Answer 3

只有另一種方法可以做到這一切。 plugin.此示例使用 jQuery插件。 在這里，所有表單域都會自動驗證。 from here :從這里下載 ：

https://ajax.googleapis.com/ajax/libs/jquery/1.8.1/jquery.min.js

from here :從這里下載：

http://jquery.bassistance.de/validate/jquery-validation-1.10.0.zip

    <script type="text/javascript" src="jquery-1.7.1.min.js"></script>
    <script type="text/javascript" src="jquery.validate.1.7.min.js"></script>
    <script type="text/javascript">
    $(document).ready(function(){
      $("#customForm").validate({
    debug: false,
    rules: {
      name: {required:true,minlength:8,maxlength:8},
      email: {required:true,email:true},
    },
        submitHandler: function(form) {
     // do other stuff for a valid form 
     $('#rsltx').html('<img src="WhiteLoading.gif"> Processing... please wait');
     $.post('post.php', $("#customForm").serialize(), function(data) {
                        $('#rsltx').html(data);
         });
        }
  });
});
</script>

<form method="post" id="customForm" action="">
<div>
 <label for="name">Name</label>
 <input id="name" name="name" type="text" autocomplete="off" required/>
</div>
<div>
 <label for="email">E-mail</label>
 <input id="email" name="email" type="email" autocomplete="off" required/>
</div>
<div>
 <input id="send" name="send" type="submit" value="Send" />
</div>
</form>

Answer 4

 var form = $('#mob_ticket_form')[0]; var data = new FormData(form); $.ajax({ type: "POST", enctype: 'multipart/form-data', url: '{{route("submit_ticket")}}', data: data, processData: false, contentType: false, cache: false, timeout: 800000, success: function(response) { if (response.status == 200) { $("#exampleModalCenter").modal('show'); } else { alert(response.message); } }, error: function(response) { // console.log(response) notify('Something went Wrong!', 'danger') } });