带有PHP的jQuery Ajax POST示例

我正在尝试将数据从表单发送到数据库。这是我使用的表格:

<form name="foo" action="form.php" method="POST" id="foo">
    <label for="bar">A bar</label>
    <input id="bar" name="bar" type="text" value="" />
    <input type="submit" value="Send" />
</form>

典型的方法是提交表单,但这会导致浏览器重定向。使用jQuery和Ajax,是否可以捕获表单的所有数据并将其提交给PHP脚本(例如form.php)?

AMandy2020/03/11 15:16:13

This is a very good article that contains everything that you need to know about jQuery form submission.

Article summary:

Simple HTML Form Submit

HTML:

<form action="path/to/server/script" method="post" id="my_form">
    <label>Name</label>
    <input type="text" name="name" />
    <label>Email</label>
    <input type="email" name="email" />
    <label>Website</label>
    <input type="url" name="website" />
    <input type="submit" name="submit" value="Submit Form" />
    <div id="server-results"><!-- For server results --></div>
</form>

JavaScript:

$("#my_form").submit(function(event){
    event.preventDefault(); // Prevent default action
    var post_url = $(this).attr("action"); // Get the form action URL
    var request_method = $(this).attr("method"); // Get form GET/POST method
    var form_data = $(this).serialize(); // Encode form elements for submission

    $.ajax({
        url : post_url,
        type: request_method,
        data : form_data
    }).done(function(response){ //
        $("#server-results").html(response);
    });
});

HTML Multipart/form-data Form Submit

To upload files to the server, we can use FormData interface available to XMLHttpRequest2, which constructs a FormData object and can be sent to server easily using the jQuery Ajax.

HTML:

<form action="path/to/server/script" method="post" id="my_form">
    <label>Name</label>
    <input type="text" name="name" />
    <label>Email</label>
    <input type="email" name="email" />
    <label>Website</label>
    <input type="url" name="website" />
    <input type="file" name="my_file[]" /> <!-- File Field Added -->
    <input type="submit" name="submit" value="Submit Form" />
    <div id="server-results"><!-- For server results --></div>
</form>

JavaScript:

$("#my_form").submit(function(event){
    event.preventDefault(); // Prevent default action
    var post_url = $(this).attr("action"); // Get form action URL
    var request_method = $(this).attr("method"); // Get form GET/POST method
    var form_data = new FormData(this); // Creates new FormData object
    $.ajax({
        url : post_url,
        type: request_method,
        data : form_data,
        contentType: false,
        cache: false,
        processData: false
    }).done(function(response){ //
        $("#server-results").html(response);
    });
});

I hope this helps.

JinJin村村2020/03/11 15:16:13

HTML

    <form name="foo" action="form.php" method="POST" id="foo">
        <label for="bar">A bar</label>
        <input id="bar" class="inputs" name="bar" type="text" value="" />
        <input type="submit" value="Send" onclick="submitform(); return false;" />
    </form>

JavaScript

   function submitform()
   {
       var inputs = document.getElementsByClassName("inputs");
       var formdata = new FormData();
       for(var i=0; i<inputs.length; i++)
       {
           formdata.append(inputs[i].name, inputs[i].value);
       }
       var xmlhttp;
       if(window.XMLHttpRequest)
       {
           xmlhttp = new XMLHttpRequest;
       }
       else
       {
           xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
       }
       xmlhttp.onreadystatechange = function()
       {
          if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
          {

          }
       }
       xmlhttp.open("POST", "insert.php");
       xmlhttp.send(formdata);
   }