使用 Javascript 从 html FORM 发送帖子数据

Send post data from html FORM with Javascript?

本文关键字：数据 FORM Javascript html 使用      更新时间：2023-09-26

我正在尝试从空操作发送帖子数据，并使用javascript将信息发送到必须处理帖子信息的文件。

这是我的代码：

<HTML>
<head>
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
</head>
            <script>    
    function post_to_url(path, params, method) {
        method = method || "post";
        var form = document.createElement("form");
        form._submit_function_ = form.submit;
        form.setAttribute("method", method);
        form.setAttribute("action", path);
        for(var key in params) {
            var hiddenField = document.createElement("input");
            hiddenField.setAttribute("type", "hidden");
            hiddenField.setAttribute("name", key);
            hiddenField.setAttribute("value", params[key]);
            form.appendChild(hiddenField);
        }
        document.body.appendChild(form);
        form._submit_function_();
    }
    post_to_url("./postinfo.php", { submit: "submit" } );
        </script>   
<form method="post" onsubmit="return post_to_url()">
<input type="text" name="ime">
<input type="submit" id="submit" name="submit" value="Send">
</form>

那么我如何在我的 html 表单中使用空操作将帖子数据发送到带有 javascript 的postinfo.php呢？

提前感谢！

你很幸运，因为有一个名为"Ajax"的JQuery函数可以为你处理这个问题！

您需要做的就是调用 JQuery，并使用如下代码：

$('#form_id').on('submit', function(e){
    e.preventDefault();
    $.ajax({
       type: "POST",
       url: "/postinfo.php",
       data: $(this).serialize(),
       success: function() {
         alert('success');
       }
    });
});

function post_to_url(path, params, method) {
    method = method || "post";
    var form = document.createElement("form");

   _submit_function_ = form.submit;
    form.setAttribute("method", method);
    form.setAttribute("action", path);
    for(var key in params) {
        var hiddenField = document.createElement("input");
        hiddenField.setAttribute("type", "hidden");
        hiddenField.setAttribute("name", key);
        hiddenField.setAttribute("value", params[key]);
        form.appendChild(hiddenField);
    }
    document.body.appendChild(form);
    form._submit_function_();
}
post_to_url("./postinfo.php", { submit: "submit" } );

更改为 

post_to_url("/postinfo.php", { submit: "submit" } );

试试这段代码。

    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script>
<script type="text/javascript">
    function validate() {
        var ra = document.getElementById("uname").value;
        var rag = document.getElementById("pwd").value;
        $.ajax({
       type: "POST",
        url: "/login",
            contentType: "application/json",
         dataType: 'json',
       data:JSON.stringify({
           username:ra,
           password:rag
       }),
       success: function() {
         alert('success');
       }
    });
        console.log(ra, rag)
    }
</script>
<html>
<form name="myform">
<p>ENTER USER NAME <input id="uname" type="text" name="username"></p>
    <p>ENTER PASSWORD <input type="password" id="pwd" name="pwd"></p>
<input type="button" value="Check In" name="Submit" onclick= "validate()">

</form>
</html>

<html>
<form action="{{ url_for('details') }}" method="post">
<p>ENTER NAME<input id="uname" type="text" name="username"></p>
    <p>ENTER DESIGNATION<input type="text" id="pwd" name="pwd"></p>
    <!--<P>ENTER EMAIL-ID</EMAIL-ID><input id="email" type="text" name="email-id"></P>-->
<input type="submit" value="Submit" name="Submit">

</form>
</html>`

 __author__ = 'raghavendra'
from flask import Flask, render_template, request, jsonify
import os
import sys
# os.environ.setdefault("DJANGO_SETTINGS_MODULE", "your_project.settings")
# sys.path.append(('/path/to/django/project'))
app = Flask(__name__)
@app.route('/')
def index():
    return render_template('temp.html')
@app.route('/save',methods=['GET','POST'])
def details():
    a = request.form.get('username')
    b = request.form.get('pwd')
    print a, b
if __name__ == '__main__':
    app.run()
相关文章：