使用 Ajax 使用 php 将记录插入到 mysql 数据库
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时间：2020-08-26 00:15:17 来源:igfitidea点击:
Insert records to mysql database with php using Ajax
提问by Ashish Kumar
How to do coding of this Code With Using Ajax.Please Help.I am Bignner here and i have written this code it's working but i want to use with ajax this because don't want to reload the page...?
如何使用 Ajax 编写此代码。请帮助。我在这里是 Bignner，我已经编写了这段代码，它正在运行，但我想与 ajax 一起使用，因为不想重新加载页面......？
PHP File
PHP文件
//Code For Making Form And getting Data…..
<html>
<body>
Fill -ID,NAME,EMAIL_ID,PASSWORD,CREDITS,
<form action="Form_Data.php" method="post">
ID: <input type="text" name="ID"><br><br>
NAME: <input type="text" name="NAME"><br><br>
PASSWORD: <input type="text" name="PASSWORD"><br><br>
CREDITS: <input type="text" name="CREDITS"><br><br>
E_mail: <input type="text" name="EMAIL_ID"><br><br>
CREATED_ON:<input type="text" name="CREATED_ON"><br><br>
MODIFIED_ON:<input type="text" name="MODIFIED_ON"><br><br>
<input type="submit">
</form>
</body>
</html>
//code for taking data from form data.
//从表单数据中获取数据的代码。
<html>
<?php
include 'connnect.php';
mysql_set_charset('utf8');
//query for insert data into tables
$ID = $_POST['ID'];
$NAME =$_POST['NAME'];
$EMAIL_ID =$_POST['EMAIL_ID'];
$PASSWORD =$_POST['PASSWORD'];
$CREDITS =$_POST['CREDITS'];
$CREATED_ON=$_POST['CREATED_ON'];
$MODIFIED_ON=$_POST['MODIFIED_ON'];
$query = "INSERT INTO `user_table`
(`ID`,`NAME`,`EMAIL_ID`,`PASSWORD`,`CREDITS`,`CREATED_ON`,`MODIFIED_ON`)
VALUES
('$ID','$NAME','$EMAIL_ID','$PASSWORD','$CREDITS','$CREATED_ON','$MODIFIED_ON')";
$query_run= mysql_query($query);
$retval=mysql_query($query,$conn);
if ($query_run)
{ echo 'It is working';
}
mysql_close($conn);
?>
</html>
I have Tried Yet ... Is Blewo...
我已经试过了......是Blewo......
file for html and ajax
<html>
<HEAD>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
</HEAD>
<body>
<div id="status_text">
Fill -ID,NAME,EMAIL_ID,PASSWORD,CREDITS,
<form onsubmit="return false" method="post">
ID: <input type="text" id="ID" name="ID"><br><br>
NAME: <input type="text" id="NMAE" name="NAME"><br><br>
PASSWORD: <input type="text" id= "PASSWORD"name="PASSWORD"><br><br>
CREDITS: <input type="text" Id= "CREDITS"name="CREDITS"><br><br>
Email_ID: <input type="text" id="Email_ID"name="EMAIL_ID"><br><br>
CREATED_ON:<input type="text" id="CREATED_ON" name="CREATED_ON"><br><br>
MODIFIED_ON:<input type="text" id="MODIFIED_ON" name="MODIFIED_ON"><br><br>
<input type="submit" id="btn_submit" name="submit" value="Send">
</div>
<script>
//on the click of the submit button
$("#btn_submit").click(function(){
//get the form values
var ID = $('#ID').val();
var NAME = $('#NAME').val();
var PASSWORD = $('#PASSWORD').val();
var CREDITS = $('#CREDITS').val();
var EMAIL_ID = $('#EMAIL_ID').val();
var CREATED_ON = $('#CREATED_ON').val();
var MODIFIED_ON = $('#MODIFIED_ON').val();
//make the postdata
var postData = '&ID='+ID+'&NAME='+NAME+'&PASSWORD='+PASSWORD+'&REDITS'+CREDITS+'&EMAIL_ID'+EMAIL_ID+'&CREATED_ON'+CREATED_ON+'&MODIFIED_ON'+MODIFIED_ON;
//call your .php script in the background,
//when it returns it will call the success function if the request was successful or
//the error one if there was an issue (like a 404, 500 or any other error status)
});
$.ajax({
url : "Form_Data.php",
type: "POST",
data : postData,
success: function(data,status, xhr)
{
//if success then just output the text to the status div then clear the form inputs to prepare for new data
$("#status_text").html(data);
$('#ID').val();
$('#NAME').val('');
$('#PASSWORD').val('');
$('#EMAIL_ID').val('');
$('#CREATED_ON').val('');
$('#MODIFIED_ON').val('');
}
});
</script>
</form>
</body>
</div>
</html>
code for query...
查询代码...
<html>
<?php
include 'connnect.php';
mysql_set_charset('utf8');
//query for insert data into tables
$ID = $_POST['ID'];
$NAME =$_POST['NAME'];
$EMAIL_ID =$_POST['EMAIL_ID'];
$PASSWORD =$_POST['PASSWORD'];
$CREDITS =$_POST['CREDITS'];
$CREATED_ON=$_POST['CREATED_ON'];
$MODIFIED_ON=$_POST['MODIFIED_ON'];
$query = "INSERT INTO `user_table`
(`ID`,`NAME`,`EMAIL_ID`,`PASSWORD`,`CREDITS`,`CREATED_ON`,`MODIFIED_ON`)
VALUES
('$ID','$NAME','$EMAIL_ID','$PASSWORD','$CREDITS','$CREATED_ON','$MODIFIED_ON')";
$query_run= mysql_query($query);
$retval=mysql_query($query,$conn);
if ($query_run)
{ echo 'It is working';
}
mysql_close($conn);
?>
</html>
回答by Ashish Kumar
I have Solved It ... How To Use Ajax and MYSQL...PHP CODE
我已经解决了...如何使用 Ajax 和 MYSQL...PHP 代码
<?php
include 'connnect.php';
mysql_set_charset('utf8');
//query for insert data into tables
$ID = $_POST['ID'];
$NAME =$_POST['NAME'];
$EMAIL_ID =$_POST['EMAIL_ID'];
$PASSWORD =$_POST['PASSWORD'];
$CREDITS =$_POST['CREDITS'];
$CREATED_ON=$_POST['CREATED_ON'];
$MODIFIED_ON=$_POST['MODIFIED_ON'];
$query = "INSERT INTO `user_table`
(`NAME`,`EMAIL_ID`,`PASSWORD`,`CREDITS`,`CREATED_ON`,`MODIFIED_ON`)
VALUES
('$NAME','$EMAIL_ID','$PASSWORD','$CREDITS','$CREATED_ON','$MODIFIED_ON')";
$query_run= mysql_query($query);
// $retval=mysql_query($query,$conn);
if ($query_run)
{
echo 'It is working';
}
mysql_close($conn);
?>
HTML FILE....
HTML 文件....
<html>
<HEAD>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
</HEAD>
<body>
<div id="status_text">
Fill -ID,NAME,EMAIL_ID,PASSWORD,CREDITS,
ID: <input type="text" id="ID" name="ID"><br><br>
NAME: <input type="text" id="NAME" name="NAME"><br><br>
PASSWORD: <input type="text" id= "PASSWORD"name="PASSWORD"><br><br>
CREDITS: <input type="text" Id= "CREDITS"name="CREDITS"><br><br>
Email_ID: <input type="text" id="EMAIL_ID"name="EMAIL_ID"><br><br>
CREATED_ON:<input type="text" id="CREATED_ON" name="CREATED_ON"><br><br>
MODIFIED_ON:<input type="text" id="MODIFIED_ON" name="MODIFIED_ON"><br><br>
<input type="submit" id="btn_submit" name="submit" value="Send"/>
</div>
<script>
<!--
//Browser Support Code
function ajaxFunction(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
}catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
} }
//on the click of the submit button
$("#btn_submit").click(function(){
//get the form values
var ID = $('#ID').val();
var NAME = $('#NAME').val();
var PASSWORD = $('#PASSWORD').val();
var CREDITS = $('#CREDITS').val();
var EMAIL_ID = $('#EMAIL_ID').val();
var CREATED_ON = $('#CREATED_ON').val();
var MODIFIED_ON = $('#MODIFIED_ON').val();
// make the postdata
// var postData = '&ID='+ID+'&NAME='+NAME+'&PASSWORD='+PASSWORD+'&CREDITS'+CREDITS+'&EMAIL_ID'+EMAIL_ID+'&CREATED_ON'+CREATED_ON+'&MODIFIED_ON'+MODIFIED_ON;
// alert(postData);
var myData={"ID":ID,"NAME":NAME,"PASSWORD":PASSWORD,"CREDITS":CREDITS,"EMAIL_ID":EMAIL_ID,"CREATED_ON":CREATED_ON,"MODIFIED_ON":MODIFIED_ON};
//call your .php script in the background,
//when it returns it will call the success function if the request was successful or
//the error one if there was an issue (like a 404, 500 or any other error status)
$.ajax({
url : "Form_Data.php",
type: "POST",
data : myData,
success: function(data,status,xhr)
{
//if success then just output the text to the status div then clear the form inputs to prepare for new data
$("#status_text").html(data);
$('#ID').val();
$('#NAME').val('');
$('#PASSWORD').val('');
$('#EMAIL_ID').val('');
$('#CREATED_ON').val('');
$('#MODIFIED_ON').val('');
}
});
});
</script>
</body>
</div>
</html>
回答by Tariq hussain
change your script because your ajax was outside the click function
更改您的脚本，因为您的 ajax 在点击功能之外
//on the click of the submit button
$("#btn_submit").click(function(){
//get the form values
var ID = $('#ID').val();
var NAME = $('#NAME').val();
var PASSWORD = $('#PASSWORD').val();
var CREDITS = $('#CREDITS').val();
var EMAIL_ID = $('#EMAIL_ID').val();
var CREATED_ON = $('#CREATED_ON').val();
var MODIFIED_ON = $('#MODIFIED_ON').val();
//make the postdata
var postData = '&ID='+ID+'&NAME='+NAME+'&PASSWORD='+PASSWORD+'&REDITS'+CREDITS+'&EMAIL_ID'+EMAIL_ID+'&CREATED_ON'+CREATED_ON+'&MODIFIED_ON'+MODIFIED_ON;
//call your .php script in the background,
//when it returns it will call the success function if the request was successful or
//the error one if there was an issue (like a 404, 500 or any other error status)
$.ajax({
url : "Form_Data.php",
type: "POST",
data : postData,
success: function(data,status, xhr)
{
//if success then just output the text to the status div then clear the form inputs to prepare for new data
$("#status_text").html(data);
$('#ID').val();
$('#NAME').val('');
$('#PASSWORD').val('');
$('#EMAIL_ID').val('');
$('#CREATED_ON').val('');
$('#MODIFIED_ON').val('');
}
});
});
</script>
and change your php code to this
并将您的 php 代码更改为此
<?php
include 'connnect.php';
mysql_set_charset('utf8');
//query for insert data into tables
if(isset($_POST['NAME'])){
$ID = $_POST['ID'];
$NAME =$_POST['NAME'];
$EMAIL_ID =$_POST['EMAIL_ID'];
$PASSWORD =$_POST['PASSWORD'];
$CREDITS =$_POST['CREDITS'];
$CREATED_ON=$_POST['CREATED_ON'];
$MODIFIED_ON=$_POST['MODIFIED_ON'];
$query = "INSERT INTO `user_table`
(`ID`,`NAME`,`EMAIL_ID`,`PASSWORD`,`CREDITS`,`CREATED_ON`,`MODIFIED_ON`)
VALUES
('$ID','$NAME','$EMAIL_ID','$PASSWORD','$CREDITS','$CREATED_ON','$MODIFIED_ON')";
$query_run= mysql_query($query);
$retval=mysql_query($query,$conn);
if ($query_run)
{ echo 'It is working';
}
}
mysql_close($conn);
?>
回答by Tariq hussain
In this code I'm just submitting your two input fields, the rest you can add by yourself. Try this:
在这段代码中，我只是提交了您的两个输入字段，其余的您可以自己添加。尝试这个：
<html>
<body>
Fill -ID,NAME,EMAIL_ID,PASSWORD,CREDITS,
<form action="Form_Data.php" method="post">
NAME: <input id="name" type="text" name="NAME"><br><br>
PASSWORD: <input id="password" type="text" name="PASSWORD"><br><br>
<input type="submit" id="submit">
</form>
</body>
</html>
$("#submit").click(function() {
var name= $("#name").val();
var password= $("#password").val();
$.ajax({
type: "POST",
url: "your_php_path.php",
data: 'name=' + name+ '&password=' + password,
success: function(result) {
alert(result);
}
});
});
