Php 在第一个输入值之后填充另一个输入字段值
我有两个剧本 1。HTML部分:Php 在第一个输入值之后填充另一个输入字段值,php,jquery,html,Php,Jquery,Html,我有两个剧本 1。HTML部分: <html> <head> <script src="//code.jquery.com/jquery-1.9.1.js"></script> <script> function loaddata() { var email=document.getElementById( "email" ); var email1 = $(email).val(); if(email1) { $.ajax(
<html>
<head>
<script src="//code.jquery.com/jquery-1.9.1.js"></script>
<script>
function loaddata()
{
var email=document.getElementById( "email" );
var email1 = $(email).val();
if(email1)
{
$.ajax({
type: 'post',
url: 'form-inputs.php',
data: {
user_email:email1,
},
success: function (response) {
// We get the element having id of display_info and put the response inside it
$( '#display_info' ).html(response);
}
});
}
else
{
$( '#display_info' ).value("Please Enter Some Words");
}
}
</script>
</head>
<body>
<input name="email" id="email" onkeyup="loaddata();" required/>
<br><br>
<input name="udf1" id="udf1" readonly="readonly"/>
<br>
<div id="display_info" ></div>
</body>
</html>
<?php
$con = mysql_connect("localhost","username","password");
if (!$con)
{
die("Could not connect: " . mysql_error());
}
mysql_select_db("databasename", $con);
##############################
if( isset( $_POST['user_email'] ) ){
$emailfetch=$_POST['user_email'];
$sqlraw = "select username, email from registration where email = '$emailfetch' ";
$sql=mysql_query($sqlraw);
while($row=mysql_fetch_array($sql)){
if(isset($row['email'])){
echo $row['username']; //shown only if post email exists in database
}else{
echo "Check Email Again";
}
}
}
?>
函数loaddata()
{
var email=document.getElementById(“电子邮件”);
var email1=$(email.val();
如有需要(电邮1)
{
$.ajax({
键入:“post”,
url:'form inputs.php',
数据:{
用户\电子邮件:email1,
},
成功:功能(响应){
//我们获取id为display_info的元素,并将响应放入其中
$('display#u info').html(响应);
}
});
}
其他的
{
$(“#显示信息”).value(“请输入一些单词”);
}
}
2。form-inputs.php:
<html>
<head>
<script src="//code.jquery.com/jquery-1.9.1.js"></script>
<script>
function loaddata()
{
var email=document.getElementById( "email" );
var email1 = $(email).val();
if(email1)
{
$.ajax({
type: 'post',
url: 'form-inputs.php',
data: {
user_email:email1,
},
success: function (response) {
// We get the element having id of display_info and put the response inside it
$( '#display_info' ).html(response);
}
});
}
else
{
$( '#display_info' ).value("Please Enter Some Words");
}
}
</script>
</head>
<body>
<input name="email" id="email" onkeyup="loaddata();" required/>
<br><br>
<input name="udf1" id="udf1" readonly="readonly"/>
<br>
<div id="display_info" ></div>
</body>
</html>
<?php
$con = mysql_connect("localhost","username","password");
if (!$con)
{
die("Could not connect: " . mysql_error());
}
mysql_select_db("databasename", $con);
##############################
if( isset( $_POST['user_email'] ) ){
$emailfetch=$_POST['user_email'];
$sqlraw = "select username, email from registration where email = '$emailfetch' ";
$sql=mysql_query($sqlraw);
while($row=mysql_fetch_array($sql)){
if(isset($row['email'])){
echo $row['username']; //shown only if post email exists in database
}else{
echo "Check Email Again";
}
}
}
?>
使用此
echo '<input type="text" name="user" value="' . $row['username'] . '">';
你试过这个吗
$('#udf1').val(response);
之后
这可能会有所帮助。无法清楚地知道您真正想要什么。你能详细解释一下吗?@JyothiBabuAraja我在div中显示用户名,id显示信息。。我想在输入字段udf1中将用户名显示为值。。。