我目前正在做登录表,我有用户名和密码的数据库。我已经完成了用户名和密码是否与用户名匹配比较,但我希望在单击submit按钮时显示用户名不存在,并使用来自ajax响应的变量。我已经这样做了,但是变量存储在html标签中。
这是我以前的代码
function checkUsn(){
var usn = document.getElementById("usn").value;
if(usn){
$.ajax({
type: 'post',
url: 'checkdata.php',
data: {
emp_username: usn,
},
success: function(response){
$('#status').html(receiveUsn);
if (response == "OK"){
return true;
}else{
return false;
}
}
});
}else{
$('#status').html("");
return false;
}
}
我想要这样的东西,但下面的代码不起作用
function checkUsn(receiveUsn){
var usn = document.getElementById("usn").value;
if(usn){
$.ajax({
type: 'post',
url: 'checkdata.php',
data: {
emp_username: usn,
},
success: function(response){
var receiveUsn = response; //I want to use this for other function
$('#status').html(receiveUsn);
if (response == "OK"){
return true;
}else{
return false;
}
}
});
}else{
$('#status').html("");
return false;
}
}
然后,我将在此函数上使用来自ajax的变量
function checkall(receivePw, receiveUsn)
{
if(receiveUsn == "OK" && receivePw == "MATCH"){
return true;
}else{
return false;
}
}
php代码
<?php
include 'db_config.php';
$conn = new mysqli($db_servername, $db_username, $db_password, $db_name);
if(isset($_POST['emp_username'])){
$usn = $_POST['emp_username'];
$checkdata = "SELECT emp_username FROM emp_details where emp_username='$usn'";
$query = mysqli_query($conn, $checkdata);
if(mysqli_num_rows($query) > 0){
echo "OK";
}else{
echo "Your Username not exist";
}
exit();
}
if(isset($_POST['emp_pw']) && isset($_POST['emp_usn'])){
$pw = $_POST['emp_pw'];
$usn = $_POST['emp_usn'];
$get_pw = "SELECT emp_password FROM emp_details where emp_username='$usn'";
$query = mysqli_query($conn, $get_pw);
//$get_num_rows = mysqli_num_rows($query);
//echo $get_num_rows;
$row = mysqli_fetch_assoc($query);
//echo $row["emp_password"];
// check if password is match with username
if($pw == $row["emp_password"]){
echo "MATCH";
}else{
echo "Wrong password";
}
exit();
}
?>
登录表单
<form class="modal-content animate" action="/login_action.php" method="post" onsubmit="return checkall();">
<div class="container">
<span onclick="document.getElementById('id01').style.display='none'" class="close" title="Close Modal">×</span>
<div class="col-lg-3 col-md-3 col-sm-3 col-xs-3"></div>
<img class="avatar img-responsive col-lg-6 col-md-6 col-sm-6 col-xs-6" src="img/employee_avatar.png" alt="Avatar">
<div class="col-lg-3 col-md-3 col-sm-3 col-xs-3"></div>
</div>
<div class="container">
<label for="usn"><b>Username</b></label>
<input id="usn" type="text" placeholder="Enter Username" name="usn" onkeyup="getPw();checkUsn();" required>
<label for="pw"><b>Password</b></label>
<input id="pw" type="password" placeholder="Enter Password" name="pw" onkeyup="getPw();checkUsn();" required>
<button type="submit">Login</button>
<label>
<input type="checkbox" checked="checked" name="remember"> Remember me
</label>
</div>
<div class="container" style="background-color:#f1f1f1">
<button type="button" onclick="document.getElementById('id01').style.display='none'" class="cancelbtn">Cancel</button>
<span class="psw">Forgot <a href="#">password?</a></span>
</div>
<span id="status">status</span><br>
<span id="status2">status</span><br>
<span id="status3">status</span>
</form>
请帮帮忙谢谢!
发布于 2018-08-11 07:05:27
在<script>
标记中,您需要在根级别声明一个变量,这将创建全局变量。不污染全局空间被认为是最佳实践,因此在全局级别声明一个对象并使用它来存储全局变量。
<script>
var myGlobalContainer = {};
</script>
现在,在页面上的JavaScript代码中,可以使用myGlobalContainer
对象。
// set a value
myGlobalContainer.hello = "world";
// read a value
console.log(myGlobalContainer.hello);
// "world"
发布于 2018-08-11 06:58:29
你需要在你的java脚本开始时声明你的变量' receiveUsn‘,这样它就会变成一个全局变量。问题是你在ajax成功函数中声明了var receiveUsn .the,所以它现在是一个局部变量,不能在其他函数中使用,因为它是全局的,.make会解决你的问题
发布于 2018-08-11 07:12:29
在Javascript上初始化一个变量:
myglobalvariable = '';
现在将ajax输入存储在variable中,它可以在任何地方访问:
......
success: function(response){
myglobalvariable = response;
$('#status').html(receiveUsn);
......
https://stackoverflow.com/questions/51797133
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