[英]Storing Ajax success data into javascript variable to use in other js function
我目前正在執行登錄表單,並且我的數據庫中包含用戶名和密碼。 我已經比較了用戶名和密碼是否與該用戶名匹配,但是當我使用來自ajax響應的變量單擊“提交”按鈕時,我想顯示用戶名不存在。 我已經完成了,但是變量存儲在html標簽中。
這是我的舊代碼
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;
}
}
我想要這樣的東西,但是下面的代碼不起作用
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;
}
}
然后我將在此函數上使用來自ajax的變量
function checkall(receivePw, receiveUsn)
{
if(receiveUsn == "OK" && receivePw == "MATCH"){
return true;
}else{
return false;
}
}
PHP代碼
<?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();
}
?>
登錄表單
<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>
請幫忙謝謝!!!
您需要在Java腳本開始時聲明變量'receiveUsn',這樣它將成為全局變量。問題是您已經在ajax成功函數中聲明了var receiveUsn,因此它現在是局部變量,不能在其他變量中使用函數.global可以解決您的問題
在<script>
標記中,您需要在根級別聲明一個變量,這將創建全局變量。 最好的做法是不污染全局空間,因此在全局級別聲明一個對象並將其用於存儲全局變量。
<script>
var myGlobalContainer = {};
</script>
現在,在頁面上的JavaScript代碼中,您可以使用myGlobalContainer
對象。
// set a value
myGlobalContainer.hello = "world";
// read a value
console.log(myGlobalContainer.hello);
// "world"
在Javascript之上初始化變量:
myglobalvariable = '';
現在將ajax輸入存儲在變量中,並且可以在任何地方訪問它:
......
success: function(response){
myglobalvariable = response;
$('#status').html(receiveUsn);
......
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