[已解決]：此代碼是有效的，我在瀏覽器中顯示舊版本的 .php 頁面時犯了一個錯誤，該瀏覽器沒有下面的更新代碼。我想將我的 php 代碼產生的用于驗證登錄表單的錯誤“錯誤”回顯到登錄表單的頁面中，錯誤僅出現在 URL 中，例如：signin.php?error = 用戶名不正確。所以我希望在登錄表單本身中出現相同的訊息，我嘗試使用 $_GET['error'] 并在我的表單中回顯它，
<form method="post" onsubmit=" return formSubmit() " action="signinphp.php">
<div class="userimage">
<img class="userlogo" src="image/userlogo.png" alt="Picture- User Profile picture">
</div><br>
<?php if (isset($_GET['error'])){?>
<p class="error"><?php echo $_GET['error'];?></p>
<?php } ?>
<div class="error" id= "errorMsg"></div> <br>
<div class="error" id= "errorMsg1"></div>
<div class="field">
<label class="stafflabel"> Staff Name </label>
<input class="area" placeholder="staffmember or admin" onclick=" return userValidation()" onchange=" return userValidation()" id="staff" name="staffname" type="text" >
</div> <br>
<div class="error" id= "errorMsg2"></div>
<div class="field">
<label class="passlabel"> Password </label>
<input class="area" placeholder="password" onclick=" return userValidation()" onchange=" return userValidation()" id="pass" name="password" type="password" >
</div><br>
<div class="checkbox">
<input type="checkbox" class="remember-me">
<label class="remember" for="remember-me">Remember me </label>
<a class="pass-link" href="#"> Forgot password?</a>
</div><br><br><br>
<div class="field">
<input class="btn" onclick="check(this.form)" type="submit" value="Sign in">
</div> <br>
<div class="account-link">
Didn't create an account yet? <a href="#">Create Account</a>
</div>
</form>
PHP
<?php
if (isset($_POST['staffname'])&& isset($_POST['password'])){
function validate($data){
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
$staffname = validate($_POST['staffname']);
$pass= validate($_POST['password']);
if (empty($staffname)){
header("Location:signin2.php?error=Staff name and password are required!");
exit();
} else if (empty($pass)){
header ("Location:signin2.php?error=Staff name and password are required!");
exit();
} else {
if ($staffname == "staffmember" && $pass== "letmein!123"){
echo "Logged in!";
header("Location: log-it-reportsbeta.php");
exit();
}
else if ($staffname == "admin" && $pass== "heretohelp!456"){
echo "Logged in!";
header("Location: sql_select_updated.php");
exit();
}
}
}
else{
header("Location: signin2.php");
exit();
}
但它不起作用，我在下面提供了我的表單代碼和表單驗證的 php 代碼，請看一下，謝謝。
uj5u.com熱心網友回復：
您可能想使用額外的條件來獲取錯誤訊息，而不是在 URL 中傳遞整個錯誤字串。
在您的 php 腳本中：
if (empty($staffname)){
header("Location:signin2.php?error=1");
exit();
} else if (empty($pass)){
header ("Location:signin2.php?error=2");
exit();
}
在您的表格中：
if (isset($_GET['error'])) {
if ($_GET['error'] == 1) {
$message = "Staff name and password are required!";
}
else if ($_GET['error'] == 2) {
$message = "Another error message";
}
echo '<p >'.$message.'</p>';
}
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