Question

我有一份注册表格，其中我检查了电子邮件ID和状态，以便一个人不能再使用同一封电子邮件注册。表单和脚本都在一个页面上，页面是.php格式。问题是虽然检查工作正常，但我无法在电子邮箱前显示错误消息。如果有人能帮助我，我将不胜感激

<form role="form" action="" autocomplete="off" method="POST">

    <div class="input-group">
        <span class="input-group-addon"><i class="fa fa-user"></i></span>
            <input type="name" class="form-control" name="name" placeholder="Company Name" required>
    </div>
    <span class="help-block"></span>

    <div class="input-group">
        <span class="input-group-addon"><i class="fa fa-user"></i></span>
            <input type="email" class="form-control" name="email" placeholder="email address" required>
    </div>
    <span><?php echo $errormsg;?></span>

    <div class="input-group">
        <span class="input-group-addon"><i class="fa fa-lock"></i></span>
            <input  type="password" class="form-control" name="password" placeholder="Password" required>
    </div>
    <span class="help-block"></span>

    <div class="input-group">
        <span class="input-group-addon"><i class="fa fa-user"></i></span>
            <input type="tel" id="tel" class="form-control" name="phoneno" pattern="\d{10}" placeholder="Please enter a ten digit phone number" required />
    </div>
    <span class="help-block"></span>
    <input class="form-control" value="" type="hidden" name="status">

    <button class="btn btn-lg btn-primary btn-block" name="submit" type="submit">Sign Up</button>

</form>

<?php
$con=mysqli_connect("localhost","root","","db");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// escape variables for security
$name = mysqli_real_escape_string($con, $_POST['name']);
$email = mysqli_real_escape_string($con, $_POST['email']);
$password = mysqli_real_escape_string($con, $_POST['password']);
$phoneno = mysqli_real_escape_string($con, $_POST['phoneno']);
$status = mysqli_real_escape_string($con, $_POST['status']);

$sql = "SELECT * FROM register WHERE email='".$email."' AND status='1'";
$result = mysqli_query($con, $sql); 
if (mysqli_num_rows($result) > 0)
{
    $errormsg="email already exist";
}
else
{
    $sql="INSERT INTO register (name,email,password,phoneno,status) VALUES ('$name','$email','$password','$phoneno','1')";
    if (!mysqli_query($con,$sql)) 
    {
        die('Error: ' . mysqli_error($con));
    }
    exit;
    mysqli_close($con);
}
?>

Answer 1

如果所有代码都在单个文件中，那么你将PHP代码转移到html代码之上，为什么在执行PHP代码之前错误没有显示html首先呈现？即：

<?php
$con=mysqli_connect("localhost","root","","db");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// escape variables for security
$name = mysqli_real_escape_string($con, $_POST['name']);
$email = mysqli_real_escape_string($con, $_POST['email']);
$password = mysqli_real_escape_string($con, $_POST['password']);
$phoneno = mysqli_real_escape_string($con, $_POST['phoneno']);
$status = mysqli_real_escape_string($con, $_POST['status']);

$sql = "SELECT * FROM register WHERE email='".$email."' AND status='1'";
$result = mysqli_query($con, $sql); 
if (mysqli_num_rows($result) > 0)
    {
        $errormsg="email already exist";
    }
else{
$sql="INSERT INTO register (name,email,password,phoneno,status) VALUES ('$name','$email','$password','$phoneno','1')";
        if (!mysqli_query($con,$sql)) 
            {
                die('Error: ' . mysqli_error($con));
            }
        exit;
mysqli_close($con);
}
?>
<form role="form" action="" autocomplete="off" method="POST">

    <div class="input-group">
        <span class="input-group-addon"><i class="fa fa-user"></i></span>
            <input type="name" class="form-control" name="name" placeholder="Company Name" required>
    </div>
    <span class="help-block"></span>

    <div class="input-group">
        <span class="input-group-addon"><i class="fa fa-user"></i></span>
            <input type="email" class="form-control" name="email" placeholder="email address" required>
    </div>
    <?php echo $errormsg;?></span>

    <div class="input-group">
        <span class="input-group-addon"><i class="fa fa-lock"></i></span>
            <input  type="password" class="form-control" name="password" placeholder="Password" required>
    </div>
    <span class="help-block"></span>

    <div class="input-group">
        <span class="input-group-addon"><i class="fa fa-user"></i></span>
            <input type="tel" id="tel" class="form-control" name="phoneno" pattern="\d{10}" placeholder="Please enter a ten digit phone number" required />
    </div>
    <span class="help-block"></span>
    <input class="form-control" value="" type="hidden" name="status">

    <button class="btn btn-lg btn-primary btn-block" name="submit" type="submit">Sign Up</button>

</form>

Answer 2

您应该设置/初始化上面的错误容器，并首先在html上面进行检查。

不要忘记使用准备好的陈述！

<?php

$errormsg = '';
$con = mysqli_connect("localhost","root","","db_name");
// Check connection
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$name = $_POST['name'];
$email =  $_POST['email'];
$password = $_POST['password'];
$phoneno = $_POST['phoneno'];
$status = $_POST['status'];

$sql = "SELECT * FROM register WHERE email = ? AND status = '1' ";
$check = $con->prepare($sql);
$check->bind_param('s', $email);
$check->execute();

if ($check->num_rows > 0) {
    $errormsg = "email already exist";
}
else{

    $sql = "INSERT INTO register (name,email,password,phoneno,status) VALUES (?, ?, ?, ?, '1')";
    $insert = $con->prepare($sql);
    $insert->bind_param('ssss', $name, $email, $password, $phoneno);
    $insert->execute();

}
?>
<form role="form" action="" autocomplete="off" method="POST">

    <div class="input-group">
        <span class="input-group-addon"><i class="fa fa-user"></i></span>
            <input type="name" class="form-control" name="name" placeholder="Company Name" required>
    </div>
    <span class="help-block"></span>

    <div class="input-group">
        <span class="input-group-addon"><i class="fa fa-user"></i></span>
            <input type="email" class="form-control" name="email" placeholder="email address" required>
    </div>
    <?php echo $errormsg; ?></span>

    <div class="input-group">
        <span class="input-group-addon"><i class="fa fa-lock"></i></span>
            <input  type="password" class="form-control" name="password" placeholder="Password" required>
    </div>
    <span class="help-block"></span>

    <div class="input-group">
        <span class="input-group-addon"><i class="fa fa-user"></i></span>
            <input type="tel" id="tel" class="form-control" name="phoneno" pattern="\d{10}" placeholder="Please enter a ten digit phone number" required />
    </div>
    <span class="help-block"></span>
    <input class="form-control" value="" type="hidden" name="status">

    <button class="btn btn-lg btn-primary btn-block" name="submit" type="submit">Sign Up</button>

</form>

Answer 3

<div class="input-group">
        <span class="input-group-addon"><i class="fa fa-user"></i></span>
            <input type="email" class="form-control" name="email" placeholder="email address" required>
      <br><span><?php echo $errormsg;?></span>
    </div>

提交表单后显示错误消息

3 个答案: