当名称存在或已添加时获取消息

Question

在您按下信息提交后，我尝试显示一条消息。

1.此名称已经存在，请使用另一个
或
2。名字加上去了！

html

<form action="add.php" method="post">
<input type="text" name="name" id="name" class="form-control" required> 
<input class="btn btn-default" type="submit" value="Submit">
</form>

php

$connectie=mysqli_connect("localhost","usernam","pw","db");

$name = $_POST['name'];
$name = stripslashes($name);
$name = mysql_real_escape_string($name);

$check_if_exists="SELECT * FROM names WHERE name = '$name'";

if($data[0] > 1) {
echo"already exists";
} else {

$newUser="INSERT INTO users (name) values('$name')";
    if (mysqli_query($connectie,$newUser))
    {
        echo "name is registerd";
    }
    else
    {
        echo "error<br/>";
    }

现在，它将回声发布在add.php上，而不是表单所在的页面上。我怎么把它送到那里？

Answer 1

您应该考虑为此使用会话。将session\u start（）添加到两个文件的顶部。

if(mysqli_query($connectie, $uewUser))
{
    $_SESSION['flash'] = 'name is registered';
}
else
{
    $_SESSION['flash'] = 'error<br/>';
}

然后，在带有表单的页面中，在想要打印数据的任何地方执行以下检查。

if(isset($_SESSION['flash']))
{
    echo $_SESSION['flash'];
    unset($_SESSION['flash']);
}

这将确保仅为下一个请求保留会话消息。

Answer 2

您可以使用会话/会话闪存数据。

首先，您没有正确运行选择查询。它应该是：

$check_if_exists="SELECT * FROM names WHERE name = '$name'";

$count = mysql_num_rows($check_if_exists);
if($count != 0) { // this means that the name already exists:

 $_SESSION['msg']="The name already exists. Please try another name";
  

} else {
  
  //perform the insert query and have the session msg
   $_SESSION['msg']="The name was inserted successfully";
  
}
  
  header('Location: form.php');

// Now in your form page just above the form, have a div to display this message as:

<div class="msg"> 
  <?php if(isset($_SESSION['msg']))

{
echo $_SESSION['msg'];
unset($_SESSION['msg']);
}
?>
</div>

NOTE: you must start the session on both the pages. Have this code on top of both the pages

<?php
session_start();
?>