Question

当用户访问“访客日志”页面时，他们应该能够看到提示要求他们输入姓名的提示。提交表单后，同一页面应显示完全不同的消息，欢迎用户访问该网页。当用户刷新页面时，该过程重新开始。
到目前为止，这是我尝试过的方法，它可以工作，但是我仍然不明白如何在输入后显示一条新消息。
这是我需要的代码，仅使用PHP即可获得正确的所需结果

尝试

<?php
// define variables and set to empty values
$nameErr = "";
$name = "";

if ($_SERVER["REQUEST_METHOD"] == "POST") {
  if (empty($_POST["name"])) {
    $nameErr = "Name is required";
  } 


  else {
    $name = test_input($_POST["name"]);
    // check if name only contains letters and whitespace
    if (!preg_match("/^[a-zA-Z ]*$/",$name)) {
      $nameErr = "Only letters and white space allowed"; 
    }

  }

}

function test_input($data) {
  $data = trim($data);
  $data = stripslashes($data);
  $data = htmlspecialchars($data);
  return $data;
}
?>

<p2 id="example-id-name" class="centered-text "></p>
<p><span class="error"></span></p>
<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">  
   <input type="text" name="name" value="<?php echo $name;?>">

   <span class="error"> <?php echo $nameErr;?></span>
  <br> <br>

 <input type="submit" name="submit" value="Submit">


</form>

<?php

echo "$name";
echo "<br>";

?>

Answer 1

在回显姓名的地方，可以检查是否有姓名，然后选择要显示的消息

<?php
if($name) {
  echo "Hi $name!\n Welcome to our store!"
}
else {
  echo "Please enter your name"
}
echo "<br>";
?>

Answer 2

您可以编写内联php和函数。

代码：

WITH song_count as 
    SELECT extract(year from date) AS song_year, song_code, count(*) as play_count
    FROM singles NATURAL JOIN plays
    group by extract(year from date),song_code
select * from song_count
where
 (song_year,play_count) in (select song_year,min(play_count) from song_count group by song_year)

<?php
# filter input
function filter($var) {
    return htmlspecialchars(stripslashes(trim($var)));
}

# validate name
function validate_name(&$name, &$err){
    if(empty($name)){
        $err = "Name is required";
        return;
    }
    $name = filter($name);
    if (!preg_match("/^[a-zA-Z ]*$/",$name)) {
        $err = "Only letters and white space allowed";
    }
}

$method = filter_input(INPUT_SERVER, 'REQUEST_METHOD');
$err = "";

# If client post a name, then validate the name
if ($method === "POST"){
    $name = $_POST["name"] ?? "";
    validate_name($name, $err);
}
?>

使用php使用条件语句在同一页面上显示欢迎消息

尝试

2 个答案: