当“回显”$_GET 行时，什么都没有显示 - Nothing shows up when "echo" $_GET rows

Nothing shows up when "echo" $_GET rows

本文关键字：什么显示行时 GET 回显 | 更新日期: 2023-09-27

我正在尝试创建一个用户配置文件，它应该从URL中获取id，然后从具有该ID的用户那里获取所有信息。

<?php
  session_start();
  $userID = $_GET["id"];
  $getUserNames = "SELECT * FROM users where id=$userID";
  $result = $conn->query("getUserNames");
  $row = mysqli_fetch_assoc($result);
  if (isset($_GET["id"])) {
    echo "<h1>";
    echo $row["username"];
    echo "</h1>";
    echo "<p><b>Name: </b>";
    echo $row["username"];
    echo "</p>";
   echo "<p><b>Password: </b>***** (<a href='#'>Change Password</a>)</p>";
  } else {
if (isset($_SESSION["username"])) {
  echo "<h1>";
  echo $_SESSION["username"];
  echo "</h1>";
  echo "<p><b>Name: </b>";
  echo $_SESSION["username"];
  echo "</p>";
  echo "<p><b>Password: </b>***** (<a href='#'>Change Password</a>)</p>";
} else {
    echo "<p>You need to be logged in too see your profile!";
}
}?>

我从回声中没有得到任何东西，也没有错误。

请帮忙！

您在$conn->query("getUserNames");处有一个错误，应该是：$conn->query($getUserNames);

带有修复的完整新代码如下所示：

<?php
  session_start();
  $userID = $_GET["id"];
  $getUserNames = "SELECT * FROM users where id=$userID";
  $result = $conn->query($getUserNames);
  $row = mysqli_fetch_assoc($result);
  if (isset($_GET["id"])) {
    echo "<h1>";
    echo $row["username"];
    echo "</h1>";
    echo "<p><b>Name: </b>";
    echo $row["username"];
    echo "</p>";
   echo "<p><b>Password: </b>***** (<a href='#'>Change Password</a>)</p>";
  } else {
if (isset($_SESSION["username"])) {
  echo "<h1>";
  echo $_SESSION["username"];
  echo "</h1>";
  echo "<p><b>Name: </b>";
  echo $_SESSION["username"];
  echo "</p>";
  echo "<p><b>Password: </b>***** (<a href='#'>Change Password</a>)</p>";
} else {
    echo "<p>You need to be logged in too see your profile!";
}
}?>