Question

我尝试创建用户个人资料时，可以从网址获取ID，然后从该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!";
}
}?>

我从回声中得不到任何错误。

请帮忙！

Answer 1

$conn->query("getUserNames");出现错误，该错误应为：$conn->query($getUserNames);

修复程序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!";
}
}?>

当＆＃34; echo＆＃34; $ _GET行

1 个答案: