Question

大家好，我正在尝试为我的用户上传创建我的个人资料照片。我已经让它为用户工作，所以你可以登录上传等，然后注销，一切正常，但是目前它显示每个用户个人资料图片而不是当时登录的特定用户< / p>

id 
user_name
password 
profilePic set it as text

我的PHP：

<?php
mysql_connect("localhost","root","") or die ("could not connect");
mysql_select_db("reg") or die ("could not find db");
session_start();

if(!$_SESSION['username']){
  header("location: login.php");
}

$favs = array();
$links = array();
$sql = "SELECT * FROM recipe WHERE fav='yes'";
$records = mysql_query($sql);
//while($result =mysql_fetch_assoc($records)){
//$favs[] = $result['recipeName'];
//$links[] =    $result['url'];
//}
$user_name =    $_SESSION['username'];

if(isset($_POST['submit'])){
  move_uploaded_file($_FILES['file']['tmp_name'],
      "Image/userImages/".$_FILES['file']['name']);
  $con = mysqli_connect("localhost","root","","reg");
  $q = mysqli_query($con,"UPDATE users SET profilePic = '".
      $_FILES['file']['name']."' WHERE user_name = '".$_SESSION['username']."'");
}

?>

HTML：

<div class = "pic">
  <?php
  $con = mysqli_connect("localhost","root","","reg");
  $q = mysqli_query($con,"SELECT * FROM users");
  while($row = mysqli_fetch_assoc($q)){
    if($row['profilePic'] == ""){
      echo "<img width='100' height='100' 
        src='Image/userImages/profile.png' alt='Default Profile Pic'>";
    } else {
      echo "<img width='100' height='100' 
        src='Image/userImages/".$row['profilePic']."' alt='Profile Pic'>";
    }
    echo "<br>";
  }
  ?>

  <div class="userNameText">
    <?php echo $_SESSION['username'];?>
  </div>
</div>

这与此有关：$q = mysqli_query($con,"SELECT * FROM users");我尝试将其更改为$q = mysqli_query($con,"SELECT * FROM users WHERE user_name = $user_name");，但没有运气，所以回到了几乎有用的东西。无论如何，任何有关此事的帮助都会令人惊讶

Answer 1

更改

$q = mysqli_query($con,"SELECT * FROM users");

到

$q = mysqli_query($con,"SELECT * FROM users where user_name = '".$_SESSION['username']."'");

你忘了将' '添加到哪里，如果那个html与php在同一个文件中，那么你可以使用$ user_name，如果没有，你必须使用session

配置文件显示所有用户而不是一个用户

1 个答案: