Question

我正在创建一个profile.php页面，我希望它向用户显示他所有的项目，这是我第一次做这样的事情，我无法找到它的解决方案用于显示项目的代码：

$username = $_SESSION['username'];
            if ($_SESSION['type'] = "developer"){
              $q = "SELECT * FROM `projects` WHERE `developer` = '$username'";
              $result = mysqli_query($con,$q);
              $row = mysqli_fetch_array($result);
              $numrows = mysqli_num_rows($result);
              if(empty($numrows)){
                echo'
                <div class="row">
                  <div class="col-lg-12 newp">
                    <a href = "new_project.php"><p><span class="glyphicon glyphicon-plus plus"></span>Add a new project</p></a>
                  </div>
                </div>';
              }else{
                $p_id = $row['project_id'];
                $p_name = $row['project_name'];
                $p_owner = $row['owner'];
                $p_developer = $row['developer'];
                $p_price = $row['price'];
                $p_date_started = $row['date_started'];
                $p_date_end = $row['date_end'];
                $p_paid = $row['paid'];
              //foreach project the user has do this :
              echo"
              <div class=\"row\">
                <div class=\"col-lg-12\">
                  <a href = \"project.php?id=$p_id\"><p>$p_name </br>owner : $p_owner, developer : $p_developer, price : $p_price$</br>started : $p_date_started, ends :$p_date_end, paid :$p_paid</p></a>
                </div>
              </div>";
            }
            }

Answer 1

} else {
    while($row = mysqli_fetch_array($result)) {
        $p_id = $row['project_id'];
        ...

Answer 2

除了给出的其他答案：

您目前assigning而不是comparing

if ($_SESSION['type'] = "developer"){...}
                      ^

以上将失败以及条件语句中的所有内容都应该读作

if ($_SESSION['type'] == "developer"){...}
                      ^^

有2个等号。

确保会话也已启动，在使用会话时需要它。

session_start();

您也可以使用SQL注入。使用准备好的声明：

https://en.wikipedia.org/wiki/Prepared_statement

显示存储在数据库中的所有用户项目

2 个答案: