我正在制作一个profile.php页面,我希望它向用户显示他的所有项目,这是我第一次做这样的事情,而我找不到用于显示项目的代码的解决方案:
$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 :$p_paid</p></a>
</div>
</div>";
}
}
} else {
while($row = mysqli_fetch_array($result)) {
$p_id = $row['project_id'];
...
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