现在,我的代码正在回显数据库中的所有用户及其个人资料图片。我的目标是仅回显会话用户名和个人资料图片。任何帮助,将不胜感激。谢谢。
<?php
$q = mysqli_query($db, "SELECT * FROM users");
while($row = mysqli_fetch_assoc($q)) {
echo $username['username'];
if($row['image'] == ""){
echo "<img width='100 height='100' src='/student/globalit/2019/GamerMedia/pages/images/ao.jpg' alt ='Default Profile Pic'>";
}
else{
echo "<img width='100' height='100' src='/student/globalit/2019/GamerMedia/pages/images/".$row['image']."' alt='Profile Pic'>";
}
echo "<br>";
}
?>
如果我理解您的问题正确,那么您必须在SQL数组中添加WHERE子句。我认为变量$ username已经全部定义好,并且是一个包含当前User数据的数组。因此,通往可行解决方案的方法可能是:
<?php
$q = mysqli_query($db, "SELECT * FROM users WHERE username='".$username['username']."' LIMIT 1");
while($row = mysqli_fetch_assoc($q)) {
echo $username['username'];
if($row['image'] == ""){
echo "<img width='100 height='100' src='/student/globalit/2019/GamerMedia/pages/images/ao.jpg' alt ='Default Profile Pic'>";
}
else{
echo "<img width='100' height='100' src='/student/globalit/2019/GamerMedia/pages/images/".$row['image']."' alt='Profile Pic'>";
}
echo "<br>";
}
?>