$sqlQuery = "SELECT username as display_name, fullname as full_name, email as email, role as administrators, status as be_on FROM admin";
记录角色：0,1,...
我希望管理员的角色会显示 1 是 admin ，如果等于 0 ，则显示用户。
谁能帮帮我？
【问题讨论】：
我认为你对 IF 语句的看法
SELECT IF(`admin`.`role`= 1,TRUE,FALSE) FROM ...
【讨论】：
我的建议是使用 mysqli 功能：
<?php
$mysqliClass = mysqli_init();
$mysqliClass->real_connect("hostname or IP", "dbUsername", "dbPassword", "databasename");
$query = "SELECT `username` AS display_name, `fullname` AS full_name, `email`, `role` as administrators, `status` as be_on FROM `admin`";
$rs = $mysqliClass->query($query); # Preform your query
$showUser = false; # Start negative
$row = []; # Set a variable for the row to go into
while ($result = mysqli_fetch_assoc($rs)) # Fetch each result one at a time
{
if ($result['administrators'] == 1) # Check the role
{
$showUser = true; # Set the show to true if it equals 1
$row = $result; # Set the row to equal the result given
}
break; # You only need the first result
}
if ($showUser)
{
/* Do your work with $row! */
}
?>
我还删除了 email AS email 的 email，因为它已经有了这个名称，所以它是不必要的混乱。
我还建议在您的查询中添加 WHERE 子句以减少（过滤）结果
如果您使用数据库中的图像，您可以执行以下操作：
<img src="<?=$row['image_column_name'];?>" height="120" width="120" />
而这个应该按照你想要的方式显示图像:)
更完整的示例，如原始答案：
<?php
$mysqliClass = mysqli_init();
$mysqliClass->real_connect("hostname or IP", "dbUsername", "dbPassword", "databasename");
$query = "SELECT `name`, `logo` AS describe , IF(`status` = 0, 'Online', 'Offline') as be_on FROM `domain`";
$rs = $mysqliClass->query($query);
$rows = []; # This will have more than one in it this time!
while ($result = mysqli_fetch_assoc($rs))
{
$rows[] = $result;
}
foreach ($rows as $r)
{
foreach ($r as $rowKey => $rowValue) # Double loop because you have an array in an array this way
{
if ($rowKey == "describe")
{
print '<img src="' . $rowValue . '" height="120" width="120" />';
}
}
}
?>
【讨论】：