PHP:如何运行此IF语句
我运行了这个脚本,但我一直收到我在其中输入的“未工作”消息。我知道数据库中的值是正确的,它一定是我的PHP中的某个东西PHP:如何运行此IF语句,php,mysql,if-statement,Php,Mysql,If Statement,我运行了这个脚本,但我一直收到我在其中输入的“未工作”消息。我知道数据库中的值是正确的,它一定是我的PHP中的某个东西 $getadmin = "SELECT role FROM user WHERE user_id=$uid"; $showadmin = @mysqli_query ($dbc, $getadmin); // Run the query. $admin = mysqli_fetch_assoc(
$getadmin = "SELECT role FROM user WHERE user_id=$uid";
$showadmin = @mysqli_query ($dbc, $getadmin); // Run the query.
$admin = mysqli_fetch_assoc($showadmin);
if($admin == 'admin'){
echo 'You are an admin!';
} else {
echo 'Did not work';
}
未经测试
$row = mysqli_fetch_row($showadmin);
if($row[0] == 'admin'){ ..
试试这个:
$getadmin = "SELECT role FROM user WHERE user_id=$uid";
$showadmin = @mysqli_query ($dbc, $getadmin); // Run the query.
$role = mysqli_fetch_assoc($showadmin);
if($role['role'] == 'admin'){
echo 'You are an admin!';
} else {
echo 'Did not work';
}
fetch_assoc返回一个数组,而不是单个字符串 您需要检查$admin['role']而不是$admin的值admin。
$getadmin=“从用户选择角色,其中用户\u id={$uid}”//在插值变量周围使用{}
$getadmin = "SELECT role FROM user WHERE user_id={$uid}"; //use {} around variables for interpolation
$showadmin = @mysqli_query ($dbc, $getadmin); // Run the query.
if(!$showadmin)
{
die('<p>ERROR: DB result is null</p>');
}
while($ROW=mysqli_fetch_assoc($showadmin))//returns $ROW associative array
{
if($ROW['role'] == 'admin')
{
echo 'You are an admin!';
}
else //$ROW['role']!='admin'
{
echo 'Did not work because it is '.$ROW['role'];
}
}
$showadmin=@mysqli_query($dbc,$getadmin);//运行查询。
如果(!$showadmin)
{
die(“错误:DB结果为null”);
}
while($ROW=mysqli\u fetch\u assoc($showadmin))//返回$ROW关联数组
{
如果($ROW['role']=='admin')
{
echo‘你是管理员!’;
}
else/$ROW['role']!='admin'
{
echo'不工作,因为它是“.$ROW['role'”;
}
}
查看mysql\u fetch\u assoc()的返回值:$admin将是一个关联数组,而不是单个值。就是这个虽然我不得不改变立场…:)
$getadmin = "SELECT role FROM user WHERE user_id={$uid}"; //use {} around variables for interpolation
$showadmin = @mysqli_query ($dbc, $getadmin); // Run the query.
if(!$showadmin)
{
die('<p>ERROR: DB result is null</p>');
}
while($ROW=mysqli_fetch_assoc($showadmin))//returns $ROW associative array
{
if($ROW['role'] == 'admin')
{
echo 'You are an admin!';
}
else //$ROW['role']!='admin'
{
echo 'Did not work because it is '.$ROW['role'];
}
}