这个php查询应该像其他的一样运行。 看不出我的错误
问题描述：
This query runs when I comment out the code above '}else{'. Have done so many like it and way more complex but I can't see where I'm tripping up.
(The page won't run like a syntax mistake. all vars match table)
Thanks. Allen
if ($version == 'prem'){
$sql ="SELECT * FROM artistInfo WHERE user_id = '$user_id' AND artist_name = '$artist_name' ";
$res = mysql_query($sql);
$num = mysql_num_rows($res);
if($num>0){
while($row = mysql_fetch_array($res)){
$artist_id = $row['artist_id'];
} else {
mysql_query("INSERT INTO artistInfo (user_id, artist_name) VALUES ('$user_id', '$artist_name')");
$row_num = mysql_insert_id();
$artist_id = $user_id."-".$row_num;
mysql_query("UPDATE artistInfo SET artist_id = '$artist_id' WHERE row_num = '$row_num' ");
}
}
}
答
I think that your last bracket got misplaced. you will also need to move the last } just before else
if ($version == 'prem'){
$sql ="SELECT * FROM artistInfo WHERE user_id = '$user_id' AND artist_name = '$artist_name' ";
$res = mysql_query($sql);
$num = mysql_num_rows($res);
if($num>0){
while($row = mysql_fetch_array($res)){
$artist_id = $row['artist_id'];
}
} else {
mysql_query("INSERT INTO artistInfo (user_id, artist_name) VALUES ('$user_id', '$artist_name')");
$row_num = mysql_insert_id();
$artist_id = $user_id."-".$row_num;
mysql_query("UPDATE artistInfo SET artist_id = '$artist_id' WHERE row_num = '$row_num' ");
}
}
答
You're missing the closing } for the while loop.
答
missing } for while loop
if($num>0){
while($row = mysql_fetch_array($res)){
$artist_id = $row['artist_id'];
}
}
else {