problems converting my php script to mysqli

Question

i recently did lots of changes to a site i was working on and went thru the entire project replacing the old mysql_ methods of interacting with the database. in this certain script im having trouble getting it to work the same.

the old code is

$checkinfo = mysql_query("SELECT * FROM `myusers` WHERE `userid` = '$uid' LIMIT 1") or die(mysql_error());

if(mysql_num_rows($checkinfo) < 1){ //log and die if user isnt in db
  die("Incident has been logged!"); }

$myinfo = mysql_fetch_assoc($checkinfo);

and my new code is

$checkinfo = $mysqli->query("SELECT * FROM `myusers` WHERE `userid` = '$uid' LIMIT 1") or die('Error : ('. $mysqli->errno .') '. $mysqli->error);  

if($checkinfo->fetch_row() < 1){
    die("Incident has been logged!"); } 

$myinfo = $checkinfo->fetch_assoc();

now its simply not setting my array for the rest of the code... please point out my stupidity! thanks

"went thru the entire project replacing the old mysql_ methods" do yourself a favour and abstract your database access into a class, then when you need to make further changes, you only have to do it in one place — Steve
– Steve, Commented Feb 11, 2015 at 10:27

Kevin · Accepted Answer · 2015-02-11 10:29:43Z

3

By using ->fetch_row(), it already does fed the first row. Since you explicitly set LIMIT 1, the next fetch invocation resulted into NULL.

Change it to ->num_rows instead:

$checkinfo = $mysqli->query("SELECT * FROM `myusers` WHERE `userid` = '$uid' LIMIT 1") or die('Error : ('. $mysqli->errno .') '. $mysqli->error);  

if($checkinfo->num_rows < 1){
    die("Incident has been logged!"); // change this to something more meaningful.
} 

$myinfo = $checkinfo->fetch_assoc();

edited Feb 11, 2015 at 10:29

answered Feb 11, 2015 at 10:23

Kevin

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