listing multiple results from a mysql query

[ldb358]

i want it to list the first 10 results

thanks in advanced

 

heres my code:

function runBrowse($search){
$qSearch = mysql_query("SELECT * FROM users WHERE username LIKE '%$search%'");
$results = mysql_fetch_array($qSearch);
if($results['username'] != ""){
echo $results['username'];
}else{
echo "no results";
}
}

[ted_chou12]

function runBrowse($search){
$qSearch = mysql_query("SELECT * FROM users WHERE username LIKE '%$search%'" LIMIT 10);
while($results = mysql_fetch_array($qSearch)) {
if($results['username'] != ""){
echo $results['username'];
}else{
echo "no results";
}
}}

[Maq]

There's no reason to check if the username is blank when you have the LIKE clause that only selects the usernames with search in them.  If they are blank, they won't be selected.

 

Try:

 

function runBrowse($search){
   $qSearch = mysql_query("SELECT * FROM users WHERE username LIKE '%$search%' LIMIT 10");
   while($results = mysql_fetch_array($qSearch))
   {
      echo $results['username'];
   }
}

 

And please learn to properly, or at least consistently, indent and format.

[ldb358]

thanks for the help

[ted_chou12]

btw, if you want to tell ur users that there are no results, change to this:

function runBrowse($search){
$qSearch = mysql_query("SELECT * FROM users WHERE username LIKE '%$search%' LIMIT 10");
$rows = mysql_num_rows($qSearch);
if ($rows == 0) {
echo "no results";}
else {	

while($results = mysql_fetch_array($qSearch)) {echo "no results";}
}}

Ted