검색 결과 오류

Question

이것은, 내가 오류를 표시하는 방법, 그것은 나를 위해 작동

$sql="SELECT * FROM course WHERE course_name LIKE '%" . $search_name . "%'"; 
       //-run the query against the mysql query function 
       $result=mysql_query($sql); 

Answer 1

$sql="SELECT * FROM course WHERE course_name LIKE '%" . $search_name . "%'"; 
       //-run the query against the mysql query function 
$result=mysql_query($sql); 

if(mysql_num_rows($result) < 1) 
{ 
    echo 'There were no results.'; 
} 

Answer 2

if(mysql_num_rows($sql)=="0") 
{ 
echo "----------------"; //your code here 
}