fastfind1
asked on
PHP MySQL Assign variable new value if not found in table
I have a PHP variable with a set value ($v = 12;)
I have a MySQL table named codes with a column named reasons.
I need to write a PHP script that looks to see if 12 is a value in ANY record in the codes table under the reasons column. If 12 is in the table, it is only in once.
If the script finds that 12 does exist in the table and column, the value of $v remains 12.
But if the script finds that 12 does NOT exist in the table and column, then $v is reassigned a value of "default"
Please help me edit my code below
I have a MySQL table named codes with a column named reasons.
I need to write a PHP script that looks to see if 12 is a value in ANY record in the codes table under the reasons column. If 12 is in the table, it is only in once.
If the script finds that 12 does exist in the table and column, the value of $v remains 12.
But if the script finds that 12 does NOT exist in the table and column, then $v is reassigned a value of "default"
Please help me edit my code below
$v = 12;
$sql = sprintf("SELECT reasons FROM codes");
$q = mysql_query($sql);
while($r = mysql_fetch_array($q))
{
//HERE IS WHERE I GET STUCK
If true, then $v = 12;
If false, then $v = 'default';
ASKER CERTIFIED SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
TOP 4 Lines below:
UBER-PEREFECTED FOR reasons AS STRING
BOTTOM 4 Lines below:
UBER-PEREFECTED FOR reasons AS NUMBER
UBER-PEREFECTED FOR reasons AS STRING
BOTTOM 4 Lines below:
UBER-PEREFECTED FOR reasons AS NUMBER
$v = '12';
$sql = "SELECT reasons FROM codes WHERE reasons = '$v'";
$q = mysql_query($sql);
$v = $q.count ? $v : 'default';
$v = 12;
$sql = "SELECT reasons FROM codes WHERE reasons = $v";
$q = mysql_query($sql);
$v = $q.count ? $v : 'default';
PEREFECTED FOR reasons AS STRING
BOTTOM 4 Lines below:
PEREFECTED FOR reasons AS NUMBER
Open in new window