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


$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';

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fastfind1Asked:
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MarkusIdCommented:
Why don't you include the $v in the WHERE-clause of the sql-statement?


$v = 12;
 
$sql = sprintf("SELECT reasons FROM codes WHERE reason=$v" );
$q = mysql_query($sql);
$x = 0
while($r = mysql_fetch_array($q)) 
{ 
  $x = 1; // If you come here there's a reason the equals $v
}
if ($x == 1) {  $v = 12; }
else {$v = 'default';}

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cr4ck3rj4ckCommented:
Hope this helps.
$sql = sprintf("SELECT reasons FROM codes WHERE reasons = 12");
$q = mysql_query($sql);
$rsSize = mysql_num_rows($q);
!$rsSize ? $v = 'default' : $v = 12;

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0
NerdsOfTechTechnology ScientistCommented:
I win on code speed
>:)
$v = 12; 
$sql = 'SELECT reasons FROM codes';
$q = mysql_query($sql);
$v = eval($q.count) ? $v : 'default';

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0
NerdsOfTechTechnology ScientistCommented:
TOP 4 Lines below:
PEREFECTED FOR reasons AS STRING

BOTTOM 4 Lines below:
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';

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0
NerdsOfTechTechnology ScientistCommented:
TOP 4 Lines below:
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';

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0
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