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PHP MySQL Assign variable new value if not found in table

Posted on 2008-10-30
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Last Modified: 2013-12-13
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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Question by:fastfind1
  • 3
5 Comments
 
LVL 9

Accepted Solution

by:
MarkusId earned 300 total points
ID: 22838881
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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LVL 6

Assisted Solution

by:cr4ck3rj4ck
cr4ck3rj4ck earned 100 total points
ID: 22839012
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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LVL 19

Assisted Solution

by:NerdsOfTech
NerdsOfTech earned 100 total points
ID: 22839200
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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Expert Comment

by:NerdsOfTech
ID: 22839222
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
 
LVL 19

Expert Comment

by:NerdsOfTech
ID: 22839226
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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