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can u spot the simple problem?

Discuss can u spot the simple problem? in the Database Help forum on Codewalkers.
can u spot the simple problem? General SQL help. Help with MySQL, MSSQL, Oracle, mSQL, PostgreSQL, etc.

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  #1  
Old November 12th, 2003, 09:06 AM
Anonymous Anonymous is offline
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can u spot the simple problem?
php Code:
Original - php Code 
  1.  
  2. <?
  3.  
  4.  
  5. $query = "SELECT * FROM table WHERE id = 'x'";
  6.  
  7. if ($query){
  8.         echo "done<br>";
  9.         };
  10.         
  11.         $result = mysql_query($query);
  12.         
  13.         if (!$result){
  14.         echo "****ed";
  15.         die();
  16.         };
  17.         
  18.         $array = mysql_fetch_array($result);
  19.         $var1 = $array['data1'];
  20.         $var2 = $array['data2'];
  21.         
  22.         echo "$x<br>";
  23.         
  24.         echo "var1 = $var1<br>";
  25.         echo "var2 = $var2<br>";
  26.         
  27.  
  28.  
  29.  
  30. ?>


the above code dont work. i cant see the problem - its a simple query as you can see and all i want to do is output the results to screen.

if i hard code it with say;

php Code:
Original - php Code 
  1.  
  2. <?
  3.  
  4.  
  5. $query = "SELECT * FROM table WHERE id = 1035";
  6.  
  7. if ($query){
  8.         echo "done<br>";
  9.         };
  10.         
  11.         $result = mysql_query($query);
  12.         
  13.         if (!$result){
  14.         echo "****ed";
  15.         die();
  16.         };
  17.         
  18.         $array = mysql_fetch_array($result);
  19.         $var1 = $array['data1'];
  20.         $var2 = $array['data2'];
  21.         
  22.         
  23.         
  24.         echo "var1 = $var1<br>";
  25.         echo "var2 = $var2<br>";
  26.         
  27.  
  28.  
  29.  
  30. ?>


it works fine.

i cant see the problem, can you?

cheers in advance,

david*
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  #2  
Old November 12th, 2003, 09:22 AM
Anonymous Anonymous is offline
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RE: can u spot the simple problem?
no need to reply! i sorted it.

i for got to add the '$'

the query should read;

php Code:
Original - php Code 
  1.  
  2. $query = "SELECT * FROM table WHERE id = '$x'";


david*
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