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Update Mysql using php

Hi I'm trying to update a record where column name is draw.

$draw = "<a href=/uploads/draw/".$_FILES['userfile']['name']." target=_blank>draw</a>";
   
 
  $conn = db_connect();
  
 $query = ("UPDATE competition SET draw = '$draw' WHERE id ='$ud_id' ");

 $result = $conn->query($query);
 echo $result;

  $conn->close();    
   

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When I echo $draw the variable is loaded and there.
When I echo $result it shows 1 so it believes the function is complete.
However the record fiield is still null.
Not sure what I'm missing.
Thanks
0
encodeme
Asked:
encodeme
3 Solutions
 
Marco GasiFreelancerCommented:
Try this:

$query = ("UPDATE competition SET draw = '$draw' WHERE id ='" . $ud_id . " ' ");
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hieloCommented:
>>When I echo $result it shows 1 so it believes the function is complete
OK, but without knowing what your db object is doing, we won't be able to help you.  My best guess is you are returning true on success, false on failure.

What I would do is:
For SELECT, SHOW :  on failure - return False; on success - return mysql_num_rows();
http://www.php.net/manual/en/function.mysql-num-rows.php

For INSERT, UPDATE, REPLACE or DELETE: on failure - return false; on success - return mysql_affected_records();
http://www.php.net/manual/en/function.mysql-affected-rows.php
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encodemeAuthor Commented:
Hi Hielo
Something like this ?

$conn = db_connect();
  $query = ("UPDATE competition SET draw = '$draw' WHERE id ='" . $ud_id . " ' ");
 //$query = ("UPDATE competition SET draw = '$draw' WHERE id ='$ud_id' ");

 $result = $conn->query($query);
 if ($result) {
      echo  $conn->affected_rows." competition ammended.";
	  
  } else {
  	  echo "An error has occurred.  The item was not ammended.";
  }
	
  $conn->close();

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When I run that it echo's 0 competition ammended
0
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Ray PaseurCommented:
You might need to use this function to prepare the data fields for use in a query:
http://php.net/manual/en/function.mysql-real-escape-string.php

You might want to print out the contents of the fully-resolved query string before you call $conn->query()

You might want to use var_dump($conn) to see the object.  Same thing about $result.  If you use these data visualization techniques you will probably find the issues very quickly.
0
 
encodemeAuthor Commented:
Curiously this line works and update the db.
$query = ("UPDATE competition SET draw = 'cat' WHERE id = '21'");

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This line did not work
$query = ("UPDATE competition SET draw = 'cat' WHERE id ='" . $ud_id . " ' ");

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But I've checked that variable $ud_id has a value..
Now I'm realyy confused.
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Marco GasiFreelancerCommented:
Sorry, cancel unuseful spaces I inserted to more readibility and delete quotes (for numerical values)

$query = ("UPDATE competition SET draw = 'cat' WHERE id =" . $ud_id);

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encodemeAuthor Commented:
Thanks for all the help.
It was a valiant joint effort of giving me clues and tips on how to track the bug.
0

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