Help me Please!!

[wizzle]

hi! i  badly need your help.. i'm having an error in my addressbook system..

 

Here's the error:

mysql_num_rows() expects parameter 1 to be resource, null given in /Applications/XAMPP/xamppfiles/htdocs/sampol3/design/edit.php on line 74.

 

 

please give the right code.. thank you..

17348_.phpFetching info...

[trq]

Firstly, I haven't looked at your code because I won't download code.

 

Anyway, your error is very common and is caused by passing the results of a call to mysql_query() straight into mysql_num_rows() without first checking to see that your query has succeeded.

 

mysql_num_rows() expects a result resource, while mysql_query() will return a result resource on success or the boolean false on failure.

 

For some reason, your query is failing. What does mysql_error have to say?

[wizzle]

here' the function for mysql_error():

 

mysql_error($con)

 

 

$con=mysql_connect("localhost",$username,$password);

if(!$con)

die("couldn't connect to MySQL: ".mysql_error());

 

mysql_select_db($db_name,$con)

or die("couldn't connect to $db_name: ".mysql_error());

 

[Porl123]

Try adding or die(mysql_error()) to the end of your mysql_fetch_array() call like this:

 

mysql_fetch_array($query) or die(mysql_error());

 

This will give you a more precise definition of the problem.

[trq]

  Quote

Try adding or die(mysql_error()) to the end of your mysql_fetch_array() call like this:

 

mysql_fetch_array($query) or die(mysql_error());

 

This will give you a more precise definition of the problem.

 

Or better still, handle your errors properly.

 

An example:

 

if ($result = mysql_query($sql)) {
  if (mysql_num_rows($result)) {
    // $result is now good to use
  } else {
    // no results found
  }
} else {
  trigger_error(mysql_error());
}

[wizzle]

switch($select)

{

case 1:

$result=mysql_query("SELECT * FROM tb_addressbook WHERE lastname LIKE'%$text%'");

break;

case 2:

$result=mysql_query("SELECT * FROM tb_addressbook WHERE firstname='$text'");

break;

case 3:

$result=mysql_query("SELECT * FROM tb_addressbook WHERE address='$text'");

break;

}

 

 

 

if (!$result=)

{

echo "Could not successfully run query ($result) from database" . mysql_error($con);

 

}

 

if (mysql_num_rows($result) == 0)  //(im having a problem in this line.

{

 

echo "<br>No rows found, Please try again! <br>

<a href='searchform.php'>search</a>";

echo "<br><br>";

exit;

}

[Porl123]

lastname LIKE'%$text%'");

 

Could be due to there not being a space between LIKE and '

Not sure though, just a punt.

[wizzle]

  Quote

lastname LIKE'%$text%'");

 

Could be due to there not being a space between LIKE and '

Not sure though, just a punt.

 

i already put a space., but still same error and problem occur..  :'( :'(

[Porl123]

Could it be that $select doesn't contain 1, 2 or 3 and seeing as your switch case doesn't contain a default clause, it doesn't set a query result to $result?

[trq]

While your problem is with your query you have a logic issue here.

 

if (!$result=)
   {
   echo "Could not successfully run query ($result) from database" . mysql_error($con);
   
   }

 

If $result is false you echo an error message. this is good, however, you never stop your code from going into the next part:

 

if (mysql_num_rows($result) == 0)  

 

If $result is false, you will always get an error here because (as I have already explained) mysql_num_rows expects a result resource. Nothing else can be passed to it.

[PFMaBiSmAd]

This code seems familiar.

 

Wizzle, you already bumped an old thread with a post to add this or very similar code, where I relied what is causing the symptom and what you needed to do to troubleshoot what is causing the problem - http://www.phpfreaks.com/forums/index.php?topic=272971.msg1658897#msg1658897

[wizzle]

i already fix the problem,... thank you guys..