[Can'T Insert Via Php]

Thanks guys for your help, in trying to solve a problem with my syntax, I copied and pasted some code from a website that had mysql_query instead of mysqli_query. While i solved the original problem, I couldnt see the problem in the pasted code. A valuable lesson learned!

Many thanks.

[Can'T Insert Via Php]

I think i am connecting ok, i also get this on my screen just before the error message.

first link = aaacnew link = aaacsuccess! were in!

 

This is from my echo statements. I declared the $dbc variable as global in the main body.

If you need any other info let me know. Many thanks.

[Can'T Insert Via Php]

Hi all,
i am trying to insert some data into a table via a php function as below
 

function save_link($new_link)
 {
   echo 'new link = '.$new_link; #just to make sure the code gets to here
          if (mysqli_connect_errno())
            {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
            }
            else
            {
            echo 'success! were in!';
            }
 
$query = "INSERT INTO college_values 
  (link_code, year_2011, year_2012, year_2013, year_2014) 
  VALUES 
  ('AAAB',0,0,0,0)";
 
$retval = mysql_query($query, $dbc);
if(! $retval )
{
die('Could not enter data: ' . mysql_error());
}
      return $result;
 }

 
but i get the following message -
Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in....
 
I have tried many different combinations of $query but the one above does work if I paste it into the query section of PHPMyAdmin.I can substitute the query for a SELECT and it works fine. I just cant see why the INSERT doest work.

Many thanks.

[Count Using Mysqli_Query]

Thank you Antonio,

it works now!

Dave

[Count Using Mysqli_Query]

hi,

I am try to learn PHP and MySQL with the help of Larry's excellent book. Most of the time, I find the book easy to follow and I can find the answer I need, however I have now hit a wall!

I want to count the amount of rows in a table that meet a certain criteria and then use this number in a PHP while loop.

My query is - 

 

$SQL_count = SELECT COUNT(*) FROM college_notes WHERE link_code = 'CHES'

and I used this

$result_count = mysqli_query($dbc, $SQL_count);

 

My problem is that the variable $result_count doesnt seem to be usable. If I use echo to dump it on the screen, i get this -

 

Catchable fatal error: Object of class mysqli_result could not be converted to string in/home/www

I have also tried using AS to get an alias and $result_count['alais'] but got a message stating that the variable wasnt an array.

 

Thanks in advance for any help!

Dave