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futumuch

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  1. 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.
  2. 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 $dbc variable as global in the main body. If you need any other info let me know. Many thanks.
  3. Hi all, i am trying to insert some data into a table via a php function as below function save_link($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.
  4. 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
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