mysqli error

[xgd]

Hello,

 

I am getting the following error when i try to run some script:

 

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given

 

The script is a bit longish, so here's the part that matters (hopefully):

 

case "content":                                    #110
      $dept_id = $_GET['dept_id'];
      $type_id = $_GET['type_id'];
      $sql = "SELECT a.name, a.type_id, b.title,
                b.description, b.content_date,
                b.create_date, b.created_by,
                b.last_upd_date, b.last_upd_by,
                c.name as dept_name, content_id
              FROM Content_Type a, Department c
              LEFT OUTER JOIN Content b on
                a.type_id = b.content_type
                and a.type_id = b.content_type
                and b.dept_id = $dept_id
                and b.content_type = $type_id
              WHERE c.dept_id = $dept_id
              ORDER BY content_date DESC";
      $results = mysqli_query($cxn, $sql);

      $body_links = "";
      $content_count = 0;
      $page["body_text"] = "";

      while($row = mysqli_fetch_assoc($results))       #132
      {
        if (!isset($area_name) && $type_id == $row["type_id"])
        {

 

I think it's something to do with the joints, and i personally don't know a thing about them (if that's the problem). I am usng Mysql version 5 something.

 

thanks

[patrickmvi]

That error indicates that your query is failing.  You should do a mysqli_error($cxn) right after your mysqli_query call to see what the error is.  It may be the way that you're mixing the different types of joins but I won't know for sure until I see that actual MySQL error.

[Adam]

I think you have the $sql and $cxn variables in the wrong order in your mysqli_query() call.

 

Edit: No you don't! Thought it was same syntax as mysql_query().

[Adam]

Ah, you need to make sure the returned result variable ($results) is not equal to false to prevent the error in future. As patrick mentioned, use the mysqli_error() function to find out why the query's failing.

[xgd]

hello,

 

I get this:

 

Unknown column 'a.type_id' in 'on clause'

 

 

[Dathremar]

Make sure the table "Content_Type" has a column named type_id

[xgd]

The mysql syntax is for mysqli 4.0 or 4.1 i think, that's what the book uses, so it needs some modification for mysql 5 i believe.

 

p.s.

it does have type_id

[Dathremar]

ELECT a.name, a.type_id, b.title,

                b.description, b.content_date,

                b.create_date, b.created_by,

                b.last_upd_date, b.last_upd_by,

                c.name as dept_name, content_id

              FROM (Content_Type a, Department c)

              LEFT OUTER JOIN Content b on

                a.type_id = b.content_type

                and a.type_id = b.content_type

                and b.dept_id = $dept_id

                and b.content_type = $type_id

              WHERE c.dept_id = $dept_id

              ORDER BY content_date DESC

 

 

True. You need to put the table names in  "( table name1 as a, table_name2 as b) LEFT JOIN ....." when you use join's.

[xgd]

thanks dathremar, that helped, though there are still issues with the app, but i think it's too complicated to discuss here.