PHP Query not working

Question

I have a problem that is staring me in the face but I cannot seem to solve it.

First of:

1) the code is parsed prior to check for SQL injection.
2) $dbc is a reference that is part of another script that is "require_once()" called in and is working perfectly as its fine for 2 other scripts depending on it
3) the query text when pumped straight into mysql calls back exactly what Im after
4) If I put a false array into $displayBlogs, the rest of script acts as it should.

The table is populated. So why do I seem to receive no results, no errors (checked with mysqli_error($dbc) ) ? Blogs just seems to be empty.

function getSnippets()
{
  // set up the query   
  $query1 = "SELECT * FROM blogs LIMIT 0, 10";

  // action the query and save the connection
  $blogs = mysqli_query($dbc, $query1);

  // blank out the variable that will be used to save the query results
  $displayBlogs = '';

  // iterate through the query results and set up the module return
  while($blog = mysqli_fetch_array($blogs))
  {
    $displayBlogs .= "<div class='article'><a href='" . $blog['link'] ."'>" .
                     "<h1>" . $blog['title'] . "</h1>" .
                     "<h2>" . $blog['date'] . "</h2>" .
                     "<p>" . $blog['body'] . "</p>" .
                     "</a></div>";
  }      
  return $displayBlogs;
}

score 3 · Accepted Answer · edited May 23 '17 at 11:48

3

You can declare global scope for $dbc :

function getSnippets()
{
    global $dbc;
    // declaring $dbc global to have it accessed outside the scope of the function
}

or, even better, pass it as an argument of the function, as using global variables is considered bad practice (check, among many, this SO question as for why):

function getSnippets($dbc)
{
   $connection = $dbc;
   //...
   $blogs = mysqli_query($connection, $query1);

  // rest of code...
}

edited May 23 '17 at 11:48

Community

1
1

answered Jun 12 '11 at 22:25

Damien Pirsy

24,544
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This this was it! I just passed the variable to the function and hey preseto! – Paul Chambers Jun 13 '11 at 05:44

score 1 · Answer 2 · answered Jun 12 '11 at 22:17

1

Even though it's required, the $dbc variable shouldn't be accesable from inside a function. To make it posible, use it as a global.

In that case, it'd be written, at the start, you should write:

function getSnippets()
{
    global $dbc;

answered Jun 12 '11 at 22:17

Lumbendil

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score 0 · Answer 3 · answered Jun 12 '11 at 22:15

0

Do you need to global $dbc;?

answered Jun 12 '11 at 22:15

Scott C Wilson

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score 0 · Answer 4 · answered Jun 12 '11 at 22:22

You should be using var_dump() on each of the variables used till you nail the culprit.

$dbc
$query1
$blogs
$blog

Also, you should be monitoring your mysql logs to see what, if anything, is actually happening at the database end.

Not solving your issue I realise (pretty sure Scott got that) but hopefully leading you to isolate and identify similar errors in the future (you will make them).

PHP Query not working

4 Answers4