Re: [PHP] strange MySQL result problem

[Marek Kilimajer]

Is your query correct?

Try:

function query($sql, $current_line)
{
        $this->Result = mysql_query($sql)
                or die($this->stop($current_line));
        if(!$this->Result) {
                echo mysql_error();
        }
        $this->Result_total_fields = mysql_num_fields($this->Result);
        $this->Result_total_rows = mysql_num_rows($this->Result);
}
Chris W. Parker wrote:
Hey peeps.

Let me be quick (or try to at least).

Here is a snippet of my db class.

function query($sql, $current_line)
{
        $this->Result = mysql_query($sql)
                or die($this->stop($current_line));
// the following two lines work if they are in the next
// method get_query_results() but not when they are in
// this method.
        $this->Result_total_fields = mysql_num_fields($this->Result);
        $this->Result_total_rows = mysql_num_rows($this->Result);
}
function get_query_results()
{
        $this->Result_Arr = array();
        if($this->Result_total_rows > 0)
        {
        // initialize counter
                while($line = mysql_fetch_array($this->Result,
MYSQL_BOTH))
                {
                                $this->Result_Arr[] = $line;
                }
        }
	mysql_free_result($this->Result);

        return $this->Result_Arr;
}
Ok. As the comments in the first method state, those two lines of code
do not work* when they are used in the first method. If I put them down
into the get_query_results() method they work fine. I can't figure out
why this is happening since my code is syntactically correct.


Chris.

* By "do not work" I mean it throws the error "supplied argument is not
a valid MySQL result resource".
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