Re: Problems with Arrays and print and echo

chris smith wrote:

On 4/7/06, Michael Felt <register@felt.demon.nl> wrote:

Slowly I am getting the output I want. Trying to use "dynamic" arrays, does creat the array I want, but getting the info is sometimes surprising. I notice a difference between arrays used locally in a function, and arrays used as a 'var' in a class function (all in PHP 4 atm). Code snippet:

                echo "ROWS returned are: $max\n";
                $this->count = $max;

                while ($max--) {
                        $row = mysql_fetch_row($result);
                        $this->name[$max] = sprintf("%s", $row[0]);
                        $Name[$max] = sprintf("%s", $row[0]);

echo "init \$this->Xame[$max] = $row[0]"; echo " $Name[$max] $this->name[$max]\n";

                        $regionID[$max] = $row[1];
                        $constellationID[$max] = $row[2];
                        $this->ID[$max] = $row[3];

printf("%d:%d/%d/%s\n",$max,$regionID[$max],$constellationID[$max],

                               $this->name[$max]);
                }

================ Line wrap is messing things up a bit. Was trying sprintf to see if the was a buffer problem coming from mysql. Problem seems to be the same, regardless. Also, the names changes ($this->name[] versus $Name[]) are deliberate, for just in case.... ================ Output (debuging): ROWS returned are: 7 init $this->Xame[6] = 8-TFDX 8-TFDX Array[6]

Is the problem that you're getting array[6] instead of the value? Explain what you see and what you expect to see. What is var $name originally set to, ie:

$row = mysql_fetch_row($result); $this->name[$max] = sprintf("%s", $row[0]); $Name[$max] = sprintf("%s", $row[0]); Is the actual assignment of variables. What surprises me is that the 'local' variable echos what I expect, but the 'class' variable does not.

        function init($id)
        {
                $this->ID[0] = ERROR;
                $this->name[0] = "";

Hope this answers your question. And yes, I am not happy the the 'Array[X]' output, I am expecting the value, not what it is. I have already tried establishing an array type early in the function...

var $name = array(); (or '' or ....) ? -- Postgresql & php tutorials http://www.designmagick.com/