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Posted by PSI_Orion on 07/19/06 12:37
Thanks for the response, I'll give it a go for both suggestions.
"PSI_Orion" <orion@psiss.com> wrote in message
news:0NUmg.5431$1g.1061@newsfe1-win.ntli.net...
>I have a table set up which has 4 columns. However, when I use the
>sizeof() function on the array, it always returns DOUBLE the number of
>columns. Here is an extract of my code...
>
> function createDefaults($tableName,$userName,$configName,$dateModified)
> {
> $query="SELECT * FROM ".$tableName." WHERE id='0';";
> $result = mysql_query($query);
> if(!$result)
> die("ERROR RETRIEVING DEFAULT DATA FROM '".$tableName."");
> $dataArray = mysql_fetch_array($result);
>
> $query="INSERT INTO ".$tableName." VALUES
> ('0','$userName','$configName','$dateModified'";
> $colCount = sizeof($dataArray) / 2; <*************** this is where
> it's returning double the column count, hence the / 2
> for($count = 0; $count < $colCount - 4; $count++) // the -4 is a
> standard set of columns put in each table
> {
> $query .= ",DEFAULT";
> }
> $query .= ");";
> $result = mysql_query($query);
> if(!$result)
> die("ERROR SAVING DEFAULT DATA IN
> '".$tableName."'<br>".mysql_error()."</p>");
> }
>
> Can anyone explain why?
>
> Oh, if I do a print out of the array for each entry returned by sizeof(ie:
> 8 for 4 columns) it does say it's an empty or undefined index.
>
> One side note, where I'm doing a lot of testing and I'm using auto
> increment on the ID field, how can I restet the values so it starts from,
> lets say, 2 instead of 50?
>
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