Posted by kennthompson on 02/27/07 15:02

The name passes -- the echo test works; it just won't work as a mysql
table. The variable is getting there, but mysql for some reason does
not recognize it. I've read something about prepare and execute in
mysql, but not sure this is the answer to my problem.

On Feb 27, 8:36 am, "Jose da Silva" <josedasi...@gmail.com> wrote:
> Hello, have you tried some debug?
>
> Try to
> echo "My table name in function: ".$tableName;
>
> Maybe the variable is not getting the value you expected
>
> Regards
> jose Silva
>
> On Feb 27, 2:25 pm, kennthomp...@gmail.com wrote:
>
>
>
> > Trouble passing mysql table name in php. If I use an existing table
> > name already defined everything works fine as the following script
> > illustrates.
>
> > <?php
> > function fms_get_info()
> > {
> > $result = mysql_query("select * from $tableInfo") ;
> > for ($i = 0; $i < mysql_num_rows($result); $i++)
> > {
> > /* do something */
> > }
>
> > }
>
> > /* Main */
> > fms_get_info();
>
> > But it won't work if I pass a variable table name to the function.
>
> > <?php
> > function fms_get_info($tableName)
> > {
> > $result = mysql_query("select * from $tableName") ;
> > for ($i = 0; $i < mysql_num_rows($result); $i++)
> > {
> > /* do something */
> > }
>
> > }
>
> > /* Main */
> > fms_get_info($tableInfo);
>
> > I need to use the same function to gather information from multiple
> > tables at will without creating a different function for each possible
> > mysql database table by name. I thought this would be easy, but I
> > have failed at several tries.- Hide quoted text -
>
> - Show quoted text -

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