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Posted by Jerry Stuckle on 02/27/07 16:27
kennthompson@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.
>
This should work fine. What do you get back as an error message? How
are you calling the function?
What happens if you do the following:
function fms_get_info($tableName)
{
$sql = "select * from $tableName";
echo $sql . "<br>\n";
$result = mysql_query($sql) ;
for ($i = 0; $i < mysql_num_rows($result); $i++)
{
/* do something */
}
--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================
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