Re: Trouble passing mysql table name to php function and using it!

In article <op.toe7s3uqqnv3q9@misant>, Rik <luiheidsgoeroe@xxxxxxxxxxx>
wrote:

<kennthompson@xxxxxxxxx> wrote:
Jerry Stuckle <jstuck...@xxxxxxxxxxxxx> wrote:
kennthomp...@xxxxxxxxx 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.
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 */
}

}
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";
}

No. That's exactly the way I was trying to do it. MySQL doesn't accept
a table name passed into a function this way. There must be some
hidden mysql code that I'm unaware of.

Jerry did ask for you to echo mysql error...

But here you go:
function fms get info($tableName){
$qry = "SELECT * FROM `{$tableName}`";
$res = mysql query($qry) or die($qry.' failed, mysql
sais:'.mysql error());
}

Why does one need the back ticks and { ?? I would have expected one
could construct a query with:

$tab = "thistable";
$query = "select * from " . $tab;
$res = mysql_query ($query);

or thereabouts. What does the extra stuff do?

Thanks,
.

Relevant Pages