mysql_fetch_array problems

[eon201]

Hi im trying to access the 'update' time from my database by using the below code.

 

				$result = mysql_query("SHOW TABLE STATUS FROM logtable;");
				while($array = mysql_fetch_array($result)) {
				print_r($array[update_time]); // Will print information about each table stored in database

 

 

The only problem is I get this error. What am I doing wrong??

 

'Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in...'

 

 

Please help, been stick on this one for a while now! Many thanks Eon201 ???

[rajivgonsalves]

i think

$result = mysql_query("SHOW TABLE STATUS FROM logtable;");

 

should be

$result = mysql_query("SHOW TABLE STATUS FROM logtable");

 

there is a semi-colin in the statement

[pranav_kavi]

$result = mysql_query("SHOW TABLE STATUS FROM logtable;");
while($array = mysql_fetch_array($result,MYSQL_ASSOC)) {
    print_r($array[update_time]); // Will print information about each table stored in database
}

[aschk]

Run that SQL through your command line as the user you are using in PHP. I bet you get an error code coming up.

[eon201]

Ok.

 

So im getting the same error still if I use either rajivgonsalves, or pranav_kavi's examples.

 

Im not quite sure what you mean aschk??

 

Can anyone see what is happening here, or have doen the same thign before??

 

Thanks. Eon201

[pranav_kavi]

try running the sql statement

SHOW TABLE STATUS FROM logtable;

in sql server directly rather than thru php.Ther may b some error bcos of wich ur result set is not getting populated.

[rajivgonsalves]

please use the following code to see what the error is

 

$result = mysql_query("SHOW TABLE STATUS FROM logtable") or die("there was an error ".mysql_error());


				while($array = mysql_fetch_array($result)) {
				print_r($array[update_time]); 

[eon201]

lol.

 

I made a stupid mistake. I had the wrong database name! idiot!

 

so the coe is now like so...

 

				$result = mysql_query("SHOW TABLE STATUS FROM pc-monitor") or die("there was an error ".mysql_error());
				while($array = mysql_fetch_array($result)) 
				{
				print_r($array[update_time]); 
				}

 

But the error now says 'there was an error You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '-monitor' at line 1'

 

I believe the problem now lie's in the '-' between pc and monitor.

Is the code misenterpreting this?

[rajivgonsalves]

this should work

 

mysql_query("SHOW TABLE STATUS FROM `pc-monitor`") or die("there was an error ".mysql_error());

[eon201]

Ok that didnt bring up an error.

 

But neither did they display any output??