mysql_fetch_array(): supplied argument is not a valid MySQL result

astoller

What is wrong with my fetch array statement in this script?
& is giving me a "not valid" error?

$Host="localhost";
$User="root";
$Password="123";
$DBName= "NewDatabase";
$TableName= "Feedback";

$Link = mysql_connect ($Host, $User, $Password);
$Query = "SELECT * from $TableName" ;
$Result = mysql_db_query ($Query, $Link);

//create a table for display

print ("<TABLE BODER=1 WIDTH=\"75%\"
CELLSPACING=2 CELLPADDING =2
ALIGN=CENTER>\n");
print ("<TR ALIGN=CENTER VALIGN=TOP>\n");

print ("<TD ALIGN=CENTER VALIGN=TOP>Company</td>\n");
print ("<TD ALIGN=CENTER VALIGN=TOP>Activity</td>\n");
print ("<TD ALIGN=CENTER VALIGN=TOP>Contact</td>\n");
print ("<TD ALIGN=CENTER VALIGN=TOP>Phone</td>\n");

print ("<TD ALIGN=CENTER VALIGN=TOP>Mobile</td>\n");
print ("<TD ALIGN=CENTER VALIGN=TOP>Email</td>\n");
print ("<TD ALIGN=CENTER VALIGN=TOP>Add1</td>\n");
print ("<TD ALIGN=CENTER VALIGN=TOP>Add2</td>\n");
print ("<TD ALIGN=CENTER VALIGN=TOP>Add3</td>\n");

print("</tr>\n");

//Fetch the results from the database
while ($Row = mysql_fetch_array ($Result))
{
print ("<TR ALIGN=CENTER VALIGN=TOP>\n");

DigitalExpl0it

here try this out for starters, you forgot a close } and some table close stuff


<?
$Host="localhost";
$User="root";
$Password="123";
$DBName= "NewDatabase";
$TableName= "Feedback";

$connect = mysql_connect($Host, $User, $Password) or die ("<b><center>".mysql_errno().": ".mysql_error()."</b></center>\n");
$db = mysql_select_db($DBName, $connect) or die ("<b><center>".mysql_errno().": ".mysql_error()."</b></center>\n");
$check_query = "SELECT * from $TableName";
$results = mysql_query($check_query, $connect);

//create a table for display
?>
<TABLE BODER=1 WIDTH="75%" CELLSPACING="2" CELLPADDING="2" ALIGN="CENTER">
	<TR ALIGN="CENTER" VALIGN="TOP">
		<TD ALIGN="CENTER" VALIGN="TOP">Company</td>
		<TD ALIGN="CENTER" VALIGN="TOP">Activity</td>
		<TD ALIGN="CENTER" VALIGN="TOP">Contact</td>
		<TD ALIGN="CENTER" VALIGN="TOP">Phone</td>
		<TD ALIGN="CENTER" VALIGN="TOP">Mobile</td>
		<TD ALIGN="CENTER" VALIGN="TOP">Email</td>
		<TD ALIGN="CENTER" VALIGN="TOP">Add1</td>
		<TD ALIGN="CENTER" VALIGN="TOP">Add2</td>
		<TD ALIGN="CENTER" VALIGN="TOP">Add3</td>
	</TR>
<?
//Fetch the results from the database
while ($row = mysql_fetch_array($results)){
?>
		<TR ALIGN="CENTER" VALIGN="TOP"></TR>
<?
}
?>
</TABLE>

tsinka

Hi,

you forgot the database parameter in mysql_db_query, e.g.

mysql_db_query ($DBName,$Query, $Link);

kevinsequeira

mysql_db_query syntax
http://ca.php.net/manual/en/function.mysql-db-query.php

reg
kevin

astoller

Whether or not I have
mysql_db_query ($DBName,$Query,$Link);
or drop the db in this statement,
the error still seems to be in the line

while ($Row=mysql_fetch_arrray($Result))

I keep getting same error message as before, i.e.

mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\program

tsinka

Did you try

DigitalExpl0it's code ? You might also want to add the error handling to your mysql_query call to see what error exactly occurs.

kevinsequeira

debug ..
after
$Result = mysql_db_query ($Query, $Link);
echo $Query."<BR>";
if(mysql_error())
echo mysql_error();

reg
kevin

astoller

Thanks Kev,

That little bit of trapping caught my bug and helped me fix the problem.
Art