mysql_fetch_array(): supplied argument is not a valid..

[princess]

hey another question from me.. still working on the guestbook thing, but started over with a new one, problems i dont understand came on my way..

 

please can you look at this and explain the mistake to me? i really dont see whats wrong.

 

This is the warning i receive while i test it:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:FoxServwwwtestguestbookindex.php on line 19

 

this is my code:

 

<?

$db = mysql_connect(\"lxxx\",\"xxx\",\"xxx\");

mysql_select_db (\"testgastenboek\", $db);

 

$sql = \'SELECT * FROM `testgastenboek`;\';

$result = mysql_query($sql,$db);

 

while ( $myrow = mysql_fetch_array($result) )

{

echo \"<b>Name:</b>\";echo \"<br>\";

echo $myrow[name];

echo \"<br>\";

echo \"<b> Email:</b>\";

echo \"<br>\";

echo \"<a href=\"mailto:\";

echo $myrow;

echo \"\">\";

echo $myrow;

echo \"</a>\";

echo \"<br>\";

echo \"<b> Homepage:</b>\";

echo \"<br>\";

echo \"<a href=\"\";

echo $myrow;

echo \"\">\";

echo $myrow;

echo \"</a>\";

echo \"<br>\";

echo \"<b>Comment:</b>\";

echo \"<br>\";

echo $myrow[comment];

echo \"<br>\";

echo \"<br>\";

echo \"<a href=changeform.php?id=$myrow[id]>[change]</a>\";

echo \"&nbsp\";

echo \"<a href=changeform.php?id=$myrow[id]>[delete]</a>\";

echo \"<hr noshadow width=\"300\" align=\"left\" color=\"#FF6600\">\";

}

?>

 

 

i hope this is enough information..

 

thanks in advance, Princess

[effigy]

the error means that the query never executed, therefore it has nothing to be identified with. this means there is a problem with any of these: (a) the database connection; (B) choosing a database; © running the query

 

i think this extra \';\' here is the problem:

$sql = \'SELECT * FROM `testgastenboek`;\';

semicolons are not needed to end a query when using php, opposed to the mysql command line.

[delamitri]

easiest way, and i guess that the last poster is right

print mysql_error();

 

after the query call.

 

 

Kevin

[munnabd_02]

progli@progli.awardspace.com here my data show 'tt' but other server side not work Why ? Please solve and mail me: munnabd_02@yahoo.com

 

<?php

//if you view 'tt' mail me(munnabd_02@yahoo.com)

$cnn=mysql_connect('db1.awardspace.com', 'progli_progli', '8013731')

or die ("Cannot connect to MySQL server: " . mysql_error());

mysql_select_db('progli_progli')

or die ("Cannot connect to db: " . mysql_error());

$result=mysql_query("select *from user");

$row=mysql_fetch_row($result);

echo $row[1];

?>

 

[artacus]

Change

$result=mysql_query("select *from user");

to

$result=mysql_query("select *from user") or die(mysql_error() . "<br>$query");

 

and

"select *from"

to

"select * from"