MySQL查询有效，但在PHP中不起作用

Question

I want to simply display a value fetched via a query.

$newquery = "SELECT fullname FROM `users` WHERE user_id=" . $row['usernumber'];
$newresult=mysql_query($newquery);
$newrow = mysql_fetch_row($newresult);
echo "<td>" . $newrow[0]. "</td>";

(Where $row['usernumber'] is fetched from a previous query.)

I get this error:

Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in /a/b/viewall.php on line 41

Which I presume, means that my query is failing because of some reason.

Here if I echo $newquery; I get

SELECT fullname FROM `users` WHERE user_id=3

And if I run this in mysql separately, it returns the fullname corresponding to user_id=3 properly.

What could be the problem?

Answer 1

For starters, put some error checking after mysql_query:

$newresult = mysql_query($newquery);
if (!$newresult) {
   die('Invalid query: ' . mysql_error());
}

Answer 2

Always write PHP code to check the return value of mysql_query() (or equivalent function in mysqli or PDO). It returns false if there's an error. That's why you get the error you did, because $newresult was false -- not a legitimate query resource.

If mysql_query() returns false, you should report the error message and find out what's wrong.

$newquery = "SELECT fullname FROM `users` WHERE user_id=" . $row['usernumber'];
$newresult=mysql_query($newquery);
if ($newresult === false) { die(mysql_error()); }
$newrow = mysql_fetch_row($newresult);

For example, other errors are possible instead of syntax errors. Perhaps you don't have the privileges needed to read that table when connecting from PHP, even though you do have the privileges when connecting in a direct query client.

Answer 3

On my system, it works best to backtick all field/table names and to single-quote variables.

The addition of mysql_error() should also help you identify the error.

Note change to query text.

Try this:

$un = $row['usernumber'];

$newquery = "SELECT `fullname` FROM `users` WHERE `user_id`= '$un'";
$newresult=mysql_query($newquery) or die(mysql_error());
$newrow = mysql_fetch_row($newresult);
echo "<td>" . $newrow[0]. "</td>";

Answer 4

It looks likely to be the connection / database selection code which precedes this code block.

As others have mentioned, Please look in to other database driver alternatives (mysqli or PDO)

Answer 5

try this

$newquery = sprintf("SELECT fullname FROM `users` WHERE user_id=%d " ,$row['usernumber']);
$result = mysql_query($newquery);
if (!$result ) 
{
   die('Invalid query: ' . mysql_error());
}
else
{
   //do something
}