如何在PHP网页上使用mySQL正确回显？

Question

I got a problem in querying an information in mySQL, here's the code:

SELECT avatar FROM amcms_users WHERE username='admin'

and the result is ' 59da6ceb5c74ac98f317a4b4af3c72f6.jpg ' which is correct.

Now when I load it on php page using these php codes...

<?php
$locAvatar = mysql_query("SELECT avatar FROM amcms_users WHERE username='admin'");
echo $locAvatar;
?>

and the result is wrong, ' Resource id #27 '

How can I echo it correctly? Thank you.

Answer 1

You need to fetch the data first, by using the mysql_fetch_* functions:

$res = mysql_query("SELECT avatar FROM amcms_users WHERE username='admin'");
$locAvatar = mysql_fetch_assoc($res)["avatar"];
echo $locAvatar;

( mysql_fetch_assoc fetches into an array)

Answer 2

You need to fetch the results and then echo them. The resource (which is a reference to the results) that mysql_query() returns is to be passed on to one of the mysql_fetch_ series of functions. These functions dereference the result and return the data in the appropriate format.

Have a look at:

• mysql_fetch_array
• mysql_fetch_assoc
• mysql_fetch_row