Echoing users in a MySQL database via PHP

Question

Possible Duplicate:
mysql_fetch_array() expects parameter 1 to be resource, boolean given in select

I have the following code

$result = mysql_query("SELECT * FROM members WHERE group = '$groupid'");

while($row = mysql_fetch_array($result))
{
  echo $row['name'];
}

which I want to echo all users who are in a certain group, but I get the error "Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\Program Files\wamp\www\state_ecotracker\group.php on line 11"

I do not know what to do.

score 3 · Accepted Answer · answered Feb 06 '11 at 23:53

3

Your PHP code is correct.

GROUP is a reserved word in MySQL as it corresponds to the GROUP BY clause of SELECT queries.

You shouldn't use it as a column name. Change your table to call group something else.

If you can't do this for some slightly absurd reason (I don't know your boss), you'll have to always enclose that column name in backticks in every SQL query that uses it.

$result = mysql_query("SELECT * FROM members WHERE `group` = '$groupid'");

answered Feb 06 '11 at 23:53

Dan Grossman

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Jesus f'ing f that's frustrating. It works now though, so THANK YOU! – ahota Feb 06 '11 at 23:57
1

Here's the list of words not to use as column names for reference: http://dev.mysql.com/doc/refman/5.1/en/reserved-words.html – Dan Grossman Feb 06 '11 at 23:59

score 0 · Answer 2 · answered Feb 06 '11 at 23:52

0

Change

$result = mysql_query("SELECT * FROM members WHERE group = '$groupid'");`

to

$result = mysql_query("SELECT * FROM members WHERE group = '$groupid'") or die(mysql_error());

You can then see what error message MySQL is throwing you, once you fix that your query should work fine.

It's also worth noting that if you really want to use those MySQL PHP functions and not something like PDO then you should ideally use mysql_fetch_assoc and not mysql_fetch_array.

answered Feb 06 '11 at 23:52

Prisoner

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@downvoter.. Please let me know what was wrong. – Prisoner Feb 06 '11 at 23:55
what would be the problem with `mysql_fetch_array`? Anyway your solution does not really solve the problem, which is using a MySQL keyword as a field name. – nico Feb 06 '11 at 23:58
I know my solution did not solve the problem, but it would have helped solve the problem. Using debugging helps you solve problems by yourself, instead of requiring you to be assisted. The difference is using array you get an array with both index and field name (larger array). – Prisoner Feb 07 '11 at 00:05
I was just pointing out that in this case debugging would not be so easy as MySQL won't tell you that it is complaining because of a bad choice in field (btw, I'm not the downvoter) – nico Feb 07 '11 at 06:55
@nico, that is true but it would have began the error with '..near group' which may have lead him to a google search and then lead to the solution. (http://www.google.co.uk/search?sourceid=chrome&ie=UTF-8&q=mysql+error+near+group - 6th link down) – Prisoner Feb 07 '11 at 11:30

score 0 · Answer 3 · answered Feb 06 '11 at 23:55

0

While using mysql reserved words as field name, enclose them in `:

$result = mysql_query("SELECT * FROM members WHERE `group` = '$groupid'");

Or rename the field name.

answered Feb 06 '11 at 23:55

a1ex07

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score 0 · Answer 4 · answered Feb 06 '11 at 23:55

First you should sanitise your input:

$groupId = (int)$groupId; // cast to integer (I suppose because of the Id)
$result = mysql_query('SELECT * FROM members WHERE `group` = ' . $groupId); 
// use backticks because GROUP is a reserved word.

while ($row = mysql_fetch_array($result)) {
  echo $row['name'];
}

You can use mysql_error() to check what is causing an error with the query.

Echoing users in a MySQL database via PHP

4 Answers4