MySQL Query not printing out data

Question

I'm trying to retrieve some data from my table only for some reason I cant get it to return anything.

<?php

$curr_uemail = mysql_query("select * from produgg_users where produgg_users.id = ".$usersClass->userID().") or die(mysql_error())");    

$arr_uemail = mysql_fetch_array($data);

while($arr_uemail = mysql_fetch_array($data)) 
{
echo $arr_uemail['email'];
}

/*For Debugging purposes
echo $usersClass->userID();*/

?>

Can anybody see anything wrong with my syntax?

what is this? ") or die(whoops)"); didn't you notice the strange highlighting? x) — reeaal
– reeaal, Commented May 31, 2011 at 21:05
haha sorry I didnt, either way with that fixed it's still not working — Liam
– Liam, Commented May 31, 2011 at 21:06

dynamic · Accepted Answer · 2011-05-31 21:14:36Z

1

There are a lot of errors.

The first is

$curr_uemail = mysql_query("select * from produgg_users where produgg_users.id = ".$usersClass->userID()) or die(whoops);

The second you shouldn't call mysql_fetch_assoc before while, because if the result contains only 1 Record it would never enter in the while

The final code is:

$curr_uemail = mysql_query("select * from produgg_users where produgg_users.id = ".$usersClass->userID()) or die('whoops');

while($arr_uemail = mysql_fetch_array($curr_uemail)) {
echo $arr_uemail['email'];
}

As stated by marc if you have only one records than this could become:

$curr_uemail = mysql_query("select * from [etc]") or die('whoops');    
$arr_uemail = mysql_fetch_array($curr_uemail);
echo $arr_uemail['email'];

edited May 31, 2011 at 21:14

answered May 31, 2011 at 21:07

dynamic

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3 Comments

Liam Over a year ago

Sorry I've amended my original question.

dynamic Over a year ago

that's because you should do mysql_fetch_array($curr_uemail) and not $data And I have edited my answer

Marc B Over a year ago

Assuming that produgg_users.id is a PK field, the query could only ever produce once record at most, so the while loop is useless. a single fetch call would take care of things.

reeaal · Accepted Answer · 2011-05-31 21:14:12Z

0

I told you the first error via my comment. But please enable error handling and do what the interpreter will tell you.

error_reporting(E_ALL);
ini_set('display_errors', 1);

What also can be erroneous, is this statement here:

while($arr_uemail = mysql_fetch_array($data))

You should enclose it with parentheses or the interpreter may warn you.

while(($arr_uemail = mysql_fetch_array($data)))

answered May 31, 2011 at 21:14

reeaal

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2 Comments

Marc B Over a year ago

The extra brackets are useless. Omitting them does not cause a warning/error.

reeaal Over a year ago

No, it may cause a warning, see: stackoverflow.com/questions/718415/…

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