For Each loop not echoing data (mysql_fetch_assoc problem?) - php

Question

Hello and Good Morning,
I am still learning PHP and for some reason my script will not post any data in my foreach loop. Any Idea why? The emailRow Echos out fine but I am going to remove My code is below:
<?php
include 'includes/header.php';
$accountUser = array();
$upgradeEmail = $_GET['currEmail'];
$emailQuery = "SELECT fbID, firstName, lastName FROM users WHERE emailOne='".$upgradeEmail."' AND authLevel=0";
<?php echo $emailRow['fbID']; ?>
<?php echo $emailRow['firstName']; ?>
<?php echo $emailRow['lastName']; ?>
while($emailRow = mysql_fetch_assoc($emailQuery, $conn))
{
$accountUser[]=$emailRow;
}
?>
<table>
<?php foreach($accountUser as $emailData) { ?>
<tr><td> <?php emailData['fbID']; ?> </td><td><?php emailData['firstName']; ?></td><td><?php emailData['lastName']; ?></td></tr>
<?php } ?>
</table>

Answer 1

You have constructed your SQL query in $emailQuery, but never executed it. Call mysql_query(), and pass its result resource to mysql_fetch_assoc().
$emailQuery = "SELECT fbID, firstName, lastName FROM users WHERE emailOne='".$upgradeEmail."' AND authLevel=0";
$result = mysql_query($emailQuery);
if ($result)
{
while($emailRow = mysql_fetch_assoc($result, $conn))
{
$accountUser[]=$emailRow;
}
}
else // your query failed
{
// handle the failure
}
Please also be sure to protect your database from SQL injection by calling mysql_real_escape_string() on $upgradeEmail since you're receiving it from $_GET.
$upgradeEmail = mysql_real_escape_string($_GET['currEmail']);

Answer 2

You don't actually echo anything.
as well as not running the query.
and there are some other methods to do things, much cleaner than usual uglyPHP.
a function
function sqlArr($sql){
$ret = array();
$res = mysql_query($sql) or trigger_error(mysql_error()." ".$sql);
if ($res) {
while($row = mysql_fetch_array($res)){
$ret[] = $row;
}
}
return $ret;
}
a code
$email = mysql_real_escape_string($_GET['currEmail']);
$data = sqlArr("SELECT * FROM users WHERE emailOne='$email' AND authLevel=0");
include 'template.php';
a template
<? include 'includes/header.php' ?>
<table>
<? foreach($data as $row) { ?>
<tr>
<td><?=$row['fbID']?></td>
<td><?=$row['firstName']?></td>
<td><?=$row['lastName']?></td>
</tr>
<? } ?>
</table>

Answer 3

YOU HAVE A SYNTAX ERROR. You can't open a new php tag within an existing php tag. You have to close the already open tag first.
As far as getting the query to work,
First you have to fetch data before printing it or echoing it...
while($emailRow = mysql_fetch_assoc($emailQuery, $conn))
{
$accountUser[]=$emailRow;
}
then you may write statements..
echo $emailRow['fbID']; etc. code.
Secondly you have not fired a query, just written the query statement. Use mysql_query to fire it.
Your code would be something like this..
<?php include 'includes/header.php';
$accountUser = array();
$upgradeEmail = $_GET['currEmail'];
$emailQuery = mysql_query("SELECT fbID, firstName, lastName FROM users WHERE emailOne='".$upgradeEmail."' AND authLevel=0") or die (mysql_error());
while($emailRow = mysql_fetch_assoc($emailQuery, $conn))
{
$accountUser[]=$emailRow;
}
echo $emailRow['fbID'];
echo $emailRow['firstName'];
echo $emailRow['lastName'];
print '<table>';
foreach($accountUser as $emailData) {
print '<tr><td>'$.emailData['fbID'].'</td><td>'.$emailData['firstName'].'</td><td>'.$emailData['lastName'].'</td></tr>';
}
print '</table';
?>
Feel free to use this code, modifying it to fit your needs.