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@NogDogOct 29.2007 — #It's telling you that your mysql_query() failed, most likely due to a syntax error or invalid identifier in your SQL. (mysql_query() returns a resource ID unless it fails, so when you use that result in another function which is expecting a MySQL resource ID but you are instead sending it a boolean FALSE, you get that error message. To do some quick debugging, change the query line to:
Then if the query fails, you will see the query as it was actually sent to MySQL in the parentheses, followed by the error message returned by MySQL.
Getting parse error can’t figure out what I’m doing wrong. Need your expertise.
Hello everyone.
I am developing a site. I am no stranger to HTML code, as I have been writing it for years. But, this is my first foray into PHP and everytime I upload my site to my web server and run the pages, pretty much every page is coming up with a parse error. This page especially is coming up with a ‘Parse error: syntax error, unexpected T_VARIABLE, expecting ‘(‘ in /home/crescen9/public_html/logintop.php on line 9’. Here is the code for that page. I have been racking my brain trying to figure it out, it’s probably something stupid, but:
<?
session_start();
if (!isset($_SESSION[’email’])) {
mysql_connect(“localhost” , “*****“, “
or die(‘Database is not responding.’);
mysql_select_db(“crescen9_Bandusers”) or die(‘Unable to connect to database’);
$query= “SELECT first_name, band_name, cdsalecount FROM Userinfo”;
$result= ‘mysql_query($query)’;
while $row= ‘mysql_fetch_assoc($result)’;
{
echo “Name :$row[‘first_name’]}<br>” .
“Band Name : $row[‘band_name’] <br>” .
“CDs Sold To Date : $row[‘cdsalecount’] <br><br>”;
}
} else {
print'<html><body bgcolor=”#888888″><center><font size=”2″ face=”verdana”>You are not logged in! Click <a href=”login.php”>here to log in!<br>
Don’t have an account? <a href=”bandusersignup.php”>Sign Up Now!!!</a></font></center></body></html>’;
}
?>
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@NogDogOct 29.2007 — #It's telling you that your mysql_query() failed, most likely due to a syntax error or invalid identifier in your SQL. (mysql_query() returns a resource ID unless it fails, so when you use that result in another function which is expecting a MySQL resource ID but you are instead sending it a boolean FALSE, you get that error message. To do some quick debugging, change the query line to:[code=php]
$result= mysql_query($query) or die("query failed ($query) " . mysql_error());
[/code]Then if the query fails, you will see the query as it was actually sent to MySQL in the parentheses, followed by the error message returned by MySQL.