Error in SQL Syntax HELP!!!


Posted on 16th Feb 2014 07:03 pm by admin

I have this page:

Code: <?php

Did you know?Explore Trending and Topic pages for more stories like this.
session_start();

//connect to server and select database
$conn = mysql_connect("localhost", "root", "")
or die(mysql_error());
$db = mysql_select_db("smrpg", $conn) or die(mysql_error());

//show scouts characters
$get_scouts = "select * from scouts where username = '".$_SESSION['userName']."'";
$get_scouts_res = mysql_query($get_scouts, $conn) or die(mysql_error());
while ($list_scouts = mysql_fetch_array($get_scouts_res)) {
$identity = ucwords($list_scouts['identity']);
$topic_id = $list_scouts['id'];
echo "<ul class="character_list"><li><a href="fight.php?identity=$identity">$identity</li></ul> ";
}
?>

And it goes to this page:

Code: <?php

session_start();

//connect to server and select database
$conn = mysql_connect("localhost", "root", "")
or die(mysql_error());
$db = mysql_select_db("smrpg", $conn) or die(mysql_error());

//check for required info from the query string
if (!$_GET['identity']) {
header("Location: train_fight.php");
exit;
}

//get derived values

$derived = "select * from derived_values where identity = $_GET[identity]";
$derived_res = mysql_query($derived, $conn) or die(mysql_error());

$display_block = "<ul>";

while ($derived_info = mysql_fetch_array($derived_res)) {
$derived_id = $derived_info['id'];
$derived_identity = $derived_info['identity'];
$derived_health = $derived_info['health'];
$derived_energy = $derived_info['energy'];
$derived_acv1 = $derived_info['acv1'];
$derived_acv2 = $derived_info['acv2'];
$derived_dcv1 = $derived_info['dcv1'];
$derived_dcv2 = $derived_info['dcv2'];
$derived_total_cp = $derived_info['total_cp'];

$display_block .= "<li>$derived_identity</li>";

}

$display_block .= "</ul>";

?>

But I am getting this error:

You have an error in your SQL syntax; check the manual that correspondsto your MySQL server version for the right syntax to use atline 1

What am I doing wrong here? If I change my where statement or take it out, it is displaying the information, but I can't figure out what's wrong with my where statement or where I'm getting my "identity" from. Something's wrong but I can't find it. Can anyone help?
No comments posted yet

Your Answer:

Login to answer
136 Like 33 Dislike
Previous forums Next forums
Other forums

Problems with returning true or false in eval()'d code
Hi guys,

Would appreciate some help with a problem when running eval() on a function that sho

Problem with the Update command used with a sqldataadapter
I'm connected to a database on an SQL Server and I'm using a sqldataadapter, sqlconnection, sqldatas

cPanel API
Hi Guys,

Need a little guidance. I'm trying to get my script to communicate with cPanel (or W

How to get all server headers like Live http Headers does
Hey all, like many of you I use the Firefox addon "Live http Headers". I'm trying to write

cURL proxy
Okay here is the thing... i know how to add proxys

Code: curl_setopt($ch, CURLOPT_PROXYTYPE,

Code clarification
Hi

In the following code what could be the "search_print()" and where it could be

Perplexing problem showing a .jpg
Please disregard..........I figured it out

Problem Dereferencing
With these types and tables:

CREATE TYPE MANAGER AS OBJECT (
MGR_ID INTEGER,

MySql timezone
Code: $sql = "SELECT *, date_format(date, '%m/%d/%Y at %I:%i %p' )as date FROM comments WHERE t

ORA-01017: invalid username/password; logon denied
Dear All,

I am facing problem in taken backup from db13 it comes up with the following l

Sign up to write
Sign up now if you have flare of writing..
Login   |   Register
Follow Us
Indyaspeak @ Facebook Indyaspeak @ Twitter Indyaspeak @ Pinterest RSS



Play Free Quiz and Win Cash