Converting Array Value to Variable

[Capt Spaghetti]

Thanks to a forum member I have moved a little forward but now I am stumped on how to convert a passed value to a variable so it can be used to search a MySQL database. When I use the passed variable as the value I receive the following error:

"SELECT * FROM courses WHERE org_id = 

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1"

If I "hard code" the actual value passed I am able to select the correct record. Here is php code:

<?php

// connect include
require ("connect.php");

$once = print_r($_GET, true);
echo "<p>" . $once. "<p>";

$twice= var_export($_GET, true);
echo "<p>" . $twice. "<p>";


print_r($_GET);
if($_GET["varID"] === "") echo "varID is an empty string\n";
if($_GET["varID"] === false) echo "varID is false\n";
if($_GET["varID"] === null) echo "varID is null\n";
if(isset($_GET["varID"])) echo "varID is set\n";
if(!empty($_GET["varID"])) echo "varID is not empty";

/* $query = "SELECT * FROM organizations WHERE org_id = {$_GET['varID']}"; */
$query = "SELECT * FROM organizations WHERE org_id = 9661";

/* Temporary ECHO of the $sql string */ 
echo "<p>" . $query . "</p>";

$result = mysql_query($query)or die(mysql_error());
if (!$result) { 
  echo("<p>Error performing query: " . mysql_error() . "</p>"); 
  exit(); 
} 

$result = mysql_query($query)or die(mysql_error());
if (!$result) { 
  echo("<p>Error performing query: " . mysql_error() . "</p>"); 
  exit(); 
} 

$num = mysql_num_rows($result);

while ($row = mysql_fetch_assoc($result))
{

    echo("<tr>\n<td>" . $row["cityname"] . "</td>"); 
    echo("<td>" . $row["statename"] . "</td>"); 
    echo("<td>" . $row["orgname"] . "</td>");
    echo "<td>" .$row['org_id']."</td>";  

}
/* Closes Connection to MySQL server */ 

mysql_close ($connect); 
?> 

The output of this code is as follows:

Array ( [varID_] => 9661 ) 


array ( 'varID_' => '9661', )

Array ( [varID_] => 9661 ) varID is null 

SELECT * FROM organizations WHERE org_id = 9661

BILLINGSMONTANAAssociation of State Grazing Districts9661 

I have tried reading up on the php functions for arrays but can't find what I need. Any help would be appreciated.

Hi Capt Spaghetti (love the name btw),

Are you sure the key 'varID' is being passed correctly in the URL? It looks like 'varID_' is the key that is being displayed in your echo statements ($once and $twice).

Zagga

[Capt Spaghetti]

I down select using a MySQL statement and the user picks their choice based on the org_id. I am passing it as follows:

echo "<a href = 'graborginfo9.php?varID =".$row['org_id']."'>".$row['org_id']."</td></a>";

Should I be doing something differently?

Capt Spaghetti

Hi again,

you need to remove the space where you declare varID.

Change

echo "<a href = 'graborginfo9.php?varID =".$row['org_id']."'>".$row['org_id']."</td></a>";

to

echo "<a href = 'graborginfo9.php?varID=".$row['org_id']."'>".$row['org_id']."</td></a>";

and try that.

Zagga

[Capt Spaghetti]

Zagga,

My father use to kid me by asking me if I majored in space(i.e. taking up space). Now I see how important space is. That did the trick. Thank you very much for the assistance.

Thanks,
Capt Spaghetti