[SOLVED] SELECT FROM failure....

[George Botley]

I must once again ask for the PHP Freak communites help,

 

Here is the code:

 

<?
//Variable Setup
$name = $_POST["name"];
$dayphone = $_POST["dayphone"];
$evephone = $_POST["evephone"];
$mobphone = $_POST["mobphone"];
$email = $_POST["email"];
$q5 = $_POST["q5"];
$assignment = $_POST["q5a"];
$q5b = $_POST["q5b"];
$q6 = $_POST["q6"];
$q5c = "$q5b $q5";
$q5d = "Enquiry about $q5"; 

//Script Execute

echo "<h2 class='greenh2'>Message Sent</h2>

$name,<br/><br />

Your message has been sent to the relevant person within the club. Please bare in mind that our managers run teams as well as there jobs. As a result there may be a long delay before you recieve a response.

<br /><br />

Kindest Regards,<br/>
Pen Mill FC";

//Check for Relevant Person
$query = "SELECT * FROM team_members WHERE assignment='$assignment'"; 

$qry = mysql_query($query)
or die ("Could not match data because ".mysql_error());

while($row = mysql_fetch_array($qry))
  {
  $to_contact = $row['name'];
  $to_email = $row['email'];
  $to_subject = "$q5d";
  $to_message = "$q6";
  $today = date("Y-m-d");
  $time = date("g:i a");
  }
  
  echo "$to_contact";
  
//Send Message To Relevant Person

//$con = mysql_connect("mysql.interdns.co.uk","penmillfc","derrick2009");
//if (!$con)
//  {
//  die('Could not connect: ' . mysql_error());
//  }
//
//mysql_select_db("penmillfc", $con);
//
//$sql="INSERT INTO team_messages (to, from, from_email, subject, message, date, time) 
//VALUES ('$to_contact','$_POST[name]','$_POST[email]', '$to_subject', '$to_message', '$today', '$time')";
//
//if (!mysql_query($sql,$con))
//  {
//  die('Error: ' . mysql_error());
//  }
//echo "1 record added";

?>

 

Ignore the commented out section but when I run this script and try to extract the person from the database it doesn't extract them with the variable in the

WHERE assignment='$assignment'

section. It will however extract a person when I manually write their assignment for example

WHERE assignment='Payment'

.

 

My variables are present and correct which was proven after echoing them onto the page.

 

Thanks

George

[ignace]

WHERE assignment LIKE '%$assignment%'

[rhodesa]

What happens when you echo out the $query:

$query = "SELECT * FROM team_members WHERE assignment='$assignment'";
echo $query;

what is the output?

[George Botley]

@ignace

 

Thanks but that didn't seem to work...

 

 

@rhodesa

 

The output is SELECT * FROM team_members WHERE assignment='%Pen Mill U14s%' 

 

The % follows the suggestion of ignace.

[George Botley]

Hmmm.

 

I may have found my own solution to this quesiton.... It is outputting the team not the assignment, something wrong with my variables me thinks.

 

I will update you shortly.

[George Botley]

I have sorted the intial fault but now when adding a AND to the equation it spits out

 

Could not match data because 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 ''Pen Mill U14s' at line 1

 

The new query is

 

$query = "SELECT * FROM team_members WHERE assignment='$assignment' AND team = '$team";

[rhodesa]

I would check your html form and make sure your input names are assigned correctly. also, at the top of this script, you can put a print_r($_POST); and make sure all key/values are what you expect