Solved

Message in Email Not Passing

Posted on 2013-06-25
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154 Views
Last Modified: 2013-06-26
I have a form that sends E'mails.  This works fine.  The problem is that I can't getthe message portion to show.

<?php
include("config.php");
$Lid = $_POST['Lid'];
$Lid = mysql_real_escape_string($Lid);  
mysql_query("SET CHARACTER SET utf8");
$sql=mysql_query("SELECT tblRestaurants.RestName,tblLocations.StreetNumber,tblLocations.Street,
	tblLocations.CrossOne, tblLocations.CrossTwo, tblCities.CityName,tblStates.StateName,tblZipCodes.ZipCodeName,
	tblLocations.Phone,tblLocations.Email, tblLocations.Fax, tblLocations.SMS, tblLocations.LocationID
	FROM tblLocations
	INNER JOIN tblRestaurants ON tblRestaurants.RestID = tblLocations.RestID
	INNER JOIN tblCities ON tblCities.CityID = tblLocations.CityID
	INNER JOIN tblStates ON tblStates.StateID = tblLocations.StateID
	INNER JOIN tblZipCodes ON tblZipCodes.ZipCodeID = tblLocations.ZipCodeID
	WHERE tblLocations.LocationID = '$Lid'");
	while($row=mysql_fetch_array($sql))
	{
	$RestName = $row['RestName'];
	$StreetNumber = $row['StreetNumber'];
	$Street = $row['Street'];
	$CrossOne = $row['CrossOne'];
	$CrossTwo = $row['CrossTwo'];
	$City = $row['CityName'];
	$State = $row['StateName'];
	$ZipCode = $row['ZipCodeName'];
	$Phone = $row['Phone'];
	$Email = $row['Email'];
	$Fax = $row['Fax'];
	$SMS = $row['SMS'];
	$Lid = $row['LocationID'];
	}
 
    $to = $_POST['ESend'];
 	$subject = "Menuhead.com";
 	$RestName = $_POST['RestName'];
	$Email = $_POST['Email'];
	$message="
	Name: " . $_POST['RestName']." \n
	Emai: $Email \n";
 	$from = "Menuhead";
 	$headers = "From:" . $from;
 	mail($to,$subject,$message,$headers);
 	echo "Mail Sent.";
 ?>

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Question by:DS928
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11 Comments
 
LVL 58

Expert Comment

by:Gary
ID: 39276784
Not sure what you are trying to do here
	$message="
	Name: " . $_POST['RestName']." \n
	Emai: $Email \n";

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But try
$message='Name: "' . $_POST['RestName'].'" \nEmail: $Email \n';

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0
 

Author Comment

by:DS928
ID: 39276824
Not quite.  I should be getting back:

Name: Name of Restaurant
Email: Email of Restaurant

What I am getting is.

Name:''\nEmail:$Email\n

Is there a way to check if the first part is grabbing the values?
0
 
LVL 83

Expert Comment

by:Dave Baldwin
ID: 39276837
First, if you are letting users put in the 'To:' email address, you are inviting them to spam the world.  You don't appear to be doing any filtering or checking of your inputs either.

Other than that... try this.
 	$RestName = $_POST['RestName'];
	$Email = $_POST['Email'];
	$message="Name: $RestName\r\n Email: $Email \r\n";
 

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Author Comment

by:DS928
ID: 39276845
All that comes back are the labels.

Name:
Email:

I know the values are there because I did an echo on the above PHP query and all of the values appeared.  This is just very basic for now, once this works I can go onto the next step.
0
 
LVL 83

Expert Comment

by:Dave Baldwin
ID: 39276895
I'm wondering if you are confusing a form POST with 'rows' from the database because you are using the same variable names in two different ways.  Try this..
 	//$RestName = $_POST['RestName'];
	//$Email = $_POST['Email'];
	$message="Name: $RestName\r\n Email: $Email \r\n";
 

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0
 

Author Comment

by:DS928
ID: 39276909
Still Labels.  Not quit sure why these values aren't passing.  Should some variable names be changed?
0
 
LVL 56

Accepted Solution

by:
Julian Hansen earned 500 total points
ID: 39277080
Change
while($row=mysql_fetch_array($sql))
	{
	$RestName = $row['RestName'];
	$StreetNumber = $row['StreetNumber'];
	$Street = $row['Street'];
	$CrossOne = $row['CrossOne'];
	$CrossTwo = $row['CrossTwo'];
	$City = $row['CityName'];
	$State = $row['StateName'];
	$ZipCode = $row['ZipCodeName'];
	$Phone = $row['Phone'];
	$Email = $row['Email'];
	$Fax = $row['Fax'];
	$SMS = $row['SMS'];
	$Lid = $row['LocationID'];
	}
 
    $to = $_POST['ESend'];
 	$subject = "Menuhead.com";
 	$RestName = $_POST['RestName'];
	$Email = $_POST['Email'];
	$message="
	Name: " . $_POST['RestName']." \n
	Emai: $Email \n";
 	$from = "Menuhead";
 	$headers = "From:" . $from;
 	mail($to,$subject,$message,$headers);
 	echo "Mail Sent.";

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TO
$row=mysql_fetch_array($sql);
if ($row) {
	$RestName = $row['RestName'];
	$StreetNumber = $row['StreetNumber'];
	$Street = $row['Street'];
	$CrossOne = $row['CrossOne'];
	$CrossTwo = $row['CrossTwo'];
	$City = $row['CityName'];
	$State = $row['StateName'];
	$ZipCode = $row['ZipCodeName'];
	$Phone = $row['Phone'];
	$Email = $row['Email'];
	$Fax = $row['Fax'];
	$SMS = $row['SMS'];
	$Lid = $row['LocationID'];
 
        $to = $_POST['ESend']; // Is this posted 
 	$subject = "Menuhead.com";
	$message="Name:  $RestName\n";
        $message += "Emai: $Email \n";
 	$headers = "From: <info@menuhead.com Menuhead";
 	mail($to,$subject,$message,$headers);
 	echo "Mail Sent.";
}
else {
 // process error here
}

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Your original code reads from the database and then spins its wheels setting values that are overwritten if there is more than one record - which their most likely won't be because you are filtering by an ID. But in the second section some of those values are overwritten by POST values - which does not make sense. In the context of your earlier question where you were only POSTING the LiD not sure where the ESend is comming in unless something changed.
0
 
LVL 83

Expert Comment

by:Dave Baldwin
ID: 39277091
Don't you mean...
$message .= "Email: $Email \n";

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0
 
LVL 56

Expert Comment

by:Julian Hansen
ID: 39277191
um (blush) yes - was in JScript mode with my project ...
0
 
LVL 83

Expert Comment

by:Dave Baldwin
ID: 39277262
I understand that!
0
 

Author Closing Comment

by:DS928
ID: 39278368
Bingo! That worked!  Thank you!
0

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