Adding $id to end of url

bulldog317

As a follow up to this thread , I now have my page that will display the information depending on what the $id is.

Once the form is filled out by the user and "posted", how do I pass the $id number assigned to the information the user just posted to the end of the url to display what they just added once they post the information so I can automatically send the user an e-mail with a link to the page displaying their information? Here is the code for the page (view.php) that displays the information:

<?php
include ("include/header.inc.php");

include ("include/dbconnect.php");

if ($id) {

$result = mysql_query("SELECT * FROM $table WHERE id=$id",$db);

$links = mysql_fetch_array($result);

$id = $links["id"];
$staff = $links["staff"];
$cnum = $links["cnum"];
$boss = $links["boss"];

if(strlen($links["opend"]))
$opend.=$links["opend"];

$cname = $links["cname"];
$client = $links["client"];

if(strlen($links["ref"]))
$ref.=$links["ref"];

if(strlen($links["other"]))
$other.=$links["other"];

if(strlen($links["comments"]))
$comments.=$links["comments"];
echo "<table width='800'>";
 echo "<tr><td width='150'>Prepared by:</td><td width='650'>$staff</td></tr>";
echo "<tr><td width='150'>Case No.:</td><td width='650'>$cnum</td></tr>";
echo "<tr><td width='150'>Boss:</td><td width='650'>$boss</td></tr>";
echo "<tr><td width='150'>Date Opened:</td><td width='650'>$opend</td></tr>";
echo "<tr><td width='150'>Case:</td><td width='650'>$cname</td></tr>";
echo "<tr><td width='150'>Client:</td><td width='650'>$client</td></tr>";
echo "<tr><td width='150'>Ref. No.:</td><td width='650'>$ref</td></tr></table>";
echo "<table width='800'>";
echo "<tr><td width='800'>Other:</td></tr>";
echo "<tr><td width='800'>$other</td></tr>";
echo "<tr><td width='150'>Comments:</td></tr>";
echo "<tr><td width='800'>$comments</td></tr>";
} else {
echo "No ID number exists";

}
include ("include/footer.inc.php");

?>

<html>
<head><title></title></head>
<body> 
If you have a problem with the above, please contact <? echo "$boss"; ?>. Otherwise, please acknowledge you understand by selecting the button beside your name and selecting OK. 

</body></html>

paul088

view.php?id=$id

kevinsequeira

reg
kevin

bulldog317

Maybe I am making things more complicated than they are. I understand that the page would be view.php?id=$id but how do you pass the id number assigned to the information that was just submitted to replace $id in the URL?

kevinsequeira

hi bulldog,

well im not sure what it is that you need. i read the earlier thread and this is what i understood.

user submits data on the form.
the data is collected, checked and submitted to a database and a mail sent to you.

now do you want a mail to be sent to the user at the same time with a link to the view.php page ???

reg
kevin

paul088

Q:
1. Where is $id from?
2. What url and on what page do you want to have $id appended to it?

kevinsequeira

if thats what you want to do then in process.php add this code after you insert into the database

// your code
$query="insert into conflicts (staff,cnum,boss,jury,rate,comments) values ('".$staff."','".$cnum."','".$boss."','".$jury."','".$rate."','".$comments."')";
mysql_query($query);

//followed by this

$id_query = "select id from conflicts where ";
$id_query .= "staff = '".$staff."',;
$id_query .= "cnum = '".$cnum."',;
$id_query .= "boss = '".$boss."',;
$id_query .= "jury = '".$jury."',;
$id_query .= "rate = '".$rate."',;
$id_query .= "comments = '".$comments."'";

$id_result = mysql_query($id_query);
if(mysql_num_rows($id_result)==1){
$id_row = mysql_fetch_array($id_result);
$id = $id_row['id'];// assuming that id is the name in the table
$link_to_user = "<a href=/view.php?id=$id>view link</a>
// mail this to the user.
}

reg
kevin

bulldog317

You are correct Kevin. I now want an e-mail sent to the user with a link to the view.php page. Thanks for the code!!

I have placed the code as you mentioned and now I am getting a parse error as follows:

Parse error: parse error, unexpected T_STRING in c:\process.php on line 60

Line 60 is: $id_query .= "cnum = '".$cnum."',;

Here is my code with the code you have provided:

<?
include("global.inc.php");
$errors=0;
$error="The following errors occured while processing your form input.<ul>";
pt_register('POST','staff');
pt_register('POST','cnum');
pt_register('POST','boss');
pt_register('POST','jury');
pt_register('POST','rate');
pt_register('POST','comments');
$comments=preg_replace("/(\015\012)|(\015)|(\012)/"," ", $comments);if($staff=="" || $cnum=="" || $boss=="" ){
$errors=1;
$error.="<li>You did not enter one or more of the required fields. Please go back and try again.";
}
if($errors==1) echo $error;
else{
$where_form_is="http".($HTTP_SERVER_VARS["HTTPS"]=="on"?"s":"")."://".$SERVER_NAME.strrev(strstr(strrev($PHP_SELF),"/"));
$message="Prepared By: ".$staff."
Case No.: ".$cnum."
Boss: ".$boss."
Jury: ".$jury."
Rate: ".$rate."
Comments: ".$comments."
";
mail("me@here.com","Memo - $cnum",$message,"From: $staff");
$link = mysql_connect("blah","blah","blah");
mysql_select_db("memos",$link);
$query="insert into conflicts (staff,cnum,boss,jury,rate,comments) values ('".$staff."','".$cnum."','".$boss."','".$jury."','".$rate."','".$comments."')";
mysql_query($query);

}

$id_query = "select id from conflicts where ";
$id_query .= "staff = '".$staff."',;
$id_query .= "cnum = '".$cnum."',;
$id_query .= "boss = '".$boss."',;
$id_query .= "jury = '".$jury."',;
$id_query .= "rate = '".$rate."',;
$id_query .= "comments = '".$comments."'";

$id_result = mysql_query($id_query);
if(mysql_num_rows($id_result)==1){
$id_row = mysql_fetch_array($id_result);
$id = $id_row['id'];// assuming that id is the name in the table
$link_to_user = "<a href=/view.php?id=$id>view link</a>
// mail this to the user.
}

?>
<HTML>
<HEAD><TITLE>Memo Confirmation</TITLE></HEAD>
<BODY>
<? $time = date("H:i"); ?><? $date = date("n/d/y"); ?>

MEMO SENT SUCCESSFULLY... 



Thanks <? echo "$staff"; ?>. The conflict check re: <? echo "$cnum"; ?> was sent successfully on <? echo "$date"; ?> at <? echo "$time"; ?>.
 If you need to create a new conflict check, hit the back button to return to the previous
form and hit the 'reset form' button.
 Click <a href="mailto:me@here.com">here</a> to e-mail any questions or problems you have. 

</BODY>
</HTML>

kevinsequeira

sorry abt that, i didnt close the quotes at the end of the lines and i have t use AND , so change
$id_query = "select id from conflicts where ";
$id_query .= "staff = '".$staff."',;
$id_query .= "cnum = '".$cnum."',;
$id_query .= "boss = '".$boss."',;
$id_query .= "jury = '".$jury."',;
$id_query .= "rate = '".$rate."',;
$id_query .= "comments = '".$comments."'";

$id_query = "select id from conflicts where ";
$id_query .= "staff = '".$staff."' AND ";
$id_query .= "cnum = '".$cnum."' AND ";
$id_query .= "boss = '".$boss."' AND ";
$id_query .= "jury = '".$jury."' AND ";
$id_query .= "rate = '".$rate."' AND ";
$id_query .= "comments = '".$comments."'";

also do a print($id_query); to see what it prints out

re
kevin

bulldog317

It is giving a parse error stating that I have an unexpected $ on line 76 which is the last line closing my php code [ ?> ]. I have not added the "print" code yet.

kevinsequeira

change $link_to_user = "<a href=/view.php?id=$id>view link</a>

$link_to_user = "<a href=/view.php?id=$id>view link</a>";

here you go, the complete code

<?
include("global.inc.php");
$errors=0;
$error="The following errors occured while processing your form input.<ul>";
pt_register('POST','staff');
pt_register('POST','cnum');
pt_register('POST','boss');
pt_register('POST','jury');
pt_register('POST','rate');
pt_register('POST','comments');
$comments=preg_replace("/(\015\012)|(\015)|(\012)/"," <br />", $comments);
if($staff=="" || $cnum=="" || $boss=="" ){
     $errors=1;
     $error.="<li>You did not enter one or more of the required fields. Please go back and try again.";
}
if($errors==1){
	echo $error;
}else{
     $where_form_is="http".($HTTP_SERVER_VARS["HTTPS"]=="on"?"s":"")."://".$SERVER_NAME.strrev(strstr(strrev($PHP_SELF),"/"));
     $message = "Prepared By: ".$staff."Case No.: ".$cnum."Boss: ".$boss."Jury: ".$jury."Rate: ".$rate."Comments: ".$comments."";
     mail("me@here.com","Memo - $cnum",$message,"From: $staff");
     $link = mysql_connect("blah","blah","blah");
     mysql_select_db("memos",$link);
     $query="insert into conflicts (staff,cnum,boss,jury,rate,comments) values ('".$staff."','".$cnum."','".$boss."','".$jury."','".$rate."','".$comments."')";
     mysql_query($query);

 $id_query = "select id from conflicts where ";
 $id_query .= "staff = '".$staff."',;
 $id_query .= "cnum = '".$cnum."',;
 $id_query .= "boss = '".$boss."',;
 $id_query .= "jury = '".$jury."',;
 $id_query .= "rate = '".$rate."',;
 $id_query .= "comments = '".$comments."'";

 $id_result = mysql_query($id_query);
 if(mysql_num_rows($id_result)==1){
      $id_row = mysql_fetch_array($id_result);
      $id = $id_row['id'];// assuming that id is the name in the table
      $link_to_user = "<a href=/view.php?id=$id>view link</a>";
			print($link_to_user);
      // mail this to the user.
 }
}

reg
kevin

bulldog317

Thanks Kevin. I can't believe I missed such a simple mistake. It works perfectly. So then I assume that all I have to do is add a reference to $link_to_user somewhere in my $message string for the e-mail and that should do it. Thanks again.

kevinsequeira

thats right, u just display the $link_to_user in your message somewhere.

could u mark this as resolved if you have no more doubts

reg
kevin