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Solved

How do I fix this MySQL UPDATE statement

Posted on 2011-02-24
3
343 Views
Last Modified: 2012-06-21
Experts,
What is the proper way to construct this MySQL UPDATE statement?

Thanks for your help!


<?php
include("global.inc3.php");
$errors=0;
$error="The following errors occured while processing your form input.<ul>";
pt_register('POST','id');
pt_register('POST','Date');
pt_register('POST','LastName');
pt_register('POST','FirstName');
pt_register('POST','Address');
pt_register('POST','Phone1');
pt_register('POST','Phone2');
pt_register('POST','Friend1Name');
pt_register('POST','Friend1Phone');
pt_register('POST','Friend2Name');
pt_register('POST','Friend2Phone');
pt_register('POST','Friend3Name');
pt_register('POST','Friend3Phone');

if($errors==1) echo $error;
else{
$where_form_is="http".($HTTP_SERVER_VARS["HTTPS"]=="on"?"s":"")."://".$SERVER_NAME.strrev(strstr(strrev($PHP_SELF),"/"));
$message="id: ".$id."
Date: ".$Date."
Last Name: ".$LastName."
Last Name: ".$FirstName."
Address: ".$Address."
Primary Phone: ".$Phone1."
Secondary Phone: ".$Phone2."
Friend/Relative Name 1: ".$Friend1Name."
Friend/Relative Phone 1: ".$Friend1Phone."
Friend/Relative Name 2: ".$Friend2Name."
Friend/Relative Phone 2: ".$Friend2Phone."
Friend/Relative Name 3: ".$Friend3Name."
Friend/Relative Phone 3: ".$Friend3Phone."
";
$message = stripslashes($message);
mail("my email","org",$message,"From: me");
$link = mysql_connect("HOST","347UN","PW");
mysql_select_db("DB",$link);

$query="UPDATE seniorslist SET Address = ".$Address.",Phone1 = ".$Phone1.",Phone2 = ".$Phone2.",Friend1Name = ".$Friend1Name.",Friend1Phone = ".$Friend1Phone.",Friend2Name = ".$Friend2Name.",Friend2Phone = ".$Friend2Phone.",Friend3Name = ".$Friend3Name.",Friend3Phone = ".$Friend3Phone." WHERE id = $id";
mysql_query($query);


header("Refresh: 0;url=http://www.url.com/NewSite/seniorwatch_view.php?id=$id"); 
?><?php 
}
?>

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0
Comment
Question by:rlb1
3 Comments
 
LVL 3

Accepted Solution

by:
sergiobg57 earned 350 total points
ID: 34971709
Just the query?
Then it's this way.(look at the code snippet or below)

"UPDATE seniorslist SET Address = '$Address', Phone1 = '$Phone1', Phone2 = '$Phone2', Friend1Name = '$Friend1Name', Friend1Phone = '$Friend1Phone', Friend2Name = '$Friend2Name', Friend2Phone = '$Friend2Phone', Friend3Name = '$Friend3Name', Friend3Phone = '$Friend3Phone' WHERE id = '$id' "

_________________________
Sergio C.





"UPDATE seniorslist SET Address = '$Address', Phone1 = '$Phone1', Phone2 = '$Phone2', Friend1Name = '$Friend1Name', Friend1Phone = '$Friend1Phone', Friend2Name = '$Friend2Name', Friend2Phone = '$Friend2Phone', Friend3Name = '$Friend3Name', Friend3Phone = '$Friend3Phone' WHERE id = '$id' "

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0
 
LVL 3

Assisted Solution

by:LFLFM
LFLFM earned 150 total points
ID: 34974072
Is all this data coming from the user?
If so, did you mysql_real_escape_string() it?
example:
$Address = mysql_real_escape_string($Address);

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If you haven't, then do so for every single string that gets sent to a database... ;-)

Here is your update code:
$query="UPDATE seniorslist SET Address = '$Address', Phone1 = '$Phone1', Phone2 = '$Phone2', Friend1Name = '$Friend1Name', Friend1Phone = '$Friend1Phone', Friend2Name = '$Friend2Name', Friend2Phone = '$Friend2Phone', Friend3Name = '$Friend3Name', Friend3Phone = '$Friend3Phone." WHERE id = '$id'";

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0
 

Author Closing Comment

by:rlb1
ID: 34976906
Thanks!
0

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