Update problem w/MYSQL- STUMPED!!

tony_l

Ok, here is the problem:

I have a site that is members only and it allows members to enter a profile that will eventually be searchable.

The members need a way to update and modify their Profiles, therefore, I have been working on getting that section up.

I have a form that re-displays the profile information and they can just click on a checkbox to indicate that they wish to modify that section.

I then have a script that the form parses to and after the script runs, it goes to a success page that displays the newly modified profile.

BUT- when I tested it, it does not update the database at all. I will give an excerpt of the code from the form and then an piece of the Parser:

FORM THAT THEY MODIFY THE PROFILE FROM:

<tr>
<td width='5%'> <font face='Arial, Helvetica, sans-serif'>
<input type='checkbox' name='chglname' value='1'>
</font></td>
<td width='49%'><font face='Arial, Helvetica, sans-serif'>Change<font color='#FF8000'>
LAST NAME</font> to:</font></td>
<td width='46%'> <font face='Arial, Helvetica, sans-serif'>
<input type='text' name='lname'value='$Array2[lname]'>
</font></td>
</tr>

PIECE OF CODE THAT CORESSPONDS TO THE ABOVE, BUT WITHIN THE PARSER-

IF (chgpass=='1') {

$Query4 = "UPDATE $TableName2 SET lname='$lname' WHERE profile.profile_id='$ID'";

}

Now the link to the database and all the other stuff that would make these scripts work are in the code, but the application is like over 700 lines each script, so this is the readers digest version of the script.

Do you guys see anything wrong with this script? I thought I did it correctly, unless my SQL is all wrong 🙁

Anon

Well, one possible error is this:

$Query4 = "UPDATE $TableName2 SET lname='$lname' WHERE profile.profile_id='$ID'";

Is $TableName2 'profile'? coz if it's not, then the query won't work as expected, try:

$Query4 = "UPDATE $TableName2 SET lname='$lname' WHERE $TableName2.profile_id='$ID'";

or even easier:

$Query4 = "UPDATE $TableName2 SET lname='$lname' WHERE profile_id='$ID'";

Also, before you call mysql_query(), print out the sql statement, to see if there's anything wrong there.

Diego

tony_l

Well I have tried that actually, and it does nothing but print out the ID Variable- to help you guys help me- here is the whole parser script-

<?php
session_start();

########################################

This Parses the Modify Profile App

########################################

###################

Include Files

###################

include ("file outside of root that has MYSQL variables");

#########################

Connect to Database

#########################

$TableName1 = "auth";
$TableName2 = "profile";
$TableName3 = "bio";

$Link1 = mysql_connect($Host, $User, $Password);

mysql_select_db($DBName, $Link1);

$MID = $varpass;
#########################################################################

GO THROUGH IF STATEMENTS TO DETERMINE WHAT DATA NEEDS TO BE UPDATED

#########################################################################

IF (chgpass=='1') {

$Query1 = "UPDATE $TableName1 SET pass='$pass' WHERE auth.auth_id='$MID'";

}

IF (chgemail=='1') {

$Query2 = "UPDATE $TableName2 SET email='$email' WHERE profile.profile_id='$MID'";

}

IF (chgpass=='1') {

$Query3 = "UPDATE $TableName2 SET fname='$fname' WHERE profile.profile_id='$MID'";

}

IF (chgpass=='1') {

$Query4 = "UPDATE $TableName2 SET lname='$lname' WHERE profile.profile_id='$MID'";

}

IF (chgpass=='1') {

$Query5 = "UPDATE $TableName2 SET occupation='$occupation' WHERE profile.profile_id='$MID'";

}

IF (chgpass=='1') {

$Query6 = "UPDATE $TableName2 SET country='$country' WHERE profile.profile_id='$MID'";

}

IF (chgpass=='1') {

$Query7 = "UPDATE $TableName2 SET city='$city' WHERE profile.profile_id='$MID'";

}

IF (chgpass=='1') {

$Query8 = "UPDATE $TableName2 SET state='$state' WHERE profile.profile_id='$MID'";

}

IF (chgpass=='1') {

$Query9 = "UPDATE $TableName2 SET timezone='$timezone' WHERE profile.profile_id='$MID'";

}

IF (chgpass=='1') {

$Query10 = "UPDATE $TableName2 SET aim='$aim' WHERE profile.profile_id='$MID'";

}

IF (chgpass=='1') {

$Query11 = "UPDATE $TableName2 SET interest='$interests' WHERE profile.profile_id='$MID'";

}

IF (chgpass=='1') {

$Query12 = "UPDATE $TableName3 SET species='$species' WHERE bio.bio_id='$MID'";

}

IF (chgpass=='1') {

$Query13 = "UPDATE $TableName3 SET cage='$cage' WHERE bio.bio_id='$MID'";

}

IF (chgpass=='1') {

$Query14 = "UPDATE $TableName3 SET cgender='$cgender' WHERE bio.bio_id='$MID'";

}

IF (chgpass=='1') {

$Query15 = "UPDATE $TableName3 SET force='$force' WHERE bio.bio_id='$MID'";

}

IF (chgpass=='1') {

$Query16 = "UPDATE $TableName3 SET pro1='$pro1' WHERE bio.bio_id='$MID'";

}

IF (chgpass=='1') {

$Query17 = "UPDATE $TableName3 SET pro2='$pro2' WHERE bio.bio_id='$MID'";

}

IF (chgpass=='1') {

$Query18 = "UPDATE $TableName3 SET imageurl='$image' WHERE bio.bio_id='$MID'";

}

IF (chgpass=='1') {

$Query19 = "UPDATE $TableName3 SET chistory='$chistory' WHERE bio.bio_id='$MID'";

}

print ("$Query1");
print ("$Query2");
print ("$Query3");
print ("$Query4");
print ("$Query5");
print ("$Query6");
print ("$Query7");
print ("$Query8");
print ("$Query9");
print ("$Query10");
print ("$Query11");
print ("$Query12");
print ("$Query13");
print ("$Query14");
print ("$Query15");
print ("$Query16");
print ("$Query17");
print ("$Query18");
print ("$Query19");
print ("MID IS : $MID");
print ("Varpass is $varpass");
?>

tony_l

I was thinking that maybe it was the way i have the form set-up I have it where each input tag has the value="parameter" to show the answers they previously placed in their profile. Would this have anything to do with it? Any help would be appreciated-

Example-

I was thinking that maybe since the value equals what is already in the database that it would somehow effect the outcome of the query, such as inserting what is already there?

But the UPDATE query would have Printed I would think. I am stumped. And i need this done like ASAP, i was think of even starting all over again, but it took me forever to construct all this 🙁