[PHP-DB] Selecting Drop Down Value From DB to Edit

I have the below code that populates a drop down list. The code will correctly insert the value $StateID into another a table. The problem I am running into is that when I select the record to edit the drop-downlist has the first option as the value instead of what the corresponding StateID in

Re: [PHP-DB] Selecting Drop Down Value From DB to Edit

StateID in the column reads. How can I correct this? The form to insert and edit have the below code and correctly insert the StateID What exactly is happening? Run the script, do a View Source on the resulting page; show us the HTML generated. -Original Message- From: Steve Fitzgerald

[PHP-DB] Select Inner Join Question

I'm trying to display a summary list of calls with one attribute being CallType. I have CallType and CallTypeID defined in a table name calltypes. The problem I am having is taking the output of CallID defined in calls and having the script match the CallTypeID to the CallType. If I run the

[PHP-DB] List Results Not Working

I'm trying to list the results of a query, but I keep getting the following error: Parse error: parse error, expecting `T_STRING' or `T_VARIABLE' or `T_NUM_STRING' I must be missing something because I can't find what's causing the

[PHP-DB] Re: UPDATE Not Updating

which worked fine. I'm not sure why the query string is not passing the value. Steve Fitzgerald [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... I belive this UPDATE should work, but it isn't. I get no errors. I have also tried using FieldName= '$F

[PHP-DB] Permissions

I'm trying to build a menu page that builds links based upon specified permissions that are associated with a username or userid. I figure it might be best to associate individual privleges to specific menu items within the usertable (using checkboxes, etc). then building the menu based upon the

[PHP-DB] Re: SQL Error

Steve Fitzgerald [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... I keep getting the following SQL error: The last fieldname is CompanyID not ContactID. You have an error in your SQL syntax near ')' at line 14 Line 14 is: '$FirstName', so the er

[PHP-DB] SQL Error

I keep getting the following SQL error: You have an error in your SQL syntax near ')' at line 14 Line 14 is: '$FirstName', so the error must correspond to a different part of the code. Any ideas? $add_contact_sql = INSERT INTO $table_name (ContactID, FirstName, LastName, Title, WorkPhone,

[PHP-DB] INSERT Won't Insert

This should be a simple insert. I must be missing something because I keep getting: Couldn't execute query. What am I missing? ? error_reporting(15); $db_name = dbname; $table_name=company; $connection = @mysql_connect(servname,username, password)or die (Couldn't connect.); $db =

[PHP-DB] Avoiding Multiple IDs

What is the best method for avoiding multiple ids for the same information. For example: I have a table name company that has unique ids. The CompanyID are tied to a table name contacts which have unique ids (ContactID). The problem seems to be that if I add a contact with an existing

[PHP-DB] UPDATE Not Updating

I belive this UPDATE should work, but it isn't. I get no errors. I have also tried using FieldName= '$Fieldname', but that does not seem to work either. $table_name = contacts; $update_contact_sql = UPDATE $table_name SET FirstName =\$FirstName\, LastName = \$LastName\, Title = \$Title\,

[PHP-DB] A Join Question

I'm trying to create a join statement that pulls out a CompanyName based on a given CompanyID that is tied to a specified ContactID. For example, if ContactID=1 then the corresponding CompanyName might be Smith, Inc. depending on what was entered. Here are the tables. table = Contacts

[PHP-DB] INSERT with a JOIN

I'm trying to create an INSERT statement that will change a field in one table based upon the id defined in another table. Here is what I tried: $sql1 = INSERT INTO $table_name1 (FirstName,LastName,WorkPhone, HomePhone,EmailName,Birthday) VALUES

[PHP-DB] multiple table query error

I've figured out (with some help) how to query two tables at the same time. In order to get a corresponding CompanyName with a specified ContactID I created a SQL query that selects both tables ($table_name1,$table_name2) and then selects contacts.CompanyID = '$CompanyID'. The result is a match

[PHP-DB] 2 Tables- 1 Insert Problem

I'm trying to insert data into two separate tables using 1 INSERT. The code below represents a first crack at it, but I can't seem to figure out how to get this to work properly. Thanks. Steve ? $db_name=testDB; $table_name1 = my_contacts; $table_name2= company; $connection = @mysql_connect

[PHP-DB] Search Query

=no which would indicate the contact is no longer active). I would then like to have the result(s) to have a link to their respective ContactID. I don't think this is that difficult, but I'm having difficulty applying the SELECT statement with the text box. Any help would be greatly appreciated. Steve