mac_gyver

$item_id = $_POST['$item_id']; should be $item_id = $_POST['item_id']; are you even trying to debug these errors?

how do you know that? what is it doing and what do you expect it to do?

you need to echo out $sql so that you can see what it actually is at the point where the error is being reported. i'll guess the value in $_POST['facility_provider_number'] is empty, resulting in sql syntax of 'some value',,'some value'

that's because you are using short opening php tag <? and your .php code is not running. use a full opening php tag <?php

i'm not sure what previous time you are referring to, but for the line of code in this thread, DavidAM told you in your previous thread that one equal sign means that you are making an ASSIGNMENT, not a comparison - an argument can be made that you need === (three equal signs, a value and type match) for this line of code, since your image uploader script isn't testing if the $_FILES array is even set/not-empty. when $_FILES is empty due to an error or an upload form has not been submitted (your code isn't testing if any form was submitted), an empty $_FILES array will match a zero using two == signs. using three === (a value and type check), will only be successful if the $_FILES array is not empty and the ['error'] is a zero value. FYI, why you need to learn the meaning of the code - one = sign is an assignment operator i.e $var = some_value;. when used in a conditional test - if($var = some_value) ... while($var = some_value) the assignment is made and the value that was assigned is used in the conditional test. two == signs is an equal value test (the types can be different i.e null/empty is == to a 0.) three === signs is an exact value test (the value and types must be the same i.e. null/empty is not === 0.)

you must bind a variable to each field that is selected. by using SELECT *, no one knows just by looking at the query how many fields there actually are and there are apparently more than the 5 fields that were listed. you need to SELECT the actual fields that you want in the query so that you will have the information in front of you to bind all the selected fields.

if you mean, how did i determine what it is doing, i traced through the execution path and read the code - if you mean, what does that mean, the answer's in your code and would require that you understand what each statement in your code is doing. do you have a specific question about what any of your code is doing?

+1 for any kind of subscription tracking, you would treat it as a deposit account, where you insert a record for each addition to a person's subscription, with at least a duration of the amount of time that was added. this will allow for things such as a person extending their existing subscription by paying for/adding any available increment of time or giving someone a free amount of time due to a problem... just add a record for any user_id. to find out if their subscription is current, just sum up all the durations from the initial (lowest) start date and compare it to the current date.

the name of the $db->loadObjectList() method hints that it returns an array (list) of objects. what is it supposed to return (array, single object, true/false if it fails...) and what is it actually returning? also, you should be using one joined query to get the related information, not two separate queries.

to prevent duplicates, you would test if any submitted food item is already in the array before trying to add it. in fact, for a nicer user experience, you could dynamically build the <option></option> list by checking if each item has already been selected and either not output it in the <option></option> list or make it disabled in the <option></option> list.

the connection details for any database connection (or how many different simultaneous database connections there are) is application level information. remembering or finding each class file that has application level information hard-coded in them to edit just because you moved where code is being ran at, is brute-force time-wasting make-work programming. by hard-coding where connection details are stored at (an include file name) or directly what the connection details are (external-variables/defined constants or local assignment statements) or using a static method of another class, inside of a class, you cannot use the class for more than one database connection at a time because multiple instances of the class all use the same database connection information. if your solution to this would be to copy/paste the class to a different name or to shuffle-around (making,overwriting,remaking) instances of a class, that's most definitely NOT the point of using classes for anything. dependency injection should be used to get application level information into any instance of a class.

unless your ajax code displays everything that is sent back from the php code or you are directly examining the response sent back to the client, you won't see the php produced errors and there may not even be any php errors if the problem is in your php logic. Ajax adds a extra layer to your code that makes debugging harder. before you can use Ajax to accomplish a task, you must be able to perform that task, error free, without using Ajax and be able to debug what your code is doing. get a normal html form working with your php code, then add Ajax.

the objects that are in the array are instances of your Photograph class, one for each row that the query matches, with the properties set to whatever data is in the row from the query, and whatever methods your class has.

your for(){} loop doesn't have any {} so the only statement being executed inside the loop is the $sql = " ... "; statement and you just end up with the last $sql statement. however, for what you are doing, just use an (one) INSERT ... SELECT query (no loop is required) -

a form field of any kind won't do anything unless it is inside a form tag or you use a js event with it. why not just use a <a href=''></a> link with a css button or an image for the button?

a url like - paintings_detail.php?name=conscious will set $_GET['name'] the easiest fix would be to add $name = isset($_GET['name']) ? $_GET['name'] : ''; to the start of your code, repeat for any other get data. if your code was also relying on register_globals for any post, cookie, files and session variables, you will need to fix more things in it to get it to work.

you probably have a non-printing character as part of your submitted form data that doesn't match anything when the query runs in the php code but when you copy/paste the echoed query statement, that non-printing character is stripped out. what is your code that is building the form?

the query you posted in your code and the query where the error is occurring at aren't even the same ones queries. condition is a reserved mysql keyword - http://dev.mysql.com/doc/refman/5.5/en/reserved-words.html either change the column name to something else or enclose `condition` in back-ticks.

you would create functions or call built in php functions for each different type of validation. since you may have different requirements for each form field, i would make an array that lists the validation function names that you want to call in the order that you want to call them. then for each field, get the list of validation functions out of the array and dynamically call them using variable functions - http://php.net/manual/en/functions.variable-functions.php your code would loop over the expected form fields, getting the field name, then access the list of validation function names from the array and dynamically call each validation function and test/store the result. you could also expand upon this so that the array holds the corresponding error message and when a validation step fails, get and store the corresponding error message that you want to display.

you need to use mysql_error() to find out why your query is failing. for debugging only, add - echo mysql_error(); on the next line after your mysql_query() line. it's likely because your data contains special sql characters that are breaking the sql syntax and you are not escaping (see the mysql_real_escape_string() function) the string data being put into the query or the programming gods are upset because you are using non-helpful variable names like $var1, $var2, $var3 that don't indicate the purpose of the variable.

echoing mysql_error() or using mysql_error() in the die() statement will tell you why the connect/select_db statement is failing.

the mysql_connect() statement will always run (assuming the function is defined.) if the connection fails, the function call will return a false value to the calling code. do you have any logic in your code to test if that function returns a false value?

the display_error and log_errors settings apply to any type of error that is being reported due to the error_reporting setting. if display_errors are truly off in your php.ini and you are getting any of the of type of php error message (fatal, warning, or notice) being displayed, it means that either display_errors is not actually turned off of that your script is turning it back on. the error_reporting setting determines what is reported. the display_errors/log_errors settings determine what happens to the report of those errors.

so it appears that these text strings have been this way for a while and you want to fix them? it looks like a version upgrade or a template/theme didn't completely install (or mixed version of files were used.) you cannot find where they are defined at because they aren't. they are either leftover from an old version of oscommerce or from a template/theme that was installed. they are used in the content but where they are defined at doesn't exist. the easiest fix would be to just define them yourself, rather than to find where all of them are used at (it's in different files depending on where on the page they are at) and change the name to what they should be (assuming they are still being used in the current version.) i would put them into catalog\includes\languages\english.php. in fact, define('HEADER_TITLE_CART_CONTENTS', 'Cart Contents'); that is in catalog\includes\languages\english.php is what NOW_IN_CART looks like it should be.

if you are still working on why only one (that last one) piece of information is being displayed, it's because your display code isn't inside of your while(){} loop. your display code is after the end of your while(){} loop and is only running once, after the while(){} loop has finished and it is using the last values fetched into the variables.