1 Correct answer
Hello osgood_ and entire contributors,
First may I apologize for my no response all these days. We were running out of time in completing the assignment for the week so I decided to hold on the code since its not a priority at that time. Please accept my apologies.
I had before now, precisely 2 years ago, written a similar code for another project. So I located the code, edited it to capture the fields of the current project and tried it several times and it worked both offline and online.
Here is
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Replies
Hi Osgood_ and all who contributed to my question,
I post to say thank you. Your contributions helped me resolve my issue.
To Osgood_, her are responses to your questions;
Q. In your database you have a column called 'taxyr' which takes the type of 'date' to check against?
Ans: Yes I have a field called ‘taxyr’ that obtains its values from a drop box with years populated by php year code from the previous year to other past years like this
<?php echo date("Y",strtotime("-1 year")); ?>,
<?php echo date("Y",strtotime("-2 year")); ?>,
<?php echo date("Y",strtotime("-3 year")); ?>,
etc
So depending on the year the person wants to pay for he will select from the drop down.
So if you posted an application on the 6th December 2016 and its stored in the database column 'taxyr' in this format: 2016-12-06 then the below query should return a positive result as the application was posted exactly a year ago today but why are you only checking an 'exact' date and not a date which is over a year old? You are only giving yourself a 24hr period to check, right?
ANS: No. It doesn’t store a full day but only the year like this 2016, 2015, 2015, etc.
See attached images
$query = sprintf("select taxyr from tbl_tableassmt where uid='".$uid."' AND taxyr = DATE_SUB(CURDATE(), INTERVAL 1 YEAR)", mysql_real_escape_string($email));
Here is the correct query
$query = sprintf("select * from tbl_taxpayerassmt where taxidno='".$taxidno."' AND taxyr='".$taxyr."'", mysql_real_escape_string($taxidno));
Q: Also where is the $taxyr variable getting its value from (a date picker?) and why are you using strtolower case? The format should be passed to the database in the date format: 2016-12-06 so it can be checked.
Ans: As earlier said, the taxyr is getting it from pull down. Please see attached images
Here is the complete code for those with similar issue
if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "add_assessment_frm")) {
$taxyr = strtolower($_POST['taxyr']);
$taxidno = $_POST['taxidno'];
$query = sprintf("select * from tbl_tableassmt where taxidno='".$taxidno."' AND taxyr='".$taxyr."'", mysql_real_escape_string($taxidno));
$result = mysql_query($query);
$num_rows = mysql_num_rows($result);
echo $num_rows;
if($num_rows > 0)
{
$message = "Error: You have already sent your assessment for the previous year.";
}
elseif($num_rows == 0)
{
$insertSQL = sprintf("INSERT INTO tbl_tableassmt (`uid`, incsrc, taxidno, salary, txxamt, gratuity, health, housing, assurance, pension, bizprft, allowances, obenefts, bonus, gaincome, conallowances, chgincome, taxyr, username) VALUES (%s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s)",
GetSQLValueString($_POST['uid'], "int"),
GetSQLValueString($_POST['incsrc'], "text"),
GetSQLValueString($_POST['taxidno'], "text"),
GetSQLValueString($_POST['salary'], "double"),
GetSQLValueString($_POST['txxamt'], "double"),
GetSQLValueString($_POST['gratuity'], "double"),
GetSQLValueString($_POST['health'], "double"),
GetSQLValueString($_POST['housing'], "double"),
GetSQLValueString($_POST['assurance'], "double"),
GetSQLValueString($_POST['pension'], "double"),
GetSQLValueString($_POST['bizprft'], "double"),
GetSQLValueString($_POST['allowances'], "double"),
GetSQLValueString($_POST['obenefts'], "double"),
GetSQLValueString($_POST['bonus'], "double"),
GetSQLValueString($_POST['gaincome'], "double"),
GetSQLValueString($_POST['conallowances'], "double"),
GetSQLValueString($_POST['chgincome'], "double"),
GetSQLValueString($_POST['taxyr'], "date"),
GetSQLValueString($_POST['username'], "text"));
mysql_select_db($database_CRIRS_DB, $CRIRS_DB);
$Result1 = mysql_query($insertSQL, $CRIRS_DB) or die(mysql_error());
$insertGoTo = "upd_my_tax_assessment";
if (isset($_SERVER['QUERY_STRING'])) {
$insertGoTo .= (strpos($insertGoTo, '?')) ? "&" : "?";
$insertGoTo .= $_SERVER['QUERY_STRING'];
}
Hello Friends,
Please I still need your help in this codes. I stated earlier that the issues has been resolved.
The code is running perfectly well offline on my localhost, I mean here my PC (see inserted image) but when I run it online, it doesn't work. It continues to posted assessments even when an assessment has been posted for the same year.
I even used fixed valued like $query = sprintf("select * from tbl_taxpayerassmt where taxidno='1007360900-0001' AND taxyr='2016'", mysql_real_escape_string($taxyr));, yet the error continues.
Please your help is appreciated
Thank you
Hello osgood_ and entire contributors,
First may I apologize for my no response all these days. We were running out of time in completing the assignment for the week so I decided to hold on the code since its not a priority at that time. Please accept my apologies.
I had before now, precisely 2 years ago, written a similar code for another project. So I located the code, edited it to capture the fields of the current project and tried it several times and it worked both offline and online.
Here is the screenshot of the page
And here is the correct code for anyone interested. You may need to edit it to fit your own project.
---------------------------------------------------------------------------------------------------------
<?php
$posted = false;
if( $_POST ) {
$posted = true;
if(isset($_POST['submit']))
{
mysql_connect('localhost','root','password');
mysql_select_db('dbname');
$incsrc = $_POST['incsrc'];
$taxidno = $_POST['taxidno'];
$salary = $_POST['salary'];
$gratuity = $_POST['gratuity'];
$health = $_POST['health'];
$housing = $_POST['housing'];
$assurance = $_POST['assurance'];
$pension = $_POST['pension'];
$allowances = $_POST['allowances'];
$bonus = $_POST['bonus'];
$taxyr = $_POST['taxyr'];
$username = $_POST['username'];
$query=mysql_query("select * from tbl_tableassmt where taxidno='".$taxidno."' AND taxyr='".$taxyr."'") or die(mysql_error());
$duplicate=mysql_fetch_row($query);
if($duplicate > 0)
{
$message = "Error: Dear user with TIN '.$taxidno.', you have already posted your assessment for the year '.$taxyr.'";
}
}
else
{
if($duplicate==0)
{
$query1=mysql_query("insert into tbl_tableassmt (incsrc, taxidno, salary, gratuity, health, housing, assurance, pension, allowances, bonus, taxyr, username)
VALUES ('$incsrc', '$taxidno', '$salary', '$gratuity', '$health', '$housing', '$assurance', '$pension', '$allowances', '$bonus', '$taxyr', '$username')") or die(mysql_error());
}
header("location: my_asssmnts");
}
}
?>
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Once more thank you so much
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