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I'm trying to determine number of days difference between the due duedate and current date and really need help. [CODE]<?php mysql_connect("localhost", "root", ""); mysql_select_db(numbersdb) or die("Unable to select database"); $query = "SELECT taxrate FROM numbdata" ; $result = mysql_fetch_assoc(mysql_query($query)); $taxrate = $result['taxrate']; mysql_connect("localhost", "root", ""); mysql_select_db(oodb) or die("Unable to select …
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I'm trying to read the [b]taxrate[/b] from the database table and use it with values from another table. the database and table are correct as is the field (taxrate) from the table, value is 0.06. Then I'm trying to multiply that value by the value of a field from another …
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the below code is to updata a database with calculated taxes and days late. It results in the message:[QUOTE]error: Maximum execution time of 60 seconds exceeded in C:\xampp\htdocs\invoice[/QUOTE]. Can someone help with this? [CODE]<?php $stat = mysql_connect("localhost","root",""); $stat = mysql_select_db("oodb"); $query = "SELECT invnum FROM oocust WHERE payrec = 'R' …
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the below code is intended to update a database file with the no of days late and tax calcs for each record. I'll understand this before I qive up. the subject is the error message. [CODE]<?php $stat = mysql_connect("localhost","root",""); $stat = mysql_select_db("oodb"); $query = "SELECT invnum FROM oocust Where WHERE … 
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this program inserts data (correctly) into fields b4 update but I get the error message when I submit. Can someone tell me what is wrong with my code? [CODE]<html><head> <!--when the paidamt is keyed in, the current date, pd & paidsum are autoinserted--> <script> function $_(IDS) { return document.getElementById(IDS); } …
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I've fought this for days - can someone help? [CODE]<?php mysql_connect(localhost,root,""); mysql_select_db(oob) or die( "Unable to select database"); $query = " INSERT INTO oocust (payrec,invnum,date,time,name,name,street,state,zip,terms,fob,ordernum,dateord,datecomp,datepaid,charges,paidamt,tax,amtdue) VALUES('$payrec','$invnum','$date','$time','$name','$street','$state','$zip','$terms','$fob','$ordernum','$dateord','$datepaid','$charges','$paidamt','$tax','$amtdue')"; echo "data inserted</font><br /><br />"; $stat = mysql_query($query) or die('Query failed: ' . mysql_error()); if (isset( $_POST['payrec']) ) { $payrec=$_POST['payrec']; $invnum=$_POST['invnum']; $date=$_POST['date']; $time=$_POST['time']; $name=$_POST['name']; …
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Could someone help me with this code? I need to compare the date completed(datecomp) to the date paid(datepaid) to establish the number of days difference(dayslate). Thanks in advance! [CODE]<?php $stat = mysql_connect("localhost","root",""); $stat = mysql_select_db("oodb"); $query = "SELECT name FROM oocust Where ordernum='$ordernum'"; $stat = @mysql_fetch_assoc(mysql_query($query)); echo $stat["name"]; $result= mysql_query("select … 
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I don't know why I get the message "unexpected $end" at the last line of code - can someone help? [CODE]<?php mysql_connect(localhost,root,""); mysql_select_db(test) or die( "Unable to select database"); if(!empty($_POST["submit"])) { $apt = $_POST['apt']; $query="SELECT * FROM payments Where apt='$apt'"; $result=mysql_query($query); if(mysql_num_rows($result)) ?> <html><head><script type="text/javascript"> if ($late = "L") {$rentdue … 
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can someone help with this? nothing works beyond the // *############# [CODE]<script> function $_(IDS) { return document.getElementById(IDS); } function calculate_paid() { var amtpaid = document.getElementById("amtpaid"); var rentdue = document.getElementById("rentdue"); var prevbal = document.getElementById("prevbal"); var hudpay = document.getElementById("hudpay"); var tentpay = document.getElementById("tentpay"); var datepaid = document.getElementById("datepaid"); var late = document.getElementById("late"); var …
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I think the below code will do the password If I can figure out how to direct this program to the the main program if the password is correct. Can someone help me with the link, please? [CODE]<?php mysql_connect(localhost,root,""); mysql_select_db(entrydb) or die( "Unable to select database"); if(!empty($_POST["submit"])) { $password = …
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