Update and Insert into three tables simultaneously - php

Question

I have a record that needs to be updated. If the update is successful, then it should insert record into three different tables. I did it with the code below,but one of the table(tab_loan_targetsave)is not inserting.I need a third eye to looked into this, as I have had a lot of pain in fathoming where the problem lies.
Pls i need assistance.Also, I welcome better approach if possible.
<?php
if(isset($_POST["savebtn"])){
$custNo = $_POST["custid"];
$transDate = $_POST["transDate"];
$grpid = $_POST["custgrp"];
$contAmount =$_POST["amtCont"];
$amount = $_POST["amount"];
$disAmount =$_POST["disbAmt"];
$savAmount =$_POST["savAmt"];
$intAmount =$_POST["intAmt"];
$postedBy = $_SESSION["staffid"];
//$preApproved =$_POST["preAmount"];
$loanRef = $_POST["refid"];
$st = "Approved";
$appDate = date("Y-m-d H:i:s");
$appBy = $_SESSION['staffid'];
$counter = 1;
$locate = $_SESSION['location'];
$insure = $_POST["insuAmt"];
$dis = $_POST["DisAmt"];
$update = mysqli_query($connection,"UPDATE tab_loan_request SET approval_status='$st',approvalDate='$appDate',approvedBy='$appBy',loanRef='$loanRef' WHERE custid='$custNo' AND RepayStatus='1'");
if($update && mysqli_affected_rows($connection)>0){
$insertTar = mysqli_query($connection,"INSERT INTO tab_loan_targetsave(custid,grpid,transactionDate,loanRef,savingAmt,status,postedBy,location,appStatus)
VALUES('$custNo','$grpid','$transDate','$loanRef,'$savAmount','Cr','$postedBy','$locate','1')");
$insertInt = mysqli_query($connection,"INSERT INTO tab_loan_interest(custid,requestAmt,transactionDate,interestFees,postedBy,loanRef,InsuranceFees,DisasterFees)VALUES(
'$custNo','$amount','$transDate','$intAmount','$postedBy','$loanRef','$insure','$dis')");
//if($insertInt){
//}if($insertTar){
$insertSav = mysqli_query($connection,"INSERT INTO tab_loan_saving(custid,grpid,transactionDate,loanRef,loanAmount,savingAmt,status,postedBy,location,appStatus)
VALUES('$custNo','$grpid','$transDate','$loanRef','$amount','0','Cr','$postedBy','$locate','1')");
}//first if
if($insertSav){
echo "<span style='font-weight:bold;color:red;'>"." Application Approval is successful!"."</span>";
}else{
//Unable to save
echo "<span style='font-weight:bold;color:black;>"."Error! Application Approval not Successful!"."</span>";
}
}else{
$custid = "";$saving=0.00;$st="";
$transDate = "";
$grpid = "";
$amount = "";
$postedBy = "";$loanRef="";
}
?>

Answer 1

"#Fred: See the error generated when i used mysqli_error($connection). Could you please interprete this: ErrorMessage: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1000.00','Cr','SPL002','Ojo','1')' at line 2 – Dave"
Seeing the error generated by the suggestion I've given you to check for errors.
You're missing a quote here '$loanRef
in your query:
VALUES('$custNo','$grpid','$transDate','$loanRef , '$savAmount'...
^ right there
I suggest to escape all of your incoming data.
I.e.:
$var = mysqli_real_escape_string($connection, $_POST['var']);
and apply that same logic to all your POST arrays.
Plus, as I stated; make sure you started the session, since there is no mention of that in your question and session_start(); wasn't included in your posted code.
The session needs to be started inside all pages using sessions.
Using a prepared statement will is better.
http://php.net/manual/en/mysqli.prepare.php
http://php.net/manual/en/pdo.prepared-statements.php
which is what you really should be using.
Additional references:
http://php.net/manual/en/mysqli.error.php
http://php.net/manual/en/function.error-reporting
Also make sure there aren't any constraints in your table(s).

Answer 2

Dude make sure you properly escape your variables http://php.net/manual/en/mysqli.prepare.php
i would check the Table Name! make sure it is case sesntive, also just wondering if you could do something to your database design? It seems a lot of duplicate data is going into your tables. Think about a better way to organise and store that data

Answer 3

I got where the error is emanting from . Just because I forgot to add a single quote to one of the values. ie missing the quote- near $loanRef. No closing string. Anyway, I was able to detect that through the error message stated parameter as adviced by Fred nad Mark. Correct
$insertTar = mysqli_query($connection,"INSERT INTO tab_loan_targetsave(custid,grpid,transactionDate,loanRef,savingAmt,status,postedBy,location,appStatus)
VALUES('$custNo','$grpid','$transDate','$loanRef','$savAmount','Cr','$postedBy','$locate','1')");
Thank you all.