PHP mutiple queries no result in database - php

Question

I'm working on an android app that requires connection to a database. I have the android part working but I'm having trouble with my php script.
<?php
$con = mysqli_connect(/This info is correct/) or die ("Unable to connect");
$gebruikersnaamOntvanger = $_POST["gebruikersnaamOntvanger"];
$idBetaler = $_POST["idBetaler"];
$bedrag = $_POST["bedrag"];
$saldoBetaler = $_POST["saldo"];
$response = array();
$response["success"] = "false";
$statement = mysqli_prepare($con, "SELECT idGebruiker, Saldo FROM Gebruikers WHERE Gebruikersnaam = ?");
mysqli_stmt_bind_param($statement, "s", $gebruikersnaamOntvanger);
mysqli_stmt_execute($statement);
mysqli_stmt_store_result($statement);
mysqli_stmt_bind_result($statement, $idGebruiker, $Saldo);
while($row = mysqli_stmt_fetch($statement)){
$idOntvanger = $idGebruiker;
$saldoOntvanger = $Saldo;
}
$saldoOntvanger += $bedrag;
$saldoBetaler -= $bedrag;
try {
$statement2 = mysqli_prepare($con, "INSERT INTO Transacties (idBetaler, idOntvanger, Bedrag, Datum, Uitgevoerd) VALUES(?, ?, ?, now(), 1)");
mysqli_stmt_bind_param($statement2, "iid", $idBetaler, $idOntvanger, $bedrag);
mysqli_stmt_execute($statement2);
$response["success"] = "success";
} catch(Exception $e) {
$response["success"] = $e->gettext;
}
echo json_encode($response);
?>
So the android part work and returns the JSON object correctly but nothing changes in the database. I have tried adding a try catch so I get what's the error but the try catch never works. The script alwayws returns success even if it didn't work. Please be aware I'm new to PHP and I have double checked the SQL queries and they should be correct. If you need more information please ask.

Answer 1

mysqli_* function dosn't throws exceptions. So your catch is never executed even on error. Change your code like this :
if(!mysqli_stmt_execute($statement2))
throw new Exception(mysqli_stmt_error($statement2));
When you'll see the error you can fix your code. May be add same error handling to mysqli_prepare and others

Answer 2

I think the issue was the use of iid in the binding for the insert query. It should either be iis or iii presumably. You tried using try/catch blocks but to my mind did not make full use of them - the creation of a prepared statement should be tested to see if it succeeds or fails ~ the outcome of which can be used in the try/catch logic to indicate faults. Excuse the mix of procedural and OOP style code
<?php
if( $_SERVER['REQUEST_METHOD']=='POST' && isset( $_POST["gebruikersnaamOntvanger"], $_POST["idBetaler"], $_POST["bedrag"], $_POST["saldo"] ) ){
try{
$con = mysqli_connect(/This info is correct/) or die ("Unable to connect");
$gebruikersnaamOntvanger = $_POST["gebruikersnaamOntvanger"];
$idBetaler = $_POST["idBetaler"];
$bedrag = $_POST["bedrag"];
$saldoBetaler = $_POST["saldo"];
$response = array('success'=>false);
$stmt=$conn->prepare( 'select `idgebruiker`, `saldo` from `gebruikers` where `gebruikersnaam` = ?' );
if( !$stmt ) throw new Exception('unable to prepare select query');
else {
$stmt->bind_param( 's', $gebruikersnaamOntvanger );
$result=$stmt->execute();
if( !$result )throw new Exception('No results from select query');
else {
$stmt->store_result();
$stmt->bind_result( $idGebruiker, $Saldo );
while( $stmt->fetch() ){
$idOntvanger = $idGebruiker;
$saldoOntvanger = $Saldo;
}
$saldoOntvanger += $bedrag;
$saldoBetaler -= $bedrag;
$stmt=$con->prepare('insert into `transacties` ( `idbetaler`, `idontvanger`, `bedrag`, `datum`, `uitgevoerd` ) values(?, ?, ?, now(), 1)');
if( !$stmt )throw new Exception('Unable to prepare insert query');
else {
/* What is the value of $bedrag? integer, string?? */
/*
The binding iid was incorrect - perhaps iis or iii
*/
$stmt->bind_param('iis', $idBetaler, $idOntvanger, $bedrag );
$result = $stmt->execute();
$response['success']=$result;
}
}
}
echo json_encode( $response );
}catch( Exception $e ){
echo $e->getMessage();
}
}
?>