Android Question PHP script with Parameters not working

[erasmusackon]

This code is able to retrieve records from the server

    $q = mysqli_query($con,"SELECT * FROM DepositTransactions");
    $rows = array();
    while($r = mysqli_fetch_assoc($q))
    {
    $rows[] = $r;
    }
    echo json_encode($rows);
    mysqli_close($con);

GetTransactions is unable to work. it returns [] when i checked the log. Am i doing anything wrong in this "GetTransactions"

    Dim DownloanTransactions As HttpJob
    DownloanTransactions.Initialize("DownT", Me)
    DownloanTransactions.Download2(Server, Array As String ("action", "GetTransactions","memberID",MemberID,"lasttrans",LastTransID))

$action = $_GET["action"];

switch ($action)

{
    case "CountPersons":
        $q = mysqli_query($con,"SELECT * FROM DepositTransactions WHERE MemberID=40001762");
        $count = mysqli_num_rows($q);
        print json_encode($count);
        mysqli_close($con);
    break;
  
    Case "GetTransactions":
        $memberid = $_GET["memberid"];
        $lasttrans = $_GET["lasttrans"];
        $q = mysqli_query($con,"SELECT * FROM DepositTransactions WHERE MemberID=$memberid AND TranNo>$lasttrans");
        $rows = array();
        while($r = mysqli_fetch_assoc($q))
        {
        $rows[] = $r;
        }
        print json_encode($rows);
        mysqli_close($con);
    break;

 

[ronell]

try this..

"SELECT * FROM 'DepositTransactions' WHERE MemberID= '$memberid' AND TranNo> '$lasttrans'"

not tested

 

[erasmusackon]

try this..

"SELECT * FROM 'DepositTransactions' WHERE MemberID= '$memberid' AND TranNo> '$lasttrans'"

not tested

Not working please

 

[tigrot]

try:

$q = mysqli_query($con,"SELECT * FROM DepositTransactions WHERE MemberID='$memberid' AND TranNo>'$lasttrans'");
Display $q to see if sql is well formed!
ECHO $q

 

[erasmusackon]

try:

$q = mysqli_query($con,"SELECT * FROM DepositTransactions WHERE MemberID='$memberid' AND TranNo>'$lasttrans'");
Display $q to see if sql is well formed!
ECHO $q

Thanks @tigrot. This is the error am getting from the error log

<b>Parse error</b>: syntax error, unexpected '$rows' (T_VARIABLE), expecting ',' or ';' in <b>/home/sikasoft/public_html/app/newCon.php</b> on line <b>39</b><br />

 

[ronell]

check the code before line 39 , you might miss a , or ; at the end of the line
--
let us see php code again

 

[erasmusackon]

I have modified the code and am now getting this "Response from server: []"
Not getting the actual data. is there anything wrong with my code?

    Dim DownloanTransactions As HttpJob
    DownloanTransactions.Initialize("DownT", Me)
    DownloanTransactions.Download2(Server, Array As String ("action", "GetTransactions","memberID",MemberID,"lasttrans",LastTransID))

$action = $_GET["action"];

switch ($action)

{
       Case "GetTransactions":
        $memberid = $_GET['memberid'];
        $lasttrans = $_GET['lasttrans'];
        $lasttrans= mysqli_real_escape_string($con, '$lasttrans');
        $memberid = mysqli_real_escape_string($con, '$memberid');
        $q = mysqli_query($con,"SELECT * FROM DepositTransactions WHERE MemberID='.$memberid.' AND TranNo>'.$lasttrans.'");
        $rows = array();
        while($r = mysqli_fetch_assoc($q))
        {
        $rows[] = $r;
        }
        print json_encode($rows);
        mysqli_close($con);
    break;
}

 

[tigrot]

why do you use square brackets in $rows? I think that you have to take them off

 

[tigrot]

I read your code again. It looks like you want to move all rows fetched in $rows. If you do so, you will get only the last row fetched in $rows.

 

[tigrot]

Put some echo here and there and simulate it from a browser...

 

[ronell]

why do you use square brackets in $rows?

$row in an array containing the result ..

 

[tigrot]

Trace with ECHO statements and you will quicly get the issue!

 

[tigrot]

$row in an array containing the result ..

Ahh Ok, out of order brain...

 

[ronell]

post the whole updated php code, also the latest error ..

 

[erasmusackon]

post the whole updated php code, also the latest error ..

This code doesn't work but

$q = mysqli_query($con,"SELECT * FROM DepositTransactions WHERE MemberID= '$memberid' AND TranNo>'$lasttrans'");

This code works and return the results. what might be the problem?

$q = mysqli_query($con,"SELECT * FROM DepositTransactions WHERE MemberID= 40001762 AND TranNo>0");

 

[ronell]

try working on the variable in the query.. that might be the problem

 

[tigrot]

Is tranNo a number? In this case you need to omit ' ' before and after variable name

 

[tigrot]

DISPLAY SQL text variable or you don't know what is the real SQL text!

 

[ronell]

learn basic php , read/watch tutorials .. the problem is just simple.. you were just lacking knowledge about the basic of PHP

 

[erasmusackon]

learn basic php , read/watch tutorials .. the problem is just simple.. you were just lacking knowledge about the basic of PHP

Learning for sure. But if you have the solution why not assist?