Variable in a mysql query

Question

for ($i=0; $i<$count; $i++) {
    $appid = $chk[$i];


    include "dbconnect.php";
    $selectquery = mysql_query("SELECT * FROM regform_admin WHERE tid = '$appid'");
    $fetch = mysql_fetch_array($selectquery);
    $tid = $fetch['tid']; $username = $fetch['username']; $c_month = $fetch['month']; $c_day =$fetch['day']; $c_year = $fetch['year'];
    $c_month2 = $fetch['month2']; $c_day2 =$fetch['day2']; $c_year2 = $fetch['year2']; 
    $pickup = "".$c_month."/".$c_day."/".$c_year."";
    $return = "".$c_month2."/".$c_day2."/".$c_year2."";
    $pickuploc = "".$fetch['pickupret']." "." ".$fetch['speclocation']."";
    $desti = "".$fetch['destination']." "." ".$fetch['location']."";
    $vehicle1 = $fetch['vehicle1'];
    $datesent = date("n j, Y; G:i"); ;
    $rand = rand(98765432,23456789);

    include "vehicledbconnect.php";
    $vquery = mysql_query("SELECT * FROM vehicletbl WHERE vehicle = '$vehicle1'");
            $getvquery = mysql_fetch_array($vquery);
            $maxcars = $getvquery['maxcars'];
            $carsleft = $getvquery['carsleft'];
            if ($carsleft == 0) {
            echo '
        <script language="JavaScript">
        alert("Cannot move reservation to Pending for payment status. No available vehicles left for this reservation.");
        </script>';

        echo "$vehicle1";

            }

Hi guys my problem here is that the $vehicle is not returning its values if it is inserted in a database query ($vquery = mysql_query("SELECT * FROM vehicletbl WHERE vehicle = '$vehicle1'");) but if it is echoed, it return its value. The logic here is that it will select all the values from vehicletbl wherein the value of any values in 'vehicle' column will be equal to the $vehicle1. Thanks for the help!

Nothing is assigned to a variable by name $vehicle? – Tobias Sep 07 '11 at 14:17 — Tobias, Sep 07 '11 at 14:17
Please show what `$query` outputs – Pekka Sep 07 '11 at 14:18 — Pekka, Sep 07 '11 at 14:18

score 0 · Answer 1 · answered Sep 07 '11 at 14:19

0

You've got ZERO error handling on your queries. Try adding some debugging to the query calls:

$result = mysql_query(...) or die(mysql_error());

The rest of the code is ugly, but looks "ok", so start looking at WHY you're not getting anything back from the queries.

Never ever assume a query succeeds.

answered Sep 07 '11 at 14:19

Marc B

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Thanks and yes the rest of the code is ugly. That's because I'm still a noob programmer. – glove Sep 07 '11 at 14:21

score 0 · Answer 2 · answered Sep 07 '11 at 14:24

0

try this to debug :

$sql = "SELECT * FROM vehicletbl WHERE vehicle = '" . $vehicle1 . "'";
$vquery = mysql_query($sql) or die(mysql_error() . "\n<br>$sql");

thats what i do to find errors in my sql.

answered Sep 07 '11 at 14:24

Pheonix

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score -1 · Answer 3 · edited May 23 '17 at 10:24

Noob programmer ? Here are some things to know :

for ($i=0; $i<$count; $i++) {
    $appid = $chk[$i];

// Replaced By ...
foreach($chk as $appid){

http://php.net/manual/en/control-structures.foreach.php

// Include the file before the loop ! You're including 20 times your file, but you just need to do it once ! Another thing to know:
include_once("dbconnect.php");

http://php.net/manual/en/function.include-once.php

$desti = "".$fetch['destination']." "." ".$fetch['location']."";
// WHY ?? Isn't that easier to do this ?
$desti = $fetch['destination']."  ".$fetch['location'];

And security :

// Don't forget to escape your variables before putting it in mysql queries
$appid = mysql_real_escape_string($appid);
$selectquery = mysql_query("SELECT * FROM regform_admin WHERE tid = '$appid'");

Best way to defend against mysql injection and cross site scripting

There are other remarks, but try to improve those points first !

Variable in a mysql query

3 Answers3