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Editorial Team
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Asked: 2026-05-27T01:10:10+00:00

I have my code below to update a my MySQL database, it’s running but

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I have my code below to update a my MySQL database, it’s running but is not updating the database when I check rcords using phpmyadmin. plae hlp me.

$database = "carzilla";

$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

$manufacturerTable = $_POST[vehicleManufacturer];
$numberToSearch = $_POST[vehicleIdNo];
$engineType = $_POST[engineType];
$engineCC = $_POST[engineCC];
$year = $_POST[year];
$numberofDoors = $_POST[numberofDoors];
$tireSize = $_POST[tireSize];
$chasisNumber = $_POST[chasisNumber];
$vehicleMake = $_POST[vehicleMake];
$price=$_POST[price];

mysql_select_db("$database", $con);

$sql = mysql_query("UPDATE $manufacturerTable SET username='vehicleMake', 
engineType='$engineType', engineCC='$engineCC', year='$year', chasisNo='$chasisNumber', numberOfDoors='$numberofDoors' ,numberOfDoors='$numberofDoors',  tireSize='$tireSize', price='$price' WHERE `index` ='$id'");

if (!mysql_query($sql,$con))
{
 die('Error: ' . mysql_error());
}
echo 'record has been successfuly';

mysql_close($con);


?>
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1 Answer

  1. Editorial Team

    Take a good look at your query. You are referring to PHP variables in several different fashions in the same statement. In the query $manufacturerTable is just $manufacturerTable, you encase a few others in single quotes, some of which you remove the $ from, others you do not. I know I preach this far too often, but you should really look into using prepared statements. They take all the guess work out of using variables in your queries, and they prevent you from being victimized by injection hacks. But the short answer here is that you are not referencing your variables correctly in the query.

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