How to update specific table row in mysql table using php [duplicate]

Question

I am beginner in php , i am trying to update specific row but does not work . i need your help: update.php DIR . '/db_connect.php'; // connecting to db $db = new DB_CONNECT();

    if (isset($_POST['id'])) {

$id = $_POST["id"];
$location = $_POST["location"];
  $sql=mysql_query("UPDATE citizenalert set location={'$location'} WHERE id ={'$id'}");
}

  ?>

db_config.php

  <?php

   define('DB_USER', "root"); // db user
   define('DB_PASSWORD', ""); // db password (mention your db password here)
     define('DB_DATABASE', "ikirenga"); // database name
      define('DB_SERVER', "localhost"); // db server
     ?>

db_connect.php

           <?php

  class DB_CONNECT {

// constructor
function __construct() {
    // connecting to database
    $this->connect();
}

// destructor
function __destruct() {
    // closing db connection
    $this->close();
}

/**
 * Function to connect with database
 */
function connect() {
    // import database connection variables
    require_once __DIR__ . '/db_config.php';

    // Connecting to mysql database
    $con = mysql_connect(DB_SERVER, DB_USER, DB_PASSWORD) or die(mysql_error());


    // Selecing database
    $db = mysql_select_db(DB_DATABASE) or die(mysql_error()) or die(mysql_error());

    // returing connection cursor
    return $con;
}

/**
 * Function to close db connection
 */
function close() {
    // closing db connection
    mysql_close();
}
    }
   ?>

request_update.php

               <!doctype html>

                <html lang="en">

                   <body>
                  <form name="updates" method="post" action="update.php" > 
                   <label>User location</label> <input id="location"  type="text" name="location" > <br><br>
                     <input type="submit" name="submit" value="Submit" /> 
                     </form>  
                     </body>
                     </html>

chk your connection ist, and your code is open for SQL Injection. — devpro
– devpro, Commented Sep 26, 2016 at 9:00
It amazes me that people are still using mysql_ functions still to this day. — Mikey
– Mikey, Commented Sep 26, 2016 at 9:11
@devpro sad thing is that they are probably commercial websites stackoverflow.com/questions/38297105/… — Drew
– Drew, Commented Sep 26, 2016 at 9:19

devpro · Accepted Answer · 2016-09-26 09:05:03Z

0

Please use below Syntax

<?php
require_once __DIR__ . '/db_connect.php';
$db = new DB_CONNECT();

if (isset($_POST['id'])) {

$id = $_POST["id"];
$location = $_POST["location"];
$sql = mysql_query(" UPDATE citizenalert set location = " . $location . "   WHERE id = " . $id . ");
  }

Thanks.

edited Sep 26, 2016 at 9:05

devpro

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answered Sep 26, 2016 at 9:04

user5248112

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13 Comments

devpro Over a year ago

who knows, $location is string or integer.

Drew Over a year ago

sql injection here

Niragire Sam Over a year ago

$location is a string

devpro Over a year ago

@NiragireSam: did u checked the others answers??

Niragire Sam Over a year ago

yes i am checking it and i don't find any change

|

Saurabh · Accepted Answer · 2016-09-26 09:05:26Z

0

Change your SQL query as below

$sql=mysql_query("UPDATE citizenalert set location='{$location}' WHERE id = '{$id}'");

answered Sep 26, 2016 at 9:05

Saurabh

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1 Comment

Drew Over a year ago

sql injection here

Community · Accepted Answer · 2017-05-23 12:19:29Z

If i test your code something like that:

<?php 
$id = 1;
$location = "test";
echo $sql="UPDATE citizenalert set location={'$location'} WHERE id ={'$id'}";
?>

It's giving this query:

UPDATE citizenalert set location={'test'} WHERE id ={'1'}

Please check here, both of values having issue.

Problem:

You are using {} outside the quotes. you need to use it inside the single quote.

Modified Query with same example:

echo $sql="UPDATE citizenalert set location='{$location}' WHERE id ='{$id}'";

Result:

UPDATE citizenalert set location='test' WHERE id ='1'

Side Notes:

Use mysqli_* or PDO, because mysql_* is deprecated and closed in PHP 7.
Your code is open for SQL Injection, you must need to protect with SQL Injection or just simply use Prepared Statement.

Some other Reference for Prepared Statement: php mysql bind-param, how to prepare statement for update query

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