How to display result of select where query in PHP -mysql

Question

1.00/5 (1 vote)

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I want to get password of sombody in table name in database mysql

HTML

<form action="GetPassword.php" method="post">
         <legend>
              <h3>Get Password by name</h3>
        </legend>
        <fieldset>
           <p>
             <label>Name : </label>
             <input type="text" name="nameofpass" placeholder="Enter name to get password">   
           </p>
           <p>
              <input type="submit" name="getpass" value="Go"> 
           </p>    
        </fieldset>
    </form>

What I have tried:

PHP

<?php
if(isset$_POST['getpass'])
{
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "testdb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
$name=$_POST['nameofpass'];
$query ="SELECT password from names where name='$name'";
   $result = mysql_query($query); 
    while ($row = mysql_fetch_array($result)) {
        echo $row['password'];
    }
    mysqli_close($conn);  
}
?>

Posted 20-Jun-17 4:07am

BasmaSH

Updated 4-Jul-22 23:30pm

Chris Maunder772.9K

v4

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Comments

Richard Deeming 22-Jun-17 17:29pm

NEVER store passwords in plain text, and NEVER display them to anyone!

Secure Password Authentication Explained Simply[^]
Salted Password Hashing - Doing it Right[^]

3 solutions

Solution 2

remove coats for php variable i,e.,
$query ="SELECT password from names where name=$name";

Posted 23-Jun-17 8:08am

Raghu112

Solution 4

$query ="SELECT password from names where name=$name";
$result = mysql_query($query ,$query);
return $result ;

Posted 4-Jul-22 23:30pm

Member 15695717

Comments

CHill60 5-Jul-22 6:54am

Apart from the fact the question is 5 years old and already has an accepted answer, you have effectively just copied solution 3 - with one major mistake - the first parameter for mysql_query() is the connection. Passing the query twice will simply just not work.
Stick to answering more recent posts where the OP might actually remember the post, and make sure your solution is both accurate and brings something new to the thread

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ThilinaMD · Accepted Answer · 2017-06-26T07:08:00

change your query as

$query ="SELECT password from names where name=$name";

Hence $name is varable no need to use quotes,

Better if you use

mysqli

fucntions in all sections

$result = mysqli_query($conn,$query); 
    while ($row = mysqli_fetch_array($result)) {
        echo $row['password'];
    }