CODEWITHSUNDEEP
phphtmlmysqldatabase
Rishabh Agarwal
Rishabh Agarwal

Reputation: 1

Why this is not updating Information

Why this code is not update information?

HTML Form:

<form>
<lable> ID# :</lable>
<input id= "ID" name= "ID" type= "text">

<p>
    <label>Select field to Edit</label>
    <select name="change">
    <option value=""></option>
    <option value="fname">First Name</option>
    <option value="lname">Last Name</option>
    <option value="email">Email</option>
    <option value="city">City</option>
    <option value="zip">Zip</option>
</select>
<lable> Enter the value to be replaced </label>
<input id = "replace" name = "replace" type = "text">
</p>

<input name="submit" type="submit" value="Submit">

PHP Code for updating information from database:

  <?php
    $servername = "localhost";
    $username = "root";
    $password = "root";

    $conn = mysql_connect($servername,$username,$password);
    if(!$conn)
    { 
      die('Error!' . mysqli_error()); 
    }
    $sql = 'SELECT * FROM users';
    mysql_select_db('mitsdatabase');

    $retval = mysql_query($sql, $conn);

    if(! $retval) 
    {
      die('Could not get data:' . mysql_error());
    }

    echo "<table width='300' cellpadding='5' border='1'>";
    echo "<tr> <td>ID#</td> <td>FirstName</td> <td>LastName</td> <td> Email </td> <td> City </td> <td> State </td> <td> Zip </td> </tr>";



    while($row = mysql_fetch_array($retval,MYSQL_ASSOC))
    {

      echo "<tr> <td>{$row['ID']}</td> . <td>{$row['fname']}</td> . <td>{$row['lname']}</td> . <td>{$row['email']}</td> . <td>{$row['city']}</td> . <td>{$row['state']} </td>. <td>{$row['zip']}</td>";
     }
    echo "</table>";
    $db_id = $_POST['ID'];
    $db_select = $_POST['change'];
    $db_replace= $_POST['replace'];

    echo " Do you want to edit any entry?";

    if(!_POST['submit'])
    {
        echo " ";
    }
    else{
        mysqli_query("UPDATE users SET db_select='$db_replace' WHERE ID = $db_id ");
    }
    mysql_close($conn);
  ?>

I want to update informate selected from select field but somehow it is not doing any thing. Can someone help me what is wrong with this code.

Upvotes: 0

Views: 69

Answers (2)

Alexander Nicholas Popa
Alexander Nicholas Popa

Reputation: 272

If you want to post to the same page just change <form> to <form method="post" action="#"> and get variables in php like this $nameofvar = $_POST['nameofinputfield'] . Each input field should have the name tag. Also try to change your mysql connect to this :

<?php
$servername = "localhost";
$username = "username";
$password = "password";

// Create connection
$conn = new mysqli($servername, $username, $password);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
echo "Connected successfully";

and after you finished the query 

$conn->close();

and the query to insert

$sql = "INSERT INTO MyGuests (firstname, lastname, email)
VALUES ('John', 'Doe', '[email protected]')";

if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

and you can modify this $sql string to update or delete

Upvotes: 0

Tommy Jinks
Tommy Jinks

Reputation: 757

Is your PHP on the same page as your HTML? If not, you are not directing to your php code within the <form> element in your HTML.

For example, if your PHP file was called 'myphpcode.php' (and in the same folder as your HTML code) then you could direct to it using the following:

<form method="post" action="myphpcode.php">

Upvotes: 2

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