How do I update the values in the table using PHP

Question

How do i actually update the values of table using PHP ? This code is not showing any error and its not updating either.

<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname = 'DB';
$conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
if(mysqli_connect_error()) 
{
die("couldn't connect" . $conn->connect_error());
}
echo ("connected successfully");
$id = $_POST['Id'];
$name = $_POST['Name'];
$dept = $_POST['Department'];
$update = "update info set Name='$name', Department='$dept' where Id='$id'";
if($conn->query(update) === TRUE) {
echo ("Data updated successfully");
} 
else
{
echo ("Data cant be updated" . $conn->error());
}
$conn->close();
?>

Have you checked the column names if they are correct? Is there a row matching with the bind `$id`? Did you execute your query? Show more code. — Logan Wayne, Apr 13 '15 at 05:42
connect_error()); } echo ("connected successfully"); $id = $_POST['Id']; $name = $_POST['Name']; $dept = $_POST['Department']; $update = "update info set Name='$name', Department='$dept' where Id='$id'"; if($conn->query(update) === TRUE) { echo ("Data updated successfully"); } else { echo ("Data cant be updated" . $conn->error()); } $conn->close(); ?> — Naga Naveen, Apr 13 '15 at 05:45
Yes. The column names are perfect. I would've tried at least 10 times. I don't know what is the problem. i'm pretty new to this. — Naga Naveen, Apr 13 '15 at 05:46
Post this to your question and not here in the comment area. — Amit Verma, Apr 13 '15 at 05:47
First of all, are you sure the DB user you use to connect from php has the update privileges? - EDIT: seen your updated question. I think your root has all the privileges, even if it's not very safe to use it. — Marco Bernardini, Apr 13 '15 at 05:49
Sorry. And yes the privileges are fine. But can you people tell me how to check those? I may have a look now. I'm using XAMPP — Naga Naveen, Apr 13 '15 at 05:51

score 4 · Answer 1 · answered Apr 13 '15 at 05:45

4

Hope this one help you!

$update = "update info set Name='".$name."', Department='".$dept."' where Id='".$id."'";

answered Apr 13 '15 at 05:45

This doesn't really solves the problem of OP. – Logan Wayne Apr 13 '15 at 05:49

score 3 · Answer 2 · edited May 23 '17 at 10:26

Check this part of your code:

if($conn->query(update) === TRUE) {

where it should be:

if($conn->query($update) === TRUE) {

Make sure that you are using the correct credentials (host, username, password, database name) according to your MySQL database.
Also your table name and column name should be correct which are being used in your query.
Make sure that there is a match with your condition part of your query (... WHERE Id='$id'). Check it by running a query in your PhpMyAdmin page, or Search the ID, which is also the one you try to input in your form.
Make sure that the name of the passed variables ($_POST[]) are correct.
Be case sensitive.

Try changing your connection into:

$conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);

/* CHECK CONNECTION */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

Other way to execute your query is to simply:

mysqli_query($conn,$update);

Recommendation:

You should escape the values of your variables before using them into your query by using mysqli_real_escape_string() function:

$name = mysqli_real_escape_string($conn,$_POST["Name"]);

Or better, so you won't need to worry about binding variables into your query and as well prevent SQL injections, you should move to mysqli_* prepared statement:

if($stmt = $conn->prepare("UPDATE info SET Name=?, Department=? WHERE Id=?")){

  $stmt->bind_param("ssi",$_POST['Name'],$_POST['Department'],$_POST['Id']);
  $stmt->execute();
  $stmt->close();

}

Care to explain the downvote? And I hope it's not because of downvote vengeance. Because I'm not the one that downvotes the answer of the other two. — Logan Wayne, Apr 13 '15 at 05:57
I tried this. It didn't workout. Nor any other suggestions. What should i do now? — Naga Naveen, Apr 13 '15 at 06:07
@NagaNaveen - execute your query by `mysqli_query($conn,$update);` — Logan Wayne, Apr 13 '15 at 06:50

Amitesh Yadav · Answer 3 · 2015-04-13T05:49:23.460

1

$update = "update info set Name='".$name."', Department='".$dept."' where Id='".$id."'";
 mysql_query($update);

edited Apr 13 '15 at 05:49

answered Apr 13 '15 at 05:48

Amitesh Yadav

202
1
10

This doesn't really solves the problem of OP. – Logan Wayne Apr 13 '15 at 05:49
1

Please check your datatype in table , And could you please show your mysql error. – Amitesh Yadav Apr 13 '15 at 05:51
Umm.. I'm not the OP – Logan Wayne Apr 13 '15 at 05:53
The problem is, it is not showing any kind of error. When i enter the values in the respective fields and click on update, the values in the fields disappear. When i check in the table there are no changes. – Naga Naveen Apr 13 '15 at 06:02

score 1 · Answer 4 · answered Apr 13 '15 at 06:35

1

$update = "update info set Name='".$name."',set Department='".$dept."' where Id='".$id."'";

if this is not help please provide form code.

answered Apr 13 '15 at 06:35

Sourabh

500
2
14

score 0 · Answer 5 · answered Apr 13 '15 at 06:31

Try this

<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname = 'DB';
$conn = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
if(!$conn) 
{
die("ERROR CONNECTING TO DATABASE!");
}
echo "Connected Successfully";
$id = $_POST['Id'];
$name = $_POST['Name'];
$dept = $_POST['Department'];
$update = "update info set Name='$name', Department='$dept' where Id='$id'";
$qry = mysqli_query($conn,$update);
if(!$qry) {
echo "Error Updating Details".mysqli_error($conn);
} 
else
{
echo "Data updated successfully";
}
mysqli_close($conn);
?>

(Optional) Use secure things. Change to this for more secure.

$id = mysqli_real_escape_string($conn,$_POST['Id']);
$name = mysqli_real_escape_string($conn,$_POST['Name']);
$dept = mysqli_real_escape_string($conn,$_POST['Department']);

How do I update the values in the table using PHP

5 Answers5

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