update function in php

Question

I have a update function that for now updates the required changes to MySQL database when I run index.php.

This is updating my password buy not the name field, ive been over the code and can not work out why. Any help is greatly appreciated.

Index that tells what id and fields to update with entered data

    <?php
require_once 'core/init.php';

$userInsert = DB::getInstance()->update('users', 1, array(
    'password'   => 'newpass',
    'name'       => 'Ben'
));

Function in different php that updated database

    public function update($table, $id, $fields) {
    $set = '';
    $x = 1;

    foreach($fields as $name => $value) {
        $set .= "{$name} = ?";
        if($x < count($fields)) {
            $set .= ',';
        }
        $x++;
    }

    $sql = "UPDATE {$table} SET {$set} = 'newpassword' WHERE id = {$id}";

    if(!$this->query($sql, $fields)->error()) {
        return true;
    }

    return false;
}

I believe it to be a small error or mistype but I can not see the problem.

As you can see bellow the password field has been changed but the name has not enter image description here

Echo out the query after you've filled in the variables, and you'll see what's wrong with it. — Jessica
– Jessica, Commented Jan 7, 2014 at 21:15
Take Jessica's approach, the problem is banging you on the nose. $sql = "UPDATE {$table} SET {$set} = 'newpassword' WHERE id = {$id}"; — Rottingham
– Rottingham, Commented Jan 7, 2014 at 21:17
You're never actually updating your username. SET username = 'blah',password = 'blahblah' — Zarathuztra
– Zarathuztra, Commented Jan 7, 2014 at 21:17
@InGodITrust Um.... apparently its banging you on the nose too — Rottingham
– Rottingham, Commented Jan 7, 2014 at 21:21
@InGodITrust Have you ever seen a query that looks like this work? UPDATE TABLE SET username = ? = 'newpassword' — Rottingham
– Rottingham, Commented Jan 7, 2014 at 21:22

Mixthos · Accepted Answer · 2014-01-07 21:19:23Z

4

public function update($table, $id, $fields) {
    $set = '';
    $x = 1;

    foreach($fields as $name => $value) {
        $set .= "{$name} = \"{$value}\"";
        if($x < count($fields)) {
            $set .= ',';
        }
        $x++;
    }

    $sql = "UPDATE {$table} SET {$set} WHERE id = {$id}";

    if(!$this->query($sql, $fields)->error()) {
        return true;
    }

    return false;
}

answered Jan 7, 2014 at 21:19

Mixthos

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2 Comments

Rottingham Over a year ago

@InGodITrust It does fix the problem. whats your deal?

Digital Chris Over a year ago

@InGodITrust this answer is pretty self-explanatory if you read the question: it's the OP's code, slightly modified to remove the illogic.

Maciej A. Czyzewski · Accepted Answer · 2014-01-07 21:23:21Z

1

Simply use of prepare and execute in PDO:

$sql = 'UPDATE '. $table .' SET username = :username, password = :password WHERE id = '. $id;
$sth = $dbh->prepare($sql);
$sth->execute(array(
    ':username' => 'ben',
    ':password' => 'newpassword'
));

answered Jan 7, 2014 at 21:23

Maciej A. Czyzewski

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2 Comments

Rottingham Over a year ago

This works but it doesn't fix his problem but showing him what was wrong

Digital Chris Over a year ago

This is a good practice in general, but if you look at his code, he is trying to make a "universal" update function that will allow for any number of fields and can be applied on any table.

Sambath Piazza · Accepted Answer · 2022-06-29 01:44:10Z

0

    private function update($table, $primaryKey, $fields) {
        $query = 'UPDATE `' . $this->table . '` SET ';
        
        foreach ($fields as $key => $value) {
            $query .= '`' . $key . '` = :' . $key . ',';
        }

        $query = rtrim($query, ',');

        $query .= ' WHERE `' . $this->primaryKey . '` = :primaryKey';

        $fields['primaryKey'] = $fields['id'];
    
        $this->query($query, $fields);

    }

An example of an update function. Attention mine is inside a class and the query is another function and passes as an object.

edited Jun 29, 2022 at 1:44

answered Jun 29, 2022 at 1:28

Sambath Piazza

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