Database not updating correctly in PHP / MySQL [duplicate] - php

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Can I mix MySQL APIs in PHP?
(4 answers)
Closed 7 years ago.
I've got a database with a Users table which I'm trying to update.
Currently I have customers.php, which displays form fields with the user information so it can be updated.
This form points to edit_customer_processor.php , which takes the new values, puts them into a MYSQL query... and then despite the query working correctly when I query the DB via the PHPMyAdmin command line, the record doesn't update.
customers.php
<?php
session_start();
if(!$_SESSION["logged_in"]){
header("location:home.php");
die;
}
?>
<?php include 'header.html'; ?>
<div id='maincontent'>
<?php
if (isset($_GET["id"])){
$customer_id = $_GET["id"];
require_once('config.php');
$customer_query = "SELECT * FROM customer WHERE customer_id = $customer_id";
$customer_results = mysql_query($customer_query, $conn);
if (!$customer_results) {
die ("Error selecting car data: " .mysql_error());
}
else {
while ($row = mysql_fetch_array($customer_results)) {
echo "<h3>Edit Customer</h3>";
echo "<FORM method='post' action='edit_customer_processor.php'>";
echo '<p> Name: <input type="text" name="name" size = "40" value=' . $row[name] . '></p>';
echo '<p> Address: <input type="text" name ="address" size="40" value=' . $row[address] . '></p>';
echo '<p> Email: <input type="text" name="email" value=' . $row[email] . '></p>';
echo '<p> Phone: <input type ="text" name="phone" size="20" value=' . $row[phone] . '></p>';
echo '<input type ="hidden" name="customer_id" value="' . $row[customer_id] . '">';
echo '<input type ="hidden" name="formtype" value="edit_customer">';
echo '<input type="submit" name="submit" value= "Update">';
echo '</form>';
}
}
} else {
// If there isn't an ID, display the New Customer form and all customers below, with links
// to their edit pages.
echo "<h3>Enter new customer information and submit.</h3>";
echo "<FORM method='post' action='new_customer_processor.php'>";
echo '<p> Name: <input type="text" name="name" size = "40"></p>';
echo '<p> Address: <input type="text" name ="address" size="40"></p>';
echo '<p> Email: <input type="text" name="email"></p>';
echo '<p> Phone: <input type ="text" name="phone" size="20"></p>';
echo '<input type ="hidden" name="formtype" value="new_customer">';
echo '<input type="submit" name="submit" value= "Submit">';
echo '<input type ="reset" name="reset" value ="Reset">';
echo '</form>';
require_once('config.php');
echo "<h3>Current Customers</h3>";
$query = "SELECT * FROM customer";
$results = mysql_query($query, $conn);
if (!$results) {
die ("Error selecting customer data: " .mysql_error());
}
else {
// In the absence of an ID, all customers will be displayed down
// the bottom of the page
while ($row = mysql_fetch_array($results)) {
echo "<a href=customers.php?id=";
echo $row[customer_id];
echo "><p> $row[name] </p></a>";
echo "<p> $row[address] </p>";
echo "<p> $row[phone] </p>";
echo "<p> $row[email] </p>";
}
}
}
?>
Back to Customers Page
</div>
<?php include 'footer.html' ?>
edit_customer_processor.php
<?php include 'header.html' ?>
<div id="maincontent">
<?php
// Pulling in hidden customer ID from post value
$mysqli = new mysqli( 'localhost', 'root', 'root', 'w_c_a' );
// Check our connection
if ( $mysqli->connect_error ) {
die( 'Connect Error: ' . $mysqli->connect_errno . ': ' . $mysqli->connect_error );
}
// Insert our data
$sql = mysql_query("UPDATE customer
SET name = '".mysql_real_escape_string($_POST['name'])."',
address = '".mysql_real_escape_string($_POST['address'])."',
phone = '".mysql_real_escape_string($_POST['phone'])."',
email = '".mysql_real_escape_string($_POST['email'])."'
WHERE customer_id='".mysql_real_escape_string($_POST['customer_id'])."'");
$update = $mysqli->query($sql);
echo "Customer updated: ";
echo "<a href=customers.php?id=" . $_POST['customer_id'] . ">";
echo "Back to Edit Customer</a>";
?>
</div>
<?php include 'footer.html' ?>
And when I echo the MYSQL query, I get:
UPDATE customer SET name = 'Kellyassdsa', address = 'ads', phone = '0260123123', email = 'asdasd' WHERE customer_id='1'
Which works when I put it in PHPMyAdmin.
I know it'll be some boneheaded little mistake, but I've been trying to get this work for ages now. Any ideas?

Answer 1

Maybe your program just can't connect to your MySQL database.
$customer_results = mysql_query($customer_query, $conn);
I can't see where you gave a value to the var $conn.
If the problem is connection problem then we might need your database info like the name of your table in PhpMyAdmin.

Answer 2

your problem is...
$sql = mysql_query(..);
$update = $mysqli->query($sql);
it should be
$sql = 'UPDATE ...';
$update = $mysqli->query($sql);

Answer 3

i think problem occurs due to line break. pleas make a query in single line without line break.
$sql = mysql_query("UPDATE customer SET name = '".mysql_real_escape_string($_POST['name'])."',address = '".mysql_real_escape_string($_POST['address'])."', phone = '".mysql_real_escape_string($_POST['phone'])."', email = '".mysql_real_escape_string($_POST['email'])."' WHERE customer_id='".mysql_real_escape_string($_POST['customer_id'])."'");
Hope this helps..