See the code below, it check if the data exist in the table, if not exist then insert it or else update the table.
As you can see it look a bit messy - is there anyway to improve the code logic or something smaller? I have a few tables that need doing same thing.
foreach ($sheet as $data) {
// Get Phone ID
$dataPhoneID = mysql_escape_string($data['handset']['phone_id']);
if (isset($stocks[$dataPhoneID])) {
$stockPhone = $stocks[$dataPhoneID ];
$phoneName = mysql_escape_string($stockPhone['description']);
$stock = mysql_escape_string($stockPhone['stock']);
$SQL = "SELECT * FROM phone_affiliate WHERE affiliate_id = 1 AND affiliate_phone_id = '$dataPhoneID'";
$q = mysql_query($SQL);
if (mysql_num_rows($q) == 0) {
$SQLInsert = "INSERT INTO phone (name) VALUE('$phoneName')";
if (mysql_query($SQLInsert)) {
$phone_id = mysql_insert_id();
$SQLInsert = "INSERT INTO phone_affiliate (phone_id, affiliate_id, affiliate_phone_id, stock) ";
$SQLInsert .= "VALUE('$phone_id', '1', '$dataPhoneID', '$stock')";
mysql_query($SQLInsert) or die(mysql_error());
}
} else {
$row = mysql_fetch_assoc($q);
$phone_id = $row['phone_id'];
$SQLUpdate = "UPDATE phone_affiliate set stock = '$stock' WHERE affiliate_id = 1 AND phone_id = $phone_id";
mysql_query($SQLUpdate) or die(mysql_error());
}
// Similar code block above for other tables.
}
}
Note: I am aware about PDO but I don't have time to replace it on existing system.
Use mysql's REPLACE INTO or INSERT... ON DUPLICATE KEY UPDATE. For example:
foreach ($sheet as $data) {
// Get Phone ID
$dataPhoneID = mysql_escape_string($data['handset']['phone_id']);
if (isset($stocks[$dataPhoneID])) {
$stockPhone = $stocks[$dataPhoneID ];
$phoneName = mysql_escape_string($stockPhone['description']);
$stock = mysql_escape_string($stockPhone['stock']);
$SQLInsert = "INSERT INTO phone_affiliate (affiliate_id, affiliate_phone_id, stock) ";
$SQLInsert .= "VALUES ('1', '$dataPhoneID', '$stock') ";
$SQLInsert .= "ON DUPLICATE KEY UPDATE stock = '$stock'";
mysql_query($SQLInsert);
if (mysql_insert_id()) {
$SQLInsert = "INSERT INTO phone (name) VALUE('$phoneName')";
mysql_query($SQLInsert);
$phone_id = mysql_insert_id();
$SQLUpdate = "UPDATE phone_affiliate set phone_id = $phone_id WHERE affiliate_id = 1 AND affiliate_phone_id = $dataPhoneID_id";
}
}
}
Also you can use INSERT IGNORE construction
Related
it just set my id into 0 after i inserted some data into the table
then, I get this kind of error:
Duplicate entry '0' for key 'PRIMARY'.
can someone help me identify the problem?
$query = mysql_query("INSERT INTO customer VALUES('', '$name', '$phone', '$address', '$email_add')") or die(mysql_error());
$i=1;
foreach($_SESSION as $namee => $value)
{
if($value > 0)
{
if(substr($namee, 0, 5) == 'cart_')
{
$id = substr($namee, 5, (strlen($namee)-5));
$get = mysql_query("SELECT * FROM product WHERE code='$id'");
while($get_row = mysql_fetch_assoc($get)){
$sub = $get_row['price'] * $value;
echo '<p>'.$i.' '.$get_row['code'].' '.$get_row['name_nl'].' '.$value.' SubTotal : RM '.$sub.'</p> ';
$getCustomer = mysql_query("SELECT customer.id_customer, customer.name, customer.address, product.code, product.name_nl FROM customer, product WHERE name='$name' AND address='$address'" ) or die(mysql_error());
$data = mysql_fetch_array($getCustomer);
$pemb = $data['id_customer'];
$na = $data['name'];
$al = $data['address'];
$ib = $get_row['code'];
$nb = $get_row['name_nl'];
$i++;
}
}
mysql_query("INSERT INTO book VALUES('', '$pemb', '$na', '$al', '$ib', '$nb', '$value', '$sub', now()) ") or die(mysql_error());
}
}
Primary key fields must have different values.
To resolve this, you must set this field to AUTO_INCREMENT, so each time a new record is entered, the primary key field is incremented automatically.
I am trying to fetch a cart array through foreach loop. I can fetch array items successfully and when i echo them it shows the correct values. But when i apply insert query for each item to enter in my database table, It does not insert any thing but it shows success. I have 4 tables named:
customers-->serial primary,name,email,password,address,phone,city
products-->productid primary,name,description,image,price
orders-->serial primary,date,customerid foreign key
order_detail-->orderid,productid,quantity,price,color,size
order_detail and orders table should be updated when i click on button.
Here is my code. Any suggestions will be appreciated.
Code
if (isset($_SESSION['login_email'])) {
$email = $_SESSION['login_email'];
//Insert values in order table. This works correctly!!
$query = mysqli_query($con, "SELECT serial FROM customers WHERE email = '$email'");
while ($row = mysqli_fetch_array($query, MYSQLI_ASSOC)) {
$cust_id = $row['serial'];
$date = date('Y-m-d');
$que = mysqli_query($con, "INSERT INTO orders VALUES('', '$date', '$cust_id')");
}
// Fetching serial from orders table. This also works correct.
$q = mysqli_query($con, "SELECT serial FROM orders WHERE customerid = '$cust_id'");
while ($row1 = mysqli_fetch_array($q, MYSQLI_ASSOC)) {
$serial = $row1['serial'];
echo $serial ."\n";
}
//Problem comes here. Here i am trying to insert values in order_details
foreach ($_SESSION['cart'] as $id => $value) {
$subtotal = $value['price'] * $value['quantity'];
$pid = $value['id'];
$quantity = $value['quantity'];
$color = $value['color'];
$size = $value['size'];
}
mysqli_query($con, "INSERT INTO order_detail VALUES ($serial, $pid, $quantity, $subtotal, $color, $size)");
if (true) {
echo "success";
} else {
echo"not success";
}
}
Did you already try this?
if( mysqli_query($con, "INSERT INTO order_detail VALUES ($serial, $pid, $quantity, $subtotal, $color, $size)") ){
echo "success";
} else {
echo"not success";
}
I Think you better can do the insert in this way:
INSERT INTO order_detail (column1, column2, column3,...)
VALUES (value1, value2, value3,...)
I'm making a form that submits a story into a MySQL table called 'work'. I want to later take the id of the newly created record and put the information into a different table.
But when I submit the story, it says:
$workid is undefined.
I can't see the problem though because I believe I've defined it?
<?php
if (!empty($_POST) && !empty($_POST['title']) && !empty($_POST['story']) && !empty($_POST['genre']) && !empty($_POST['rating'])) {
$title = strip_tags($_POST['title']);
$story = strip_tags($_POST['story']);
$title = mysqli_real_escape_string($db, $title);
$story = mysqli_real_escape_string($db, $story);
$genre = $_POST['genre'];
$rating = $_POST['rating'];
$query = "SELECT COUNT(*) AS count FROM works WHERE Title = '".$title."'";
$result = $db->query($query);
$data = $result->fetch_assoc();
if ($data['count'] > 0) {
echo "<p>Story already exists!</p>";
} else {
$query = "INSERT INTO works (author_id, login_id, Title, Story, Genre, Rating) VALUES ('".$userid."','".$authorid."','".$title."','".$story."','".$genre."','".$rating."')";
$query = "SELECT `id` FROM `works` WHERE `Title` = '".$title."'";
if ($result = $db->query($query)) {
while ($row = $result->fetch_assoc())
$workid = $row["id"]; //workid is written here but still considered undefined
}
$query = "INSERT INTO `author_work` (`author_id`) VALUES ('".$authorid."')";
$result = $db->query($query);
$query = "INSERT INTO `author_work` (`work_id`) VALUES ('".$workid."')";
$result = $db->query($query);
$query = "INSERT INTO `login_work` (`work_id`) VALUES ('".$workid."')";
$result = $db->query($query);
$query = "INSERT INTO `login_work` (`login_id`) VALUES ('".$userid."')";
$result = $db->query($query);
if ($result) {
echo "<p>Story submitted!</p>";
} else {
echo "SQL Error: " . $db->error;
}
}
}
?>
You never did a $db->query() on your INSERT INTO... query string, so it was never inserted, and was overwritten by your SELECT id ... query.
$query = "INSERT INTO works (author_id, login_id, Title, Story, Genre, Rating) VALUES ('".$userid."','".$authorid."','".$title."','".$story."','".$genre."','".$rating."')";
$db->query($query); // Missing this $db->query()
$query="SELECT `id` FROM `works` WHERE `Title` = '".$title."'";
if ($result = $db->query($query)) {
while ($row= $result->fetch_assoc())
$workid = $row["id"];}
Your $workid might not be initialized, depending on your condition and the result of your SQL query: so try to avoid next operations that will causes warnings/errors by using continue or else
I want to check whether the data is existing or not. If data exists, update the table called "user_star_rate", otherwise insert the data into the table. Inserting is working properly, but the updating is not working.
Here is my code.
$jsqla7 = mysql_query("select * from user_star_rate where product_id='$product_id' and email='$visit_email'") or die(mysql_error());
$jfeta7 = mysql_fetch_assoc($jsqla7);
if($jfeta7 != null) {
$rate = "UPDATE user_star_rate SET rate_value='$rate_value' WHERE product_id='$product_id' and email='$visit_email'" ;
} else {
$rate = "INSERT INTO user_star_rate (email, product_id, rate_value) VALUES ('$visit_email','$product_id','$rate_value')" ;
}
If I understand what your saying, you should really be using a replace into and it would also decrease your code significantly.
Your code would become:
$query = "REPLACE INTO user_star_rate(product_id, email) VALUES('$product_id', '$visit_email')";
mysql_query($query) or die(mysql_error());
If it already exists then it will update it, else it will insert it. You should pay special attention to the docs in regards to foreign keys and auto incrementing ids.
Try this :
$tot = mysql_num_rows($jfeta7);
if($tot > 0){
$rate = "UPDATE user_star_rate SET rate_value='$rate_value' WHERE product_id='$product_id' and email='$visit_email'" ;
} else {
$rate = "INSERT INTO user_star_rate (email, product_id, rate_value) VALUES ('$visit_email','$product_id','$rate_value')" ;
}
Try:
$jsqla7 = mysql_query("select count(*) from user_star_rate where product_id='$product_id' and email='$visit_email'") or die(mysql_error());
$count = mysql_num_rows();
if($count) {
$rate = "UPDATE user_star_rate SET rate_value='$rate_value' WHERE product_id='$product_id' and email='$visit_email'" ;
} else {
$rate = "INSERT INTO user_star_rate (email, product_id, rate_value) VALUES ('$visit_email','$product_id','$rate_value')" ;
}
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
create a mysql record if it doesnt exist, else update it
I have created a bit of PHP that gets data from a CSV and updates a Database Table to match the CSV's data.
My next task is, To check if the record exists, based on the unique column (sku). If it does exist update the database table dependant on the change.
Otherwise if that SKU column doesnt match do the INSERT.
The code I've got at the moment runs, However the issue I have is that if a new item is created it seems to run the update through the already existing records.
My code can be found on Pastebin
Or also here...
<?php
$con = mysql_connect("localhost", "root", "");
if(!$con)
{
die('Could not connect' . mysql_error());
}
mysql_select_db("db_lemonstand", $con);
class csvIterator extends LimitIterator
{
public function __construct($path)
{
$csv = new SplFileObject($path);
$csv->setFlags(SplFileObject::READ_CSV);
parent::__construct($csv, 1);
}
}
foreach (new csvIterator('data/catalogue.csv') as $entry) {
$name = $entry[23];
/* Replace Strings To Make URL Name */
$search_array = array(" ", "/", "+");
$replace_array = array("-");
$url_name = strtolower($name);
$url_name = str_replace($search_array, $replace_array, $url_name);
$long_desc = $entry[9];
$short_desc = $entry[23];
$manufacturer = $entry[11];
$price = $entry[15];
$sku = $entry[2];
$weight = $entry[29];
$width = $entry[30];
$height = $entry[5];
$enabled = '1';
$created_at = date('Y-m-d H:i:s');
$product_type_id = '1';
$tax_class_id = '1';
echo '<pre>';
print_r($entry);
echo '</pre>';
// Check see if products' SLU already exists or not
$product_exists = "SELECT sku FROM shop_products WHERE sku = '$sku'";
$result = mysql_query($product_exists, $con);
$exists = mysql_num_rows($result);
if($exists == 0 )
{
$insert = "INSERT INTO shop_products (name, description, short_description, url_name, price, sku, weight, width, height, enabled, created_at, tax_class_id, product_type_id) VALUES ('$name', '$long_desc', '$short_desc', '$url_name', '$price', '$sku', '$weight', '$width', '$height', '$enabled', '$created_at', '$tax_class_id', '$product_type_id')";
$insert_data = mysql_query($insert, $con);
}
else
{
$update = "UPDATE shop_products SET name = '$name', description = '$long_desc', short_description = '$short_desc', url_name = '$url_name', price = '$price', sku = '$sku', weight = '$weight', height = '$height', enabled = '$enabled', created_at = '$created_at', tax_class_id = '$tax_class_id', product_type_id = '$product_type_id'";
$update_data = mysql_query($update, $con);
if (!mysql_query($update,$con))
{
die('Error: ' . mysql_error());
}
}
}
You can use
INSERT INTO mytable on DUPLICATE KEY UPDATE...
You must set unique key for your table (one or more fields) and than you can use this ability.
In my example I have unique key (template, date)
$sql = 'INSERT INTO myTable
(`template`,`date`,`count`) VALUES
("' . $template . '","' . $curDate . '", 1)
ON DUPLICATE KEY UPDATE count = count + 1';
Store store you fields in a var as $fields or anything else that you want and execute this query for that you need a single field as reference that's why i included $Id
INSERT INTO table_name SET id = '$Id', $fields ON DUPLICATE KEY UPDATE $fields