i made module for prestashop. Just basic administration form with write into DB. But when i try to install module at Prestashop, i have this error:
[PrestaShopDatabaseException]
Duplicate entry '0' for key 'PRIMARY'
INSERT INTO `ps_module` (`name`, `active`, `version`) VALUES ('apishippingtextsource', '1', '0.0.1')
at line 635 in file classes/db/Db.php
629. WebserviceRequest::getInstance()->setError(500, '[SQL Error] '.$this->getMsgError().'. From '.(isset($dbg[3]['class']) ? $dbg[3]['class'] : '').'->'.$dbg[3]['function'].'() Query was : '.$sql, 97);
630. }
631. else if (_PS_DEBUG_SQL_ && $errno && !defined('PS_INSTALLATION_IN_PROGRESS'))
632. {
633. if ($sql)
634. throw new PrestaShopDatabaseException($this->getMsgError().'<br /><br /><pre>'.$sql.'</pre>');
635. throw new PrestaShopDatabaseException($this->getMsgError());
636. }
637. }
638.
639. /**
my install.php for sql:
$sql = array();
$sql[] = 'CREATE TABLE IF NOT EXISTS `' . _DB_PREFIX_ . 'api_shipping_text` (
`id_text` int(10) unsigned NOT NULL AUTO_INCREMENT,
PRIMARY KEY (`id_text`)
) ENGINE=' . _MYSQL_ENGINE_ . ' DEFAULT CHARSET=utf8;';
foreach ($sql as $query) {
if (Db::getInstance()->execute($query) == false) {
return false;
}
}
Version of Prestashop is 1.6.0.9.
Can you check the auto_increment value of your ps_module table structure? Maybe auto_increment is not defined or set to 0.
If that's the case, you probably had an error during Database migration or installed another module messing with auto_increment.
Related
I am creating a wordpress plugin which upon activation i create my database table. I have an .sql file which has ~2000 rows of data and i would like to execute it after the creation of my database table. Which do you think is the best way to insert those data upon the plugin activation?
I know it is going to execute on activation hook - but is it possible to execute an .sql file or do i have to find another way and somehow execute each row through $wpdb->insert()?
If you need any further infomation please ask and i will provide - i also searched but couldn't find the same case.
Below is the best way by which you can achieve this as first it will initialize custom_create_plugin_tables and next it will initialize custom_insert_data_custom_table
register_activation_hook( __FILE__, 'custom_create_plugin_tables' )
register_activation_hook( __FILE__, 'custom_insert_data_custom_table' );
function custom_create_plugin_tables()
{
global $wpdb;
$table_name = $wpdb->prefix . 'table_name';
$sql = "CREATE TABLE $table_name (
id int(11) NOT NULL AUTO_INCREMENT,
column_two varchar(255) DEFAULT NULL,
column_three int(11) NOT NULL,
UNIQUE KEY id (id)
);";
require_once( ABSPATH . 'wp-admin/includes/upgrade.php' );
dbDelta( $sql );
}
function custom_insert_data_custom_table()
{
global $wpdb;
$table_name = $wpdb->prefix . 'table_name';
$rows_affected = $wpdb->insert( $table_name, array( 'column_two' => "column_two_value", 'column_three' => 'column_three_value' ));
require_once( ABSPATH . 'wp-admin/includes/upgrade.php' );
dbDelta( $rows_affected );
}
I'm getting a strange result, where an update method (http://docs.doctrine-project.org/projects/doctrine-dbal/en/latest/reference/data-retrieval-and-manipulation.html#update) call that activates/deactivates site users accounts in my mySQL database sometimes are not updated and returns a 0. Here is the code:
$user_id = '198';
$sqlSetStmnt= [ 'activation_code' => NULL, 'active' => 0 ];
$conn = $this->get( 'database_connection' );
try {
$result = $conn->update( 'users', $sqlSetStmnt,[ 'id' => (int)$user_id ] );
}
catch( Exception $e ) {
$error = $e->getMessage();
throw new \Exception( 'update user account data function -- ' .
'error: ' . $error . ' - ' .
'unable to update your account!' );
} // End of catch( Exception ) block.
if( ( $result === FALSE ) || ( $result !== 1 ) ) {
throw new \Exception( 'update user account data function -- ' .
'update return value: ' . $result . ' - ' .
'unable to update your account!' );
} // if( ( $result === FALSE ) || ( $result !== 1 ) ) ...
The users table has an id int(11) column which is the primary key, an active tinyint(1) column, and activation_code varchar(40) NULL column.
Please note the $user_id variable contains a string value of '198', but it is cast to an int when creating the $sqlSetStmnt value.
Inspecting the users table confirms that there was a row in the users table with the id column value of 198 at the time of the update call.
The account used when running the update call has enough privileges to change the row active and activation_code column values.
There are no other users in the system or accessing the database, so there aren't any locks on the row.
I inspected the code using x-debug, and the values of the $user_id and $sqlSetStmnt variables were properly set to the values that I expected, that $result was set to 0 by the update method, and that no exceptions were thrown by the update method call.
By the way, there is no need to using variable binding because the values in the $user_id and $sqlSetStmnt variables are not input by an user, so no possibility of SQL-Injection, or Buffer-Overrun.
Is there some way to get information from DBAL about why the update method returned 0?
Thank you.
Before solving this issue I switched from:
$result = $conn->update( 'users', $sqlSetStmnt, [ 'id' => (int)$user_id ] );
to:
$sqlUpdateStmnt = 'UPDATE `users` SET field = value ' .
'WHERE `id` = ' . $user_id;
$result = $conn->executeUpdate( $sqlUpdateStmnt );
and got exactly the same result, where some updates would return 0 rows when there definitely was a row in the users table with an id matching the value of $user_id.
I got around this problem by fetching the existing row and then only updating the row when the fields in the set-clause were different than the same columns from the table.
This tells me that the 0 return value wasn't that there were no matching rows, but that the updated didn't have any effect on any rows.
So this issue isn't a bug, so much as a misunderstanding of the result. However, the problem still exists when using the update() method of how to determine when the update failed due to no matching rows and when no changes were made.
The solution that I ended up solves this at the cost of a pre-fetch to verify that the update would affect a row. In my case, with a multi-user database application, pre-fetching isn't actually a problem because the row that is to be updated could have been deleted by another user before the update tries to make its change. But, it would be nice if both update methods explained this more clearly, and returned different values: 0 for no affected rows and FALSE for no rows found.
I am inserting records in user_table with user_email as primary key.
$result = mysqli_query($db, "insert into user_table(user_full_name, user_email, user_password) values('".$fullnames."','".$emails."','".$passs."')");
How to know that insert failed due to duplicate primary key, without quering once again to user_table
mysqli_errno(), function can help you,
try this
if (!$result )
{
if(mysqli_errno($con) == 1062){
echo "duplicate entry";
}else{
echo("Error description: " . mysqli_error($con));
}
}
For More error code :https://dev.mysql.com/doc/refman/5.0/en/error-messages-server.html
I can answer my question myself.
after query
I can check
if( mysqli_errno($db) == 1062 ) {
""*DUPLICATE PRIMARY KEY**
}
I can't resolve my problem, this is the error from mysql that I'm getting:
I can edit and update my data when I've got one record in the database but when I add two rows, I get the error.
Some pictures from database
And when I change the row, row ID goes down to 0 and that's is a problem as I can't edit other rows.
CREATE TABLE `dati` (
`ID` int(11) NOT NULL AUTO_INCREMENT,
`title` varchar(255) NOT NULL,
`value1` varchar(255) NOT NULL,
`value2` varchar(255) NOT NULL,
PRIMARY KEY (`ID`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1 PACK_KEYS=1
Update Code:
<?php // Izlabot datus datubāzē!
$titletxt = $_POST['title_edit'];
$value1 = $_POST['value1_edit'];
$value2 = $_POST['value2_edit'];
if(isset($_POST['edit'])){
$con=mysqli_connect("localhost","root","","dbname");
if (mysqli_connect_errno())
{
echo "Neizdevās savienoties ar MySQL: " . mysqli_connect_error();
}
$sql="UPDATE dati SET ID='$ID',title= '$titletxt',value1='$value1',value2='$value2' WHERE 1";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo '<script>
alert(" Ieraksts ir veiksmīgi labots! ");
window.location.href = "index.php";
</script>';
mysqli_close($con);
}
?>
From form:
<?php
$con=mysqli_connect("localhost","root","","dbname");
if (mysqli_connect_errno())
{
echo "Neizdevās savienoties ar MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM dati");
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td><input id='titled' type='text' name='title_edit' value='" . $row['title'] . "'></td>";
echo "<td><input id='value1d' type='text' name='value1_edit' value='" . $row['value1'] . "'></td>";
echo "<td><input id='value2d' type='text' name='value2_edit' value='" . $row['value2'] . "'></td>";
echo "<input type='hidden' name='id' value='" . $row['ID'] . "'>";
echo "<td><button name='edit' id='edit_btn' class='frm_btns' value='" . $row['ID'] . "'>Edit</button></td>";
echo "</tr>";
}
mysqli_close($con);
?>
It couldn't read the value of ID, as 0 was returned.
For those arriving at this question because of the question title (as I did), this solved my problem:
This error can indicate that the table's PRIMARY KEY is not set to AUTO-INCREMENT, (and your insert query did not specify an ID value).
To resolve:
Check that there is a PRIMARY KEY set on your table, and that the PRIMARY KEY is set to AUTO-INCREMENT.
How to add auto-increment to column in mysql database using phpmyadmin?
The error log like (In my case), I'm using Aurora DB:
PHP message: WordPress database error Duplicate entry '0' for key 'PRIMARY' for query INSERT INTO `date173_postmeta
How to fix it using MySQL Workbench:
1- Connect at your DB, and go to the table with the issue, in my case date173_postmeta
2- Select the tools icon:
3- In the windows/tab at right, select the AI checkbox and click on Apply button:
Following the last steps my issues gone.
The problem is that your code attempts to change every row in the data changing the primary key to the value in $ID. This is not set anywhere in your code, and presumably is being cast as 0
$sql="UPDATE `dati` SET `ID`='$ID',`title`=
'$titletxt',`value1`='$value1',`value2`='$value2' WHERE 1";
The primary key value should be sent to the form and returned so it can be processed by your code, but the value should be retained, hence....
$sql="UPDATE `dati` SET `title`=
'$titletxt',`value1`='$value1',`value2`='$value2' WHERE `ID`=$ID";
You should also read up on MySQL injection - even after you've fixed the errors here, anyone can do just about anything they want with your database.
Try this:
ID int(11) PRIMARY KEY AUTOINCREMENT(1,3)
The problem in set ID = $ID
Try removing it so the code should be
$sql="UPDATE `dati` `title`= '$titletxt',`value1`='$value1',`value2`='$value2' WHERE 1";
Be sure to change this where cause it'll update ever row with these values
Just make sure that your primery keys are also A-I.
I'd been struggling to fix this. My tables had auto increment (AI) switched on
Before I started tinkering with records I tried a simple repair in phpMyAdmin.
Go to the SQL tab and run each command in turn.
REPAIR TABLE wp_options
REPAIR TABLE wp_users
REPAIR TABLE wp_usermeta
This did the trick for me and allowed me to login.
i am using phpmyadmin,
so go to db , search for wp_postmeta tabel
add AI(auto-increment) to meta_id
save the changes
I have an insert query (active record style) used to insert the form fields into a MySQL table. I want to get the last auto-incremented id for the insert operation as the return value of my query but I have some problems with it.
Inside the controller:
function add_post(){
$post_data = array(
'id' => '',
'user_id' => '11330',
'content' => $this->input->post('poster_textarea'),
'date_time' => date("Y-m-d H:i:s"),
'status' => '1'
);
return $this->blog_model->add_post($post_data);
}
And inside model:
function add_post($post_data){
$this->db->trans_start();
$this->db->insert('posts',$post_data);
$this->db->trans_complete();
return $this->db->insert_id();
}
I get nothing as the return of the add_post in model
Try this
function add_post($post_data){
$this->db->insert('posts', $post_data);
$insert_id = $this->db->insert_id();
return $insert_id;
}
In case of multiple inserts you could use
$this->db->trans_start();
$this->db->trans_complete();
A transaction isn't needed here, this should suffice:
function add_post($post_data) {
$this->db->insert('posts',$post_data);
return $this->db->insert_id();
}
$id = $this->db->insert_id();
From the documentation:
$this->db->insert_id()
The insert ID number when performing database inserts.
Therefore, you could use something like this:
$lastid = $this->db->insert_id();
Using the mysqli PHP driver, you can't get the insert_id after you commit.
The real solution is this:
function add_post($post_data){
$this->db->trans_begin();
$this->db->insert('posts',$post_data);
$item_id = $this->db->insert_id();
if( $this->db->trans_status() === FALSE )
{
$this->db->trans_rollback();
return( 0 );
}
else
{
$this->db->trans_commit();
return( $item_id );
}
}
Source for code structure: https://codeigniter.com/user_guide/database/transactions.html#running-transactions-manually
It is worth saying that the other answers relate to Codeigniter version 3. The answer in Version 4 (found https://codeigniter.com/user_guide/database/helpers.html) is to use $this->db->insertID()
because you have initiated the Transaction over the data insertion so,
The first check the transaction completed or not. once you start the transaction, it should be committed or rollback according to the status of the transaction;
function add_post($post_data){
$this->db->trans_begin()
$this->db->insert('posts',$post_data);
$this->db->trans_complete();
if ($this->db->trans_status() === FALSE){
$this->db->trans_rollback();
return 0;
}else{
$this->db->trans_commit();
return $this->db->insert_id();
}
}``
in the above, we have committed the data on the successful transaction even you get the timestamp
Just to complete this topic:
If you set up your table with primary key and auto increment you can omit the process of manually incrementing the id.
Check out this example
if (!$CI->db->table_exists(db_prefix() . 'my_table_name')) {
$CI->db->query('CREATE TABLE `' . db_prefix() . "my_table_name` (
`serviceid` int(11) NOT NULL PRIMARY KEY AUTO_INCREMENT,
`name` varchar(64) NOT NULL,
`hash` varchar(32) NOT NULL,
`url` varchar(120) NOT NULL,
`datecreated` datetime NOT NULL,
`active` tinyint(1) NOT NULL DEFAULT '1'
) ENGINE=InnoDB DEFAULT CHARSET=" . $CI->db->char_set . ';');
Now you can insert rows
$this->db->insert(db_prefix(). 'my_table_name', [
'name' => $data['name'],
'hash' => app_generate_hash(),
'url' => $data['url'],
'datecreated' => date('Y-m-d H:i:s'),
'active' => $data['active']
]);
**Inside Model**
function add_info($data){
$this->db->insert('tbl_user_info',$data);
$last_id = $this->db->insert_id();
return $last_id;
}
**Inside Controller**
public function save_user_record() {
$insertId = $this->welcome_model->save_user_info($data);
echo $insertId->id;
}
You must use $lastId = $this->db->insert_id();