This may or may not be obvious to people but perhaps it will help someone.
When running joins in SQL you may encounter a problem if you are trying to pull two columns with the same name. mysqli returns the last in the query when called by name. So to get what you need you can use an alias.
Below I am trying to join a user id with a user role. in the first table (tbl_usr), role is a number and in the second is a text name (tbl_memrole is a lookup table). If I call them both as role I get the text as it is the last "role" in the query. If I use an alias then I get both as desired as shown below.
<?php
$sql = "SELECT a.uid, a.role AS roleid, b.role,
FROM tbl_usr a
INNER JOIN tbl_memrole b
ON a.role = b.id
";
if ($result = $mysqli->query($sql)) {
while($obj = $result->fetch_object()){
$line.=$obj->uid;
$line.=$obj->role;
$line.=$obj->roleid;
}
}
$result->close();
unset($obj);
unset($sql);
unset($query);
?>
In this situation I guess I could have just renamed the role column in the first table roleid and that would have taken care of it, but it was a learning experience.
mysqli::query
mysqli_query
(PHP 5, PHP 7, PHP 8)
mysqli::query -- mysqli_query — Выполняет запрос к базе данных
Описание
Объектно-ориентированный стиль
Процедурный стиль
mysqli_query(mysqli
$mysql, string $query, int $result_mode = MYSQLI_STORE_RESULT): mysqli_result|bool
Выполняет запрос query к базе данных.
Внимание
Предупреждение безопасности: SQL-инъекция
Вместо составления строки запроса с включением переменных значений необходимо подготавливать запросы. Либо строки запроса должны быть экранированы функцией mysqli_real_escape_string() и правильно отформатированы.
Для не DML-запросов (не INSERT, UPDATE или DELETE), эта функция равносильна вызову функции mysqli_real_query(), а затем mysqli_use_result() или mysqli_store_result().
Замечание:
Если длина передаваемого в функцию mysqli_query() выражения больше значения системной переменной
max_allowed_packetсервера, будет сгенерирована ошибка. MySQL-драйвер (mysqlnd) и клиентская библиотека MySQL (libmysqlclient) выбрасывают разные коды ошибок. Поведение функции будет следующим:
mysqlndна платформе Linux возвращает код ошибки 1153. Сообщение об ошибке означаетразмер пакета превышает.max_allowed_packetбайт
mysqlndна платформе Windows возвращает код ошибки 2006. Это сообщение об ошибке означаетотказ сервера.
libmysqlclientна всех платформах возвращает код ошибки 2006. Это сообщение об ошибке означаетотказ сервера.
Список параметров
-
mysql -
Только для процедурного стиля: объект mysqli, который вернула функция mysqli_connect() или функция mysqli_init().
-
query -
Текст запроса.
-
result_mode -
Режим результата может быть одной из 3 констант, указывающих, как результат будет возвращён сервером MySQL.
MYSQLI_STORE_RESULT(по умолчанию) — возвращает объект mysqli_result с буферизованным набором результатов.MYSQLI_USE_RESULT— возвращает объект mysqli_result с небуферизованным набором результатов. Пока есть отложенные записи, ожидающие выборки, линия соединения будет занята и все последующие вызовы будут возвращать ошибкуCommands out of sync. Чтобы избежать ошибки, все записи должны быть получены с сервера или набор результатов должен быть отброшен путём вызова mysqli_free_result().MYSQLI_ASYNC(доступно с mysqlnd) — запрос выполняется асинхронно, набор результатов сразу не возвращается. Затем вызывают функцию mysqli_poll() для получения результатов по этим запросам. Можно использовать в сочетании с константойMYSQLI_STORE_RESULTилиMYSQLI_USE_RESULT.
Возвращаемые значения
Возвращает false в случае возникновения ошибки. В случае успешного выполнения запросов,
которые создают набор результатов, таких как SELECT, SHOW, DESCRIBE или
EXPLAIN, функция mysqli_query() вернёт объект mysqli_result.
Для остальных успешных запросов mysqli_query() вернёт true.
Ошибки
Если уведомления об ошибках mysqli включены (MYSQLI_REPORT_ERROR) и запрошенная операция не удалась,
выдаётся предупреждение. Если, кроме того, установлен режим MYSQLI_REPORT_STRICT,
вместо этого будет выброшено исключение mysqli_sql_exception.
Примеры
Пример #1 Пример использования mysqli::query()
Объектно-ориентированный стиль
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");
/* Создание таблицы, не возвращает набор результатов */
$mysqli->query("CREATE TEMPORARY TABLE myCity LIKE City");
printf("Таблица myCity создана.\n");
/* Запросы SELECT, возвращают набор результатов */
$result = $mysqli->query("SELECT Name FROM City LIMIT 10");
printf("Запрос SELECT вернул %d строк.\n", $result->num_rows);
/* Если нужно извлечь большой объем данных, используем MYSQLI_USE_RESULT */
$result = $mysqli->query("SELECT * FROM City", MYSQLI_USE_RESULT);
/* Важно заметить, что мы не можем вызывать функции, которые взаимодействуют
с сервером, пока не закроем набор результатов. Все подобные вызовы
будут вызывать ошибку 'out of sync' */
$mysqli->query("SET @a:='this will not work'");Процедурный стиль
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$link = mysqli_connect("localhost", "my_user", "my_password", "world");
/* Создание таблицы, не возвращает набор результатов */
mysqli_query($link, "CREATE TEMPORARY TABLE myCity LIKE City");
printf("Таблица myCity создана.\n");
/* Запросы SELECT, возвращают набор результатов */
$result = mysqli_query($link, "SELECT Name FROM City LIMIT 10");
printf("Запрос SELECT вернул %d строк.\n", mysqli_num_rows($result));
/* Если нужно извлечь большой объем данных, используем MYSQLI_USE_RESULT */
$result = mysqli_query($link, "SELECT * FROM City", MYSQLI_USE_RESULT);
/* Важно заметить, что мы не можем вызывать функции, которые взаимодействуют
с сервером, пока не закроем набор результатов. Все подобные вызовы
будут вызывать ошибку 'out of sync' */
mysqli_query($link, "SET @a:='this will not work'");Результат выполнения приведённых примеров:
Таблица myCity создана. Запрос SELECT вернул 10 строк. Fatal error: Uncaught mysqli_sql_exception: Commands out of sync; you can't run this command now in...
Смотрите также
- mysqli_execute_query() - Подготавливает, связывает параметры и выполняет SQL-запрос
- mysqli_real_query() - Выполнение SQL запроса
- mysqli_multi_query() - Выполняет один или несколько запросов к базе данных
- mysqli_prepare() - Подготавливает SQL выражение к выполнению
- mysqli_free_result() - Освобождает память, занятую результатами запроса
add a note
User Contributed Notes 21 notes
NUNTIUS ¶
16 years ago
theyranos at gmail dot com ¶
13 years ago
The cryptic "Couldn't fetch mysqli" error message can mean any number of things, including:
1. You're trying to use a database object that you've already closed (as noted by ceo at l-i-e dot com). Reopen your database connection, or find the call to <?php mysqli_close($db); ?> or <?php $db->close(); ?> and remove it.
2. Your MySQLi object has been serialized and unserialized for some reason. Define a wakeup function to re-create your database connection. http://php.net/__wakeup
3. Something besides you closed your mysqli connection (in particular, see http://bugs.php.net/bug.php?id=33772)
4. You mixed OOP and functional calls to the database object. (So, you have <?php $db->query() ?> in the same program as <?php mysqli_query($db) ?>).
petrus.jvr ¶
13 years ago
When calling multiple stored procedures, you can run into the following error: "Commands out of sync; you can't run this command now".
This can happen even when using the close() function on the result object between calls.
To fix the problem, remember to call the next_result() function on the mysqli object after each stored procedure call. See example below:
<?php
// New Connection
$db = new mysqli('localhost','user','pass','database');
// Check for errors
if(mysqli_connect_errno()){
echo mysqli_connect_error();
}
// 1st Query
$result = $db->query("call getUsers()");
if($result){
// Cycle through results
while ($row = $result->fetch_object()){
$user_arr[] = $row;
}
// Free result set
$result->close();
$db->next_result();
}
// 2nd Query
$result = $db->query("call getGroups()");
if($result){
// Cycle through results
while ($row = $result->fetch_object()){
$group_arr[] = $row;
}
// Free result set
$result->close();
$db->next_result();
}
else echo($db->error);
// Close connection
$db->close();
?>
xa at sagexa dot com ¶
2 years ago
"In PHP 8.1, the default error handling behavior of the MySQLi extension has changed from silencing errors to throw an Exception on errors. "
This is true even for the procedural use of mysqli, i.e. mysqli_query.
Hence, using if(!mysqli_query($c, $sql)) [...] is pointless.
To disable Exception throwing : mysqli_report(MYSQLI_REPORT_OFF);
Anonymous ¶
6 years ago
Here is an example of a clean query into a html table
<table>
<tr>
<th>First Name</th>
<th>Last Name</th>
<th>City</th>
</tr>
<?php while ($row = $myquery->fetch_assoc()) { ?>
<tr>
<td><?php echo $row["firstname"]; ?></td>
<td><?php echo $row["lastname"]; ?></td>
<td><?php echo $row["city"];?></td>
</tr>
<?php } ?>
</table>
registrations at jdfoxmicro dot com ¶
14 years ago
I like to save the query itself in a log file, so that I don't have to worry about whether the site is live.
For example, I might have a global function:
<?php
function UpdateLog ( $string , $logfile ) {
$fh = fopen ( $logfile , 'a' );
$fwrite ( $fh , strftime ('%F %T %z')." ".$string."\n";
fclose ( $fh );
}
?>
Then in my mysql function error trapper, something like this:
<?php
$error_msg = "Database error in [page].php / ";
$error_msg .= mysqli_error ( $link )." / ";
$error_msg .= $query;
UpdateLog ( $error_msg , DB_ERROR_LOG_FILE );
?>
I also include the remote IP, user agent, etc., but I left it out of these code samples. And have it e-mail me when an error is caught, too.
Jeff
ceo at l-i-e dot com ¶
16 years ago
Translation:
"Couldn't fetch mysqli"
You closed your connection and are trying to use it again.
It has taken me DAYS to figure out what this obscure error message means...
Beeners ¶
18 years ago
Stored Procedures.
Use mysqli_query to call a stored procedure that returns a result set.
Here is a short example:
<?php
$mysqli = new mysqli(DBURI,DBUSER,DBPASS,DBNAME);
if (mysqli_connect_errno())
{
printf("Connection failed: %s\n", mysqli_connect_error());
exit();
}
$SQL = "CALL my_procedure($something)";
if ( ($result = $mysqli->query($SQL))===false )
{
printf("Invalid query: %s\nWhole query: %s\n", $mysqli->error, $SQL);
exit();
}
while ($myrow = $result->fetch_array(MYSQLI_ASSOC))
{
$aValue[]=$myrow["a"];
$bValue[]=$myrow["b"];
}
$result->close();
$mysqli->close();
?>
I hope this saves someone some time.
info at ff dot net ¶
18 years ago
Calling Stored Procedures
Beeners' note/example will not work. Use mysqli_multi_query() to call a Stored Procedure. SP's have a second result-set which contains the status: 'OK' or 'ERR'. Using mysqli_query will not work, as there are multiple results.
<?php
$sQuery="CALL SomeSP('params')";
if(!mysqli_multi_query($sqlLink,$sQuery)) {
// your error handler
}
$sqlResult=mysqli_store_result($sqlLink);
if(mysqli_more_results($this->sqlLink))//Catch 'OK'/'ERR'
while(mysqli_next_result($this->sqlLink));
?>
You will have to rewrite/expand this a bit for more usability of course, but it's just an example.
joseph_robert_martinez at yahoo dot com ¶
10 years ago
For those using with replication enabled on their servers, add a mysqli_select_db() statement before any data modification queries. MySQL replication does not handle statements with db.table the same and will not replicate to the slaves if a scheme is not selected before.
Found out after days of resetting master and slaves on another site http://www.linuxquestions.org/questions/linux-server-73/mysql-5-1-not-replicating-properly-with-slave-823296/
hunreal at gmail dot com ¶
20 years ago
Use difference collation/character for connect, result.
You can set the collation before your query.
E.g. want to set the collation to utf8_general_ci
you can send the query "SET NAMES 'utf8'" first
<?php
$mysqli=new mysqli('localhost', 'root', 'password', 'test');
$mysqli->query("SET NAMES 'utf8'");
$q=$mysqli->query("select * from test");
while($r=$q->fetch_assoc()) {
print_r($r);
}
?>
There are many variables about character settings.
By running sql command, SHOW VARIABLES LIKE 'char%';
There are some variables control the character usage.
character_set_client
character_set_connection
character_set_database
character_set_results
character_set_server
character_set_system
Also SET NAMES can repalce with one or some settings like SET character_set_results='utf8';
jcwebb at dicoe dot com ¶
17 years ago
When building apps, i like to see the whole statement when if fails.
<?php
$q="SELECT somecolumn FROM sometable"; //some instruction
$r=mysqli_query($DBlink,$q) or die(mysqli_error($DBlink)." Q=".$q);
?>
If theres an error (like my numerous typing mistakes) this shows the entire instruction.
Good for development (not so good on production servers - simply find and replace when finished: $r=mysqli_query($DBlink,$q); )
Hope it helps. Jon
Igor ¶
15 years ago
mysqli::query() can only execute one SQL statement.
Use mysqli::multi_query() when you want to run multiple SQL statements within one query.
thomas dekker ¶
14 years ago
Building inserts can be annoying. This helper function inserts an array into a table, using the key names as column names:
<?php
private function store_array (&$data, $table, $mysqli)
{
$cols = implode(',', array_keys($data));
foreach (array_values($data) as $value)
{
isset($vals) ? $vals .= ',' : $vals = '';
$vals .= '\''.$this->mysql->real_escape_string($value).'\'';
}
$mysqli->real_query('INSERT INTO '.$table.' ('.$cols.') VALUES ('.$vals.')');
}
?>
Adapt it to your specific needs.
ahmed dot 3abdolah at gmail dot com ¶
10 years ago
Hi, i created function that add a new table using array , i work with it on my projects ...
<?PHP
/* this function was learned from PHP.net */
function array_keys_exist(&$key,array &$array){
$keys = split("\|",$key);
foreach($keys as $key_s){
if(array_key_exists($key_s, $array)) return true;
}
return false;
}
/*and this is my function */
array_create_table(array &$array){
if(is_array($array)){
$key = "table|rows|values";
$info = "";
if(array_keys_exist($key,$array)){
if(is_array($array["rows"]) and is_array($array["values"]) ){
if(count($array["rows"]) == count($array["values"])) {
for($i=0; $i<=count($array["rows"]); $i++){
$info = $info." ".$array["rows"][$i]." ".$array["values"][$i]." NOT NULL ";
if($i < count($array["rows"])-1 ) $info = $info.",";
}
$query = "CREATE TABLE ".$this->private_tables_name.$array["table"]." ";
$query .= "( id INT NOT NULL AUTO_INCREMENT PRIMARY KEY, ".$info." )";
return $query;
}
}
}else return "Error";
}
}
/* the use is simple */
$database = new database(); // connection to database i used mysqli ...
$array = array("table"=>"MRO", "rows"=>array("name","username") , "values" => array("VARCHAR (50) "," VARCHAR (50) ") );
$query = array_create_table($array); // convert and create the query ...
if($database->query($query)) echo "Work"; else echo "Error"; // result : work
?>
marcus at synchromedia dot co dot uk ¶
13 years ago
The exact type returned by a successful query is mysqli_result.
popere dot noel at yahoo dot com ¶
11 years ago
or you could just extend the class...
in my case i already had a wraper for the db so something like this was easy :
public function free($result) {
$result->free_result();
$this->link->next_result();
}
just tried it and it works like a charm ;-)
omidbahrami1990 at gmail dot com ¶
6 years ago
This Is A Secure Way To Use mysqli::query
--------------------------------------------------------
<?php
function secured_query($sql)
{
$connection = new mysqli($host,$username,$password,$name);
if ($connection->connect_error)
die("Secured");
$result=$connection->query($sql);
if($result==FALSE)
die("Secured");
$connection->close();
return $result;
}
/*
$host ---> DataBase IP Address
$username ---> DataBase Username
$password ---> DataBase Password
$name ---> DataBase Name
*/
?>
blinki bill, argodak at yahoo dot com ¶
11 years ago
Recently I had puzzling problem when performing DML queries, update in particular, each time a update query is called and then there are some more queries to follow this error will show on the page and go in the error_log:
"Fatal error: Exception thrown without a stack frame in Unknown on line 0"
The strange thing is that all queries go through just fine so it didn't make much sense:
$update = mysqli_query($connection, $query_string);
if(!$update){
echo 'Houston we have a problem '.mysqli_error($connection);
exit;
}
In the above example "$update" is "true", mysqli_error() is empty and of course the update operation goes through, however the nasty super cryptic error appears on the page.
What makes even less sense to me is how I fixed it - just called "mysqli_free_result" after the update query and the problem was gone, however because mysqli_free_result is not supposed to be called after DML queries (to free what, a boolean? lol) it needs to be wrapped in a try catch block:
try{
mysqli_free_result($update);
}catch (Exception $e){
//do nothing
}
So, I don't know why but it seems that when DML queries are responsible for:
"Fatal error: Exception thrown without a stack frame in Unknown on line 0"
calling "mysqli_free_result" after the query seems to be fixing the issue
David Marcus ¶
7 years ago
If you use the default resultmode of MYSQLI_STORE_RESULT, you can call $mysqli->close() right after $mysqli->query and before you use mysqli_result. This reduces the time the connection is open, which can help if the database server has a limit on how many connections there can be.
