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 — データベース上でクエリを実行する
説明
オブジェクト指向型
手続き型
mysqli_query(mysqli
$mysql, string $query, int $result_mode = MYSQLI_STORE_RESULT): mysqli_result|bool
データベースに対してクエリ query
を実行します。
警告
セキュリティ上の注意: SQLインジェクション
クエリに入力値を含める場合は、プリペアドステートメント を使うべきです。使わない場合、データを適切にフォーマットし、全ての文字列は mysqli_real_escape_string() を使ってエスケープしなければいけません。
DML (INSERT、UPDATE あるいは DELETE) 以外のクエリについては、 この関数は mysqli_real_query() に続けて mysqli_use_result() あるいは mysqli_store_result() をコールすることと同等です。
注意:
サーバーの
max_allowed_packetよりも長いステートメントを mysqli_query() に渡した場合、 返ってくるエラーコードは MySQL Native Driver (mysqlnd) を使っているか MySQL Client Library (libmysqlclient) を使っているかで異なります。 それぞれ、次のように振る舞います。
Linux 上の
mysqlndでは、エラーコード 1153 を返します。 エラーメッセージはgot a packet bigger thanです。max_allowed_packetbytesWindows 上の
mysqlndでは、エラーコード 2006 を返します。 エラーメッセージはserver has gone awayです。すべてのプラットフォームの
libmysqlclientでは、エラーコード 2006 を返します。エラーメッセージはserver has gone awayです。
パラメータ
-
link -
手続き型のみ: mysqli_connect() あるいは mysqli_init() が返す mysqliオブジェクト。
-
query -
クエリ文字列。
-
result_mode -
MySQLサーバーからどのように結果が返されたかを示す結果モード。 以下の3つの定数のうちのいずれかです。
MYSQLI_STORE_RESULT(デフォルト) - バッファリングされた結果セットを mysqli_result オブジェクトとして返します。MYSQLI_USE_RESULT- バッファリングせずに結果セットを mysqli_result オブジェクトとして返します。 取得されるのを待っているレコードが存在する限り、 接続はビジーとなり、Commands out of syncエラーを返します。 このエラーを回避するには、全てのレコードをサーバーから取得するか、 結果セットを mysqli_free_result() 関数をコールすることで捨てなければいけません。MYSQLI_ASYNC(mysqlnd で利用できます) - クエリを非同期実行します。よって結果セットはすぐには返されません。 このようなクエリから結果を取得するには、 mysqli_poll() を使います。MYSQLI_STORE_RESULTやMYSQLI_USE_RESULTと組み合わせて使います。
戻り値
失敗した場合に false を返します。
結果セットを生成するクエリ、
つまり SELECT, SHOW, DESCRIBE あるいは
EXPLAIN が成功した場合は、mysqli_query()
は mysqli_result オブジェクトを返します。それ以外のクエリが成功した場合は、
mysqli_query() は true を返します。
エラー / 例外
mysqli のエラー報告 (MYSQLI_REPORT_ERROR) が有効になっており、かつ要求された操作が失敗した場合は、警告が発生します。さらに、エラー報告のモードが MYSQLI_REPORT_STRICT に設定されていた場合は、mysqli_sql_exception が代わりにスローされます。
例
例1 mysqli::query() の例
オブジェクト指向型
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");
/* 結果セットを返さないテーブルを作成します。*/
$mysqli->query("CREATE TEMPORARY TABLE myCity LIKE City");
printf("Table myCity successfully created.\n");
/* 結果セットを返す SELECT 文 */
$result = $mysqli->query("SELECT Name FROM City LIMIT 10");
printf("Select returned %d rows.\n", $result->num_rows);
/* 大量のデータを取得しなければならない場合、MYSQLI_USE_RESULT を使います */
$result = $mysqli->query("SELECT * FROM City", MYSQLI_USE_RESULT);
/* 注意: 全てのレコードが取得されるか、結果セットがクローズされるまで
サーバーとやりとりを行うあらゆる関数は実行できません。あらゆる呼び出しは
'out of sync' を返します */
$mysqli->query("SET @a:='this will not work'");手続き型
<?php
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$link = mysqli_connect("localhost", "my_user", "my_password", "world");
/* 結果セットを返さないテーブルを作成します。*/
mysqli_query($link, "CREATE TEMPORARY TABLE myCity LIKE City");
printf("Table myCity successfully created.\n");
/* 結果セットを返す SELECT 文 */
$result = mysqli_query($link, "SELECT Name FROM City LIMIT 10");
printf("Select returned %d rows.\n", mysqli_num_rows($result));
/* 大量のデータを取得しなければならない場合、MYSQLI_USE_RESULT を使います */
$result = mysqli_query($link, "SELECT * FROM City", MYSQLI_USE_RESULT);
/* 注意: 全てのレコードが取得されるか、結果セットがクローズされるまで
サーバーとやりとりを行うあらゆる関数は実行できません。あらゆる呼び出しは
'out of sync' を返します */
mysqli_query($link, "SET @a:='this will not work'");上の例の出力は、 たとえば以下のようになります。
Table myCity successfully created. Select returned 10 rows. Fatal error: Uncaught mysqli_sql_exception: Commands out of sync; you can't run this command now in...
参考
- mysqli_execute_query() - SQL文を準備し、変数をバインドし、実行する
- mysqli_real_query() - SQL クエリを実行する
- mysqli_multi_query() - データベース上でひとつ以上のクエリを実行する
- mysqli_prepare() - 実行するための SQL文 を準備する
- mysqli_free_result() - 結果に関連付けられたメモリを開放する
add a note
User Contributed Notes 21 notes
NUNTIUS ¶
16 years ago
theyranos at gmail dot com ¶
13 years ago
The cryptic "Couldn't fetch mysqli" error message can mean any number of things, including:
1. You're trying to use a database object that you've already closed (as noted by ceo at l-i-e dot com). Reopen your database connection, or find the call to <?php mysqli_close($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,$q); )
Hope it helps. Jon
Igor ¶
16 years ago
mysqli::query() can only execute one SQL statement.
Use mysqli::multi_query() when you want to run multiple SQL statements within one query.
thomas dekker ¶
14 years ago
Building inserts can be annoying. This helper function inserts an array into a table, using the key names as column names:
<?php
private function store_array (&$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();
$this->link->next_result();
}
just tried it and it works like a charm ;-)
omidbahrami1990 at gmail dot com ¶
7 years ago
This Is A Secure Way To Use mysqli::query
--------------------------------------------------------
<?php
function secured_query($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.
