(PHP 4, PHP 5)
mysql_error — Возвращает текст ошибки последней операции с MySQL
Описание
mysql_error(resource $link_identifier = NULL): string
Список параметров
-
link_identifier -
Соединение MySQL. Если идентификатор соединения не был указан,
используется последнее соединение, открытое mysql_connect(). Если такое соединение не было найдено,
функция попытается создать таковое, как если бы mysql_connect() была вызвана без параметров.
Если соединение не было найдено и не смогло быть создано, генерируется ошибка уровняE_WARNING.
Возвращаемые значения
Возвращает текст ошибки выполнения последней функции MySQL,
или '' (пустую строку), если операция
выполнена успешно.
Примеры
Пример #1 Пример использования mysql_error()
<?php
$link = mysql_connect("localhost", "mysql_user", "mysql_password");mysql_select_db("nonexistentdb", $link);
echo mysql_errno($link) . ": " . mysql_error($link). "n";mysql_select_db("kossu", $link);
mysql_query("SELECT * FROM nonexistenttable", $link);
echo mysql_errno($link) . ": " . mysql_error($link) . "n";
?>
Результатом выполнения данного примера
будет что-то подобное:
1049: Unknown database 'nonexistentdb' 1146: Table 'kossu.nonexistenttable' doesn't exist
aleczapka _at) gmx dot net ¶
18 years ago
If you want to display errors like "Access denied...", when mysql_error() returns "" and mysql_errno() returns 0, use $php_errormsg. This Warning will be stored there. You need to have track_errors set to true in your php.ini.
Note. There is a bug in either documentation about error_reporting() or in mysql_error() function cause manual for mysql_error(), says: "Errors coming back from the MySQL database backend no longer issue warnings." Which is not true.
Florian Sidler ¶
13 years ago
Be aware that if you are using multiple MySQL connections you MUST support the link identifier to the mysql_error() function. Otherwise your error message will be blank.
Just spent a good 30 minutes trying to figure out why i didn't see my SQL errors.
Pendragon Castle ¶
14 years ago
Using a manipulation of josh ><>'s function, I created the following. It's purpose is to use the DB to store errors. It handles both original query, as well as the error log. Included Larry Ullman's escape_data() as well since I use it in q().
<?php
function escape_data($data){
global $dbc;
if(ini_get('magic_quotes_gpc')){
$data=stripslashes($data);
}
return mysql_real_escape_string(trim($data),$dbc);
}
function
q($page,$query){
// $page
$result = mysql_query($query);
if (mysql_errno()) {
$error = "MySQL error ".mysql_errno().": ".mysql_error()."n<br>When executing:<br>n$queryn<br>";
$log = mysql_query("INSERT INTO db_errors (error_page,error_text) VALUES ('$page','".escape_data($error)."')");
}
}
// Run the query using q()
$query = "INSERT INTO names (first, last) VALUES ('myfirst', 'mylast'");
$result = q("Sample Page Title",$query);
?>
l dot poot at twing dot nl ¶
16 years ago
When creating large applications it's quite handy to create a custom function for handling queries. Just include this function in every script. And use db_query(in this example) instead of mysql_query.
This example prompts an error in debugmode (variable $b_debugmode ). An e-mail with the error will be sent to the site operator otherwise.
The script writes a log file in directory ( in this case /log ) as well.
The system is vulnerable when database/query information is prompted to visitors. So be sure to hide this information for visitors anytime.
Regars,
Lennart Poot
http://www.twing.nl
<?php
$b_debugmode = 1; // 0 || 1$system_operator_mail = 'developer@company.com';
$system_from_mail = 'info@mywebsite.com';
function
db_query( $query ){
global $b_debugmode;// Perform Query
$result = mysql_query($query);// Check result
// This shows the actual query sent to MySQL, and the error. Useful for debugging.
if (!$result) {
if($b_debugmode){
$message = '<b>Invalid query:</b><br>' . mysql_error() . '<br><br>';
$message .= '<b>Whole query:</b><br>' . $query . '<br><br>';
die($message);
}raise_error('db_query_error: ' . $message);
}
return $result;
}
function
raise_error( $message ){
global $system_operator_mail, $system_from_mail;$serror=
"Env: " . $_SERVER['SERVER_NAME'] . "rn" .
"timestamp: " . Date('m/d/Y H:i:s') . "rn" .
"script: " . $_SERVER['PHP_SELF'] . "rn" .
"error: " . $message ."rnrn";// open a log file and write error
$fhandle = fopen( '/logs/errors'.date('Ymd').'.txt', 'a' );
if($fhandle){
fwrite( $fhandle, $serror );
fclose(( $fhandle ));
}// e-mail error to system operator
if(!$b_debugmode)
mail($system_operator_mail, 'error: '.$message, $serror, 'From: ' . $system_from_mail );
}?>
Anonymous ¶
18 years ago
My suggested implementation of mysql_error():
$result = mysql_query($query) or die("<b>A fatal MySQL error occured</b>.n<br />Query: " . $query . "<br />nError: (" . mysql_errno() . ") " . mysql_error());
This will print out something like...
A fatal MySQL error occured.
Query: SELECT * FROM table
Error: (err_no) Bla bla bla, you did everything wrong
It's very useful to see your query in order to detect problems with syntax. Most often, the output message from MySQL doesn't let you see enough of the query in the error message to let you see where your query went bad- it a missing quote, comma, or ( or ) could have occured well before the error was detected. I do -not- recomend using this procedure, however, for queries which execute on your site that are not user-specific as it has the potential to leak sensative data. Recomended use is just for debugging/building a script, and for general user-specific queries which would at the worst, leak the users own information to themself.
Good luck,
-Scott
olaf at amen-online dot de ¶
18 years ago
When dealing with user input, make sure that you use
<?php
echo htmlspecialchars (mysql_error ());
?>
instead of
<?php
echo mysql_error ();
?>
Otherwise it might be possible to crack into your system by submitting data that causes the SQL query to fail and that also contains javascript commands.
Would it make sense to change the examples in the documentation for mysql_query () and for mysql_error () accordingly?
Anonymous ¶
22 years ago
some error can't handle. Example:
ERROR 1044: Access denied for user: 'ituser@mail.ramon.intranet' to database 'itcom'
This error ocurrs when a intent of a sql insert of no authorized user. The results: mysql_errno = 0 and the mysql_error = "" .
Gianluigi_Zanettini-MegaLab.it ¶
16 years ago
"Errors coming back from the MySQL database backend no longer issue warnings." Please note, you have an error/bug here. In fact, MySQL 5.1 with PHP 5.2:
Warning: mysql_connect() [function.mysql-connect]: Unknown MySQL server host 'locallllllhost' (11001)
That's a warning, which is not trapped by mysql_error()!
scott at rocketpack dot net ¶
19 years ago
My suggested implementation of mysql_error():
$result = mysql_query($query) or die("<b>A fatal MySQL error occured</b>.n<br />Query: " . $query . "<br />nError: (" . mysql_errno() . ") " . mysql_error());
This will print out something like...
<b>A fatal MySQL error occured</b>.
Query: SELECT * FROM table
Error: (err_no) Bla bla bla, you did everything wrong
It's very useful to see your query in order to detect problems with syntax. Most often, the output message from MySQL doesn't let you see enough of the query in the error message to let you see where your query went bad- it a missing quote, comma, or ( or ) could have occured well before the error was detected. I do -not- recomend using this procedure, however, for queries which execute on your site that are not user-specific as it has the potential to leak sensative data. Recomended use is just for debugging/building a script, and for general user-specific queries which would at the worst, leak the users own information to themself.
Good luck,
-Scott
josh ><> ¶
19 years ago
Oops, the code in my previous post only works for queries that don't return data (INSERT, UPDATE, DELETE, etc.), this updated function should work for all types of queries (using $result = myquery($query);):
function myquery ($query) {
$result = mysql_query($query);
if (mysql_errno())
echo "MySQL error ".mysql_errno().": ".mysql_error()."n<br>When executing:<br>n$queryn<br>";
return $result;
}
phpnet at robzazueta dot com ¶
16 years ago
This is a big one - As of MySQL 4.1 and above, apparently, the way passwords are hashed has changed. PHP 4.x is not compatible with this change, though PHP 5.0 is. I'm still using the 4.x series for various compatibility reasons, so when I set up MySQL 5.0.x on IIS 6.0 running PHP 4.4.4 I was surpised to get this error from mysql_error():
MYSQL: Client does not support authentication protocol requested by server; consider upgrading MySQL client
According to the MySQL site (http://dev.mysql.com/doc/refman/5.0/en/old-client.html) the best fix for this is to use the OLD_PASSWORD() function for your mysql DB user. You can reset it by issuing to MySQL:
Set PASSWORD for 'user'@'host' = OLD_PASSWORD('password');
This saved my hide.
miko_il AT yahoo DOT com ¶
19 years ago
Gianluigi_Zanettini-MegaLab.it ¶
16 years ago
A friend of mine proposed a great solution.
<?php
$old_track = ini_set('track_errors', '1');
.....
if (
$this->db_handle!=FALSE && $db_selection_status!=FALSE)
{
$this->connected=1;
ini_set('track_errors', $old_track);
}
else
{
$this->connected=-1;
$mysql_warning=$php_errormsg;
ini_set('track_errors', $old_track);
throw new mysql_cns_exception(1, $mysql_warning . " " . mysql_error());
}
?>
Gerrit ¶
8 years ago
The following code returns two times the same error, even though I would have expected only one:
$ conn = mysql_connect ('localhost', 'root', '');
$ conn2 = mysql_connect ('localhost', 'root', '');
mysql_select_db ('db1', $ conn);
mysql_select_db ('db2', $ conn2);
$ result = mysql_query ("select 1 from dual", $ conn);
$ result2 = mysql_query ("select 1 from luad", $ conn2);
echo mysql_error ($ conn) "<hr>".
echo mysql_error ($ conn2) "<hr>".
The reason for this is that mysql_connect not working as expected a further connection returns. Since the parameters are equal, a further reference to the previous link is returned. So also changes the second mysql_select_db the selected DB of $conn to 'db2'.
If you change the connection parameters of the second connection to 127.0.0.1, a new connection is returned. In addition to the parameters new_link the mysql_connect() function to be forced.
mysql_error
(PHP 4, PHP 5)
mysql_error — Возвращает текст ошибки последней операции с MySQL
Описание
string mysql_error
([ resource $link_identifier = NULL
] )
Возвращает текст ошибки выполнения последней функции MySQL.
Ошибки работы с MySQL больше не вызывают сообщений в PHP. Вместо
этого используйте функцию mysql_error(), для
получения сообщения об ошибке. Учтите, что функция возвращает текст
ошибки только последней выполненной функции MySQL (исключая
mysql_error() и mysql_errno()),
поэтому убедитесь, что вы вызываете данную функцию до вызова
следующей функции MySQL.
Список параметров
-
link_identifier -
Соединение MySQL. Если идентификатор соединения не был указан,
используется последнее соединение, открытое mysql_connect(). Если такое соединение не было найдено,
функция попытается создать таковое, как если бы mysql_connect() была вызвана без параметров.
Если соединение не было найдено и не смогло быть создано, генерируется ошибка уровняE_WARNING.
Возвращаемые значения
Возвращает текст ошибки выполнения последней функции MySQL,
или » (пустую строку), если операция
выполнена успешно.
Примеры
Пример #1 Пример использования mysql_error()
<?php
$link = mysql_connect("localhost", "mysql_user", "mysql_password");mysql_select_db("nonexistentdb", $link);
echo mysql_errno($link) . ": " . mysql_error($link). "n";mysql_select_db("kossu", $link);
mysql_query("SELECT * FROM nonexistenttable", $link);
echo mysql_errno($link) . ": " . mysql_error($link) . "n";
?>
Результатом выполнения данного примера
будет что-то подобное:
1049: Unknown database 'nonexistentdb' 1146: Table 'kossu.nonexistenttable' doesn't exist
Вернуться к: MySQL
PHP | mysqli_error() Function
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The mysqli_error() function is used to return the error in the most recent MySQL function call that failed. If there are multiple MySQL function calls, the error in the last statement is the one that is pointed out by the function.
Syntax:
mysqli_error("database_name")
Parameters: This function accepts single parameter as mentioned above and described below:
- database_name: It is the database on which operations are being performed. It is a mandatory parameter.
Program 1:
<?php
$conn = mysqli_connect(
"localhost", "root", "", "Persons");
if (mysqli_connect_errno()) {
echo "Database connection failed.";
}
if (!mysqli_query($link, "SET Age=1")) {
printf("Error message: %sn", mysqli_error($conn));
}
mysqli_close($conn);
?>
Suppose the operation is being carried out on the table given below:
The output will be:
Error message: Unknown system variable 'Age'
Program 2:
<?php
$conn = mysqli_connect(
"localhost", "root", "", "Persons");
if (mysqli_connect_errno()) {
echo "Database connection failed.";
}
if (!mysqli_query($link, "SET Firstname='Arkadyuti'")) {
printf("Error message: %sn", mysqli_error());
}
mysqli_close($conn);
?>
Output:
Error message: mysqli_error() expects exactly 1 parameter, 0 given
This example also demonstrates that mysqli_error() needs a database as a parameter.
Last Updated :
23 Apr, 2020
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mysqli::$error
mysqli_error
(PHP 5, PHP 7, PHP 
mysqli::$error —mysqli_error-Возвращает строковое описание последней ошибки
Description
Object-oriented style
Procedural style
mysqli_error(mysqli $mysql): string
Возвращает последнее сообщение об ошибке при последнем вызове функции MySQLi,который может увенчаться успехом или неудачей.
Return Values
Строка,описывающая ошибку.Пустая строка,если ошибка не произошла.
Examples
Пример # 1 $mysqli->error Пример
Object-oriented style
<?php $mysqli = new mysqli("localhost", "my_user", "my_password", "world"); if ($mysqli->connect_errno) { printf("Connect failed: %sn", $mysqli->connect_error); exit(); } if (!$mysqli->query("SET a=1")) { printf("Error message: %sn", $mysqli->error); } $mysqli->close(); ?>
Procedural style
<?php $link = mysqli_connect("localhost", "my_user", "my_password", "world"); if (mysqli_connect_errno()) { printf("Connect failed: %sn", mysqli_connect_error()); exit(); } if (!mysqli_query($link, "SET a=1")) { printf("Error message: %sn", mysqli_error($link)); } mysqli_close($link); ?>
Приведенные выше примеры будут выведены на экран:
Error message: Unknown system variable 'a'
See Also
- mysqli_connect_errno () — возвращает код ошибки из последнего вызова соединения
- mysqli_connect_error () — Возвращает описание последней ошибки подключения
- mysqli_errno () — возвращает код ошибки для последнего вызова функции
- mysqli_sqlstate () — возвращает ошибку SQLSTATE из предыдущей операции MySQL
Really, it drive me crazy when I can’t detect what error that happens. I can’t handle it.
I managed to make a connection to MySQL, and check it out with:
$connection = mysqli_connect(HOST, USER, PASS, DB) or die('Could not connect!');
if($connection){
echo 'It's connected!';
}
Yeah, that say connected. Then, when I try a query, it fails without error reporting. I’ve tried do this to check if it fails:
$query = "SELECT $field FROM users WHERE id = ".$_SESSION['user_id'];
$result = mysqli_query($dbc, $query);
if($result){
echo 'Query OK';
}else{
echo 'Query failed';
}
The browser said: Query failed. So, there’s an error in my query. Then I echoed the query out with this:
echo $query;
// Printed in the browser: SELECT firstname FROM users WHERE id = 1
Copy that value and use it in phpMyAdmin. It works. So, i guess an error occured in mysqli_query function. But i can’t get the error message and so i don’t know what’s going on. I’ve tried this:
$result = mysqli_query($dbc, $query) or die(mysqli_error($dbc));
and this:
if(!$result){
echo mysqli_error($dbc);
}
Nothing happens. The browser just blank. Then, I tried to change this:
$result = mysqli_query($dbc, $query);
to this:
$result = mysqli_query($query);
Still nothing happens. What’s going on? How can I know what error occured?
I run the server in Debian with phpinfo(): PHP Version 5.4.36-0+deb7u3

