mysqli
"Zero indicates that no records where updated for an UPDATE statement, no rows matched the WHERE clause in the query or that no query has yet been executed."
what if i need to know which one of the three occurred? It's dumb that there is no distinction between the three.
mysqli::$affected_rows
mysqli_affected_rows
(PHP 5)
mysqli::$affected_rows -- mysqli_affected_rows — 直前の MySQL の操作で変更された行の数を得る
説明
オブジェクト指向型
手続き型
直近の INSERT、 UPDATE、REPLACE あるいは DELETE クエリにより変更された行の数を返します。
SELECT 文の場合は、mysqli_affected_rows() は mysqli_num_rows() と同じように動作します。
返り値
正の整数が返された場合、それは変更された行数かあるいは取得された行数を 示します。ゼロが返された場合、それは UPDATE 文でレコードが更新されなかったか WHERE 条件に当てはまる行がなかった、またはクエリが実行されなかったことを 示します。-1 は、クエリがエラーを返したことを示します。
注意:
変更された行数が整数型の最大値をこえた場合、結果の行数は 文字列として返されます。
例
例1 $mysqli->affected_rows の例
オブジェクト指向型
<?php
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");
/* 接続状況をチェックします */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
/* 行を挿入します */
$mysqli->query("CREATE TABLE Language SELECT * from CountryLanguage");
printf("Affected rows (INSERT): %d\n", $mysqli->affected_rows);
$mysqli->query("ALTER TABLE Language ADD Status int default 0");
/* 行を更新します */
$mysqli->query("UPDATE Language SET Status=1 WHERE Percentage > 50");
printf("Affected rows (UPDATE): %d\n", $mysqli->affected_rows);
/* 行を削除します */
$mysqli->query("DELETE FROM Language WHERE Percentage < 50");
printf("Affected rows (DELETE): %d\n", $mysqli->affected_rows);
/* すべての行を選択します */
$result = $mysqli->query("SELECT CountryCode FROM Language");
printf("Affected rows (SELECT): %d\n", $mysqli->affected_rows);
$result->close();
/* Language テーブルを削除します */
$mysqli->query("DROP TABLE Language");
/* 接続を閉じます */
$mysqli->close();
?>
手続き型
<?php
$link = mysqli_connect("localhost", "my_user", "my_password", "world");
if (!$link) {
printf("Can't connect to localhost. Error: %s\n", mysqli_connect_error());
exit();
}
/* 行を挿入します */
mysqli_query($link, "CREATE TABLE Language SELECT * from CountryLanguage");
printf("Affected rows (INSERT): %d\n", mysqli_affected_rows($link));
mysqli_query($link, "ALTER TABLE Language ADD Status int default 0");
/* 行を更新します */
mysqli_query($link, "UPDATE Language SET Status=1 WHERE Percentage > 50");
printf("Affected rows (UPDATE): %d\n", mysqli_affected_rows($link));
/* 行を削除します */
mysqli_query($link, "DELETE FROM Language WHERE Percentage < 50");
printf("Affected rows (DELETE): %d\n", mysqli_affected_rows($link));
/* すべての行を選択します */
$result = mysqli_query($link, "SELECT CountryCode FROM Language");
printf("Affected rows (SELECT): %d\n", mysqli_affected_rows($link));
mysqli_free_result($result);
/* Language テーブルを削除します */
mysqli_query($link, "DROP TABLE Language");
/* 接続を閉じます */
mysqli_close($link);
?>
上の例の出力は以下となります。
Affected rows (INSERT): 984 Affected rows (UPDATE): 168 Affected rows (DELETE): 815 Affected rows (SELECT): 169
add a note
User Contributed Notes
mysqli::$affected_rows
ric at xpellshop dot com
10-Jul-2011 11:54
Anonymous
18-Feb-2011 02:50
On "INSERT INTO ON DUPLICATE KEY UPDATE" queries, though one may expect affected_rows to return only 0 or 1 per row on successful queries, it may in fact return 2.
From Mysql manual: "With ON DUPLICATE KEY UPDATE, the affected-rows value per row is 1 if the row is inserted as a new row and 2 if an existing row is updated."
See: http://dev.mysql.com/doc/refman/5.0/en/insert-on-duplicate.html
Here's the sum breakdown _per row_:
+0: a row wasn't updated or inserted (likely because the row already existed, but no field values were actually changed during the UPDATE)
+1: a row was inserted
+2: a row was updated
djnigelharkness at yahoo dot com
12-May-2010 05:42
<?php
$gHostName ="localhost";
$gUserName ="root";
$gPassword ="mypassword";
$gDBName ="dbName";
$mysqli = new mysqli($gHostName, $gUserName, $gPassword,
$gDBName);
$username="username"; $email="user@xxx.edu";
$password="3442f6e94a733237a3e844f0286b92f559bf794d";
//insert
// $stmt = $mysqli->prepare("INSERT INTO j_user
(username,email,password) VALUES (?, ?, ?)");
// $stmt->bind_param('sss',$username, $email, $password);
// $stmt->execute();
// $stmt->close();
//delete
// $stmt = $mysqli->prepare("DELETE FROM j_user WHERE
username=?");
// $stmt->bind_param('s',$username);
// $stmt->execute();
// $stmt->close();
//update
// $stmt = $mysqli->prepare("UPDATE j_user SET email=? WHERE
username='wjw6349'");
// $stmt->bind_param('s',$email);
// $stmt->execute();
// $stmt->close();
if ($mysqli->connect_error){
echo("Connect failed: ".mysqli_connect_error()); exit();
}
$querySelect = "SELECT * FROM j_user";
$resultSet = $mysqli->query($querySelect);
if($resultSet->num_rows > 0){
while($row = $resultSet->fetch_assoc()){
foreach($row as $fieldValue){
$bigString .= "<em>$fieldValue</em><br />n";
}
$bigString .= "<hr />";
}
$mysqli->close();
echo $bigString;
}
function __autoload($className) {
require_once $className.'.class.php';
}
//start the session and create it
session_start();
$_SESSION['name'] = $_POST['username']."_".time();
//encrypt the password
sha1($_POST['password']) == $password
header("Location:admin.php");
//check if the session is set
session_start();
if (isset($_SESSION['name'])) { }
?>
oilpc at oilpc dot com
14-Nov-2009 10:06
For "INSERT" or "UPDATE" statement for modifying data contained in one row of one table I checked if number of affected rows equals 1 to determine success of the operation. It works fine both for errors and false value of WHERE condition (that might be generated according to specific application user acces privileges).
<?php
if ($mysqli->affected_rows==1){
echo "success";
}
else {
echo "fail";
}
?>
Checking if mysqli->affected_rows will equal -1 or not is not a good method of determining success of "INSERT IGNORE" statements. Example: Ignoring duplicate key errors while inserting some rows containing data provided by user only if they will match specified unique constraint causes returning of -1 value by mysqli->affected_rows even if rows were inserted. (checked on MySQL 5.0.85 linux and php 5.2.9-2 windows). However mysqli->sqlstate returns no error if statement was executed successfully.
<?php
if ($mysqli->affected_rows!=-1){
echo "success";// for "INSERT IGNORE" statements will not occur if there were any duplicate key errors ignored during execution of the query
}
else {
echo "fail";// "INSERT IGNORE" statements causing any duplicate key errors (however ignored) lead to mysqli->affected_rows equal -1
}
// Example below works for "INSERT IGNORE" stattements, too
if ($mysqli->sqlstate=="00000"){
echo "success";
}
else {
echo "fail";
}
?>