It appears that an UPDATE prepared statement which contains the same data as that already in the database returns 0 for affected_rows. I was expecting it to return 1, but it must be comparing the input values with the existing values and determining that no UPDATE has occurred.
mysqli_stmt::$affected_rows
mysqli_stmt_affected_rows
(PHP 5, PHP 7)
mysqli_stmt::$affected_rows -- mysqli_stmt_affected_rows — Returns the total number of rows changed, deleted, or inserted by the last executed statement
Descrierea
Stil obiect-orientat
Stil procedural
Returns the number of rows affected by INSERT,
UPDATE, or DELETE query.
This function only works with queries which update a table. In order to get the number of rows from a SELECT query, use mysqli_stmt_num_rows() instead.
Parametri
-
stmt -
Numai stilul procedural: Un identificator al declarației întors de mysqli_stmt_init().
Valorile întoarse
An integer greater than zero indicates the number of rows affected or retrieved. Zero indicates that no records where updated for an UPDATE/DELETE statement, no rows matched the WHERE clause in the query or that no query has yet been executed. -1 indicates that the query has returned an error. NULL indicates an invalid argument was supplied to the function.
Notă:
If the number of affected rows is greater than maximal PHP int value, the number of affected rows will be returned as a string value.
Exemple
Example #1 Stil obiect-orientat
<?php
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
/* create temp table */
$mysqli->query("CREATE TEMPORARY TABLE myCountry LIKE Country");
$query = "INSERT INTO myCountry SELECT * FROM Country WHERE Code LIKE ?";
/* prepare statement */
if ($stmt = $mysqli->prepare($query)) {
/* Bind variable for placeholder */
$code = 'A%';
$stmt->bind_param("s", $code);
/* execute statement */
$stmt->execute();
printf("rows inserted: %d\n", $stmt->affected_rows);
/* close statement */
$stmt->close();
}
/* close connection */
$mysqli->close();
?>
Example #2 Stil procedural
<?php
$link = mysqli_connect("localhost", "my_user", "my_password", "world");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
/* create temp table */
mysqli_query($link, "CREATE TEMPORARY TABLE myCountry LIKE Country");
$query = "INSERT INTO myCountry SELECT * FROM Country WHERE Code LIKE ?";
/* prepare statement */
if ($stmt = mysqli_prepare($link, $query)) {
/* Bind variable for placeholder */
$code = 'A%';
mysqli_stmt_bind_param($stmt, "s", $code);
/* execute statement */
mysqli_stmt_execute($stmt);
printf("rows inserted: %d\n", mysqli_stmt_affected_rows($stmt));
/* close statement */
mysqli_stmt_close($stmt);
}
/* close connection */
mysqli_close($link);
?>
Exemplele de mai sus vor afișa:
rows inserted: 17
A se vedea și
- mysqli_stmt_num_rows() - Return the number of rows in statements result set
- mysqli_prepare() - Prepare an SQL statement for execution
add a note
User Contributed Notes 2 notes
Carl Olsen ¶
18 years ago
Chuck ¶
17 years ago
I'm not sure whether or not this is the intended behavior, but I noticed through testing that if you were to use transactions and prepared statements together and you added a single record to a database using a prepared statement, but later rolled it back, mysqli_stmt_affected_rows will still return 1.
