mysqli_stmt::$affected_rows

mysqli_stmt_affected_rows

(PHP 5, PHP 7, PHP 8)

mysqli_stmt::$affected_rows -- mysqli_stmt_affected_rowsReturns the total number of rows changed, deleted, inserted, or matched by the last statement executed

Descrizione

Stile orientato agli oggetti

Stile procedurale

mysqli_stmt_affected_rows(mysqli_stmt $statement): int|string

Returns the number of rows affected by INSERT, UPDATE, or DELETE query. Works like mysqli_stmt_num_rows() for SELECT statements.

Elenco dei parametri

stmt

Solo nello stile procedurale: un identificatore di statement restituito da mysqli_stmt_init().

Valori restituiti

An integer greater than zero indicates the number of rows affected or retrieved. Zero indicates that no records were updated for an UPDATE statement, no rows matched the WHERE clause in the query or that no query has yet been executed. -1 indicates that the query returned an error or that, for a SELECT query, mysqli_stmt_affected_rows() was called prior to calling mysqli_stmt_store_result().

Nota:

If the number of affected rows is greater than maximum PHP int value, the number of affected rows will be returned as a string value.

Esempi

Example #1 mysqli_stmt_affected_rows() example

Stile orientato agli oggetti

<?php

mysqli_report
(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");

/* create temp table */
$mysqli->query("CREATE TEMPORARY TABLE myCountry LIKE Country");

$query = "INSERT INTO myCountry SELECT * FROM Country WHERE Code LIKE ?";

/* prepare statement */
$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);

Stile procedurale

<?php

mysqli_report
(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$link = mysqli_connect("localhost", "my_user", "my_password", "world");

/* create temp table */
mysqli_query($link, "CREATE TEMPORARY TABLE myCountry LIKE Country");

$query = "INSERT INTO myCountry SELECT * FROM Country WHERE Code LIKE ?";

/* prepare statement */
$stmt = mysqli_prepare($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));

I precedenti esempi visualizzeranno:

Rows inserted: 17

Vedere anche:

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User Contributed Notes 1 note

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Carl Olsen
18 years ago
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.
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