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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

Descrição

Estilo orientado à objeto

Estilo procedural

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.

Parâmetros

stmt

Somente no estilo procedural: Um recurso statement retornado por mysqli_stmt_init().

Valor Retornado

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.

Exemplos

Exemplo #1 mysqli_stmt_affected_rows() example

Estilo orientado à objeto

<?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);

Estilo procedural

<?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));

Os exemplos acima irão imprimir:

Rows inserted: 17

Veja Também

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User Contributed Notes 2 notes

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25
Carl Olsen
15 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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-9
Chuck
14 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.
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