PHP 8.1.0 Alpha 1 available for testing

mysqli_stmt::$affected_rows

mysqli_stmt_affected_rows

(PHP 5, PHP 7)

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

说明

面向对象风格

过程化风格

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.

参数

stmt

仅以过程化样式:由 mysqli_stmt_init() 返回的 statement 标识。

返回值

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().

注意:

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.

范例

示例 #1 mysqli_stmt_affected_rows() example

面向对象风格

<?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);

过程化风格

<?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));

以上例程会输出:

Rows inserted: 17

参见

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User Contributed Notes 2 notes

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26
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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-8
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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