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::$affected_rows
mysqli_stmt_affected_rows
(PHP 5, PHP 7, PHP 8)
mysqli_stmt::$affected_rows -- mysqli_stmt_affected_rows — Returns 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
Returns the number of rows affected by INSERT,
UPDATE, or DELETE query.
Works like mysqli_stmt_num_rows() for
SELECT statements.
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.
示例
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));以上示例会输出：
Rows inserted: 17
参见
- mysqli_stmt_num_rows() - Returns the number of rows fetched from the server
- mysqli_stmt_store_result() - Stores a result set in an internal buffer
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.
