mysqli_num_rows fails with prepared statement , procedural style

问题

I am learning Mysqli ( come from Mysql ).

I as far as I understood, the advantage of using prepared statements is to prevent SQL injections.

I managed to build queries with prepared statments using SELECT and INSERT.

But to achieve the equivalent of select count() , and I'm banging my head against the wall.
The PHP manual gives:

if ($result = mysqli_query($link, "SELECT Code, Name FROM Country ORDER BY Name")) {

    /* determine number of rows result set */
    $row_cnt = mysqli_num_rows($result);

    printf("Result set has %d rows.\n", $row_cnt);

    /* close result set */
    mysqli_free_result($result);
}

I trying to do this with prepared statement too. But maybe I shoudn't ?

This is what I'm trying:

$boy = 'yes';
$age   = 1;

$result = mysqli_prepare ($bdd, 'SELECT boy , age FROM photo WHERE  boy = ? AND age= ?' );

mysqli_stmt_bind_param( $result, "si", $boy , $age );
mysqli_stmt_execute( $result );

    $row_cnt = mysqli_num_rows( $result );
    printf( "Le jeu de résultats a %d lignes.\n", $row_cnt );

But I always get the same type of error whatever I'am trying

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, object given in C:\wamp\www\page.com\pic.php on line 36

Thanks for your help.

回答1:

I think you're looking for mysqli_stmt_num_rows in combination with mysqli_stmt_store_result - http://www.php.net/manual/en/mysqli-result.num-rows.php

<?php
$boy = 'yes';
$age   = 1;

$result = mysqli_prepare ($bdd, 'SELECT boy , age FROM photo WHERE  boy = ? AND age= ?' );

mysqli_stmt_bind_param( $result, "si", $boy , $age );
mysqli_stmt_execute( $result );

// You may need this too...
mysqli_stmt_store_result( $result );

$row_cnt = mysqli_stmt_num_rows( $result );
printf( "Le jeu de résultats a %d lignes.\n", $row_cnt );
?>

来源：https://stackoverflow.com/questions/14992356/mysqli-num-rows-fails-with-prepared-statement-procedural-style

标签

mysqli

prepared-statement