【发布时间】:2016-05-28 05:00:19
【问题描述】:
当我在浏览器中调用 archive.php 时出现错误: 警告:mysqli_query() 期望参数 2 是字符串,对象在...中给出。
<?php
$connection = new mysqli('localhost','root','', 'scale');
if (mysqli_connect_errno()) {
printf("Error de conexión: %s\n", mysqli_connect_error());
exit();
}
$stmt = $connection->prepare(' SELECT
R.cod_region, P.region, P.comuna, COUNT(DISTINCT(P.id)) AS sitios, COUNT(DISTINCT(A.TEXTO)) AS alarmas
FROM region R, pop P, sitios_pop SP, sitios S
LEFT JOIN log_alarm_2g A ON S.RSITE = A.RSITE AND A.CLASE = "A1"
AND A.INICIO >= "2016-02-14"
WHERE
R.cod_region = ? AND
R.region = P.region AND
S.ESTADO = "OPERATIVO" AND
S.SITIO = SP.cod_sitio AND
SP.id_pop = P.id
GROUP BY
P.region,
P.comuna
ORDER BY
R.cod_region ');
$stmt->bind_param('i', $cod_region);
$resultset = mysqli_query($connection, $stmt);
$records= array();
while($r = mysqli_fetch_assoc($resultset)){
$records[] = $r;
}
echo json_encode($records);
?>
【问题讨论】:
-
似乎您的数据库连接是面向对象的,并且您在面向过程中执行查询,所以如果您在完整的代码中只遵循其中一个,您就不会遇到此类错误。
-
@rjgodoy,使用
$stmt->execute();而不是mysqli_query($connection, $stmt); -
@RohitGoyani 已经在下面回答了!!
标签: php mysqli parameters bindparam