Question

我试图从postgresql数据库中获取数据，我收到错误：pg_last_error（）期望参数1是资源，在/path/to/query.php中给出的对象所以数据作为对象而不是资源。任何想法如何解决这一问题？

SQL使用此代码：

foreach ($conn->query($sql1) as $row) 
{
print $row["Site_ID"] . " ";
print $row["Site_name_1"] . "<br /> ";

}

但是当我使用pg_query代替时，问题就出现了这是我的代码：

<?php include 'header.php'; ?> 
<div class='container'> 
<?php include 'menu.php'; ?>
<?php include 'PDO_connect.php'; ?>

<?php

$sql1='SELECT "Site_ID", "Site_name_1" FROM "Sites" ORDER BY "Sites"."Site_ID" ASC'; 

$result1 = pg_query($conn,$sql1);
if(!$result1) {
    echo "There is an error!";
    echo pg_last_error($conn);
}
?>

我的连接信息

<?php
try {
$dbuser = 'usr';
$dbpass = 'pwd';
$host = "localhost";
$dbname="db";

$conn = new PDO('pgsql:host=localhost;dbname=db', $dbuser, $dbpass);
}catch (PDOException $e) {
echo "Error : " . $e->getMessage() . "<br/>";
die();
}
?>

Answer 1

正如评论中所说，我正在混合PDO和pg_connect，这解决了我的问题：

<?php
$servername = "localhost"; 
$username = "usr"; 
$password = "pwd";
$database = "db";
?>

<?php
    $conn = pg_connect("host=localhost dbname=$database user=$username password=$password")or die("Can't connect to database".pg_last_error());
?>

pg_query抛出错误

1 个答案: