PHP语法错误？输出资源 id #3

Question

<?php
$link = mysql_connect('localhost', 'user', 'password');
if (!$link) {
die('Failed to connect to MySQL: ' . mysql_error());
}
$db_selected = mysql_select_db('mysql', $link);
if (!$db_selected) {
  die ("Can\'t use db : " . mysql_error());
}
$query = sprintf("SELECT church_id FROM hours 
         WHERE day_of_week = DATE_FORMAT(NOW(), '%w') AND 
         CURTIME() BETWEEN open_time AND close_time",
    mysql_real_escape_string($day_of_week),
    mysql_real_escape_string($open_time),
    mysql_real_escape_string($close_time));
$result = mysql_query($query);

if (!$result) {
    $message = 'Invalid query: ' .mysql_error() . "\n";
    $message .= 'Whole query: ' .$query;
    die($message);
}

while ($row = mysql_fetch_array($result)) {
    echo $row['shop_id'];
}

mysql_free_result($result);
echo "end";
?>

我知道 SQL 查询通过复制/粘贴到 phpmyadmin 中来工作。我希望脚本只输出一个 shop_id 或一系列 shop_id。现在它输出资源 id #3。我查找了如何修复它，mysql_fetch_array 应该是答案。我做错了什么？

Answer 1

我正在查看您的查询，我只看到您选择了 Church_id 并且您想要输出 shop_id，您应该将其包含在您的选择中，如下所示:

$query = sprintf("SELECT church_id, shop_id FROM hours WHERE day_of_week = DATE_FORMAT(NOW(), '%w') AND CURTIME() BETWEEN open_time AND close_time",
    mysql_real_escape_string($day_of_week),
    mysql_real_escape_string($open_time),
    mysql_real_escape_string($close_time));
$result = mysql_query($query);