dongmeng0317 2014-04-15 03:04
浏览 42

PHP语法错了吗? 输出资源ID#3

<?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() . "
";
    $message .= 'Whole query: ' .$query;
    die($message);
}

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

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

I know SQL query works by copying/pasting into phpmyadmin. I want the script to just output a shop_id or series of shop_ids. Right now it outputs Resource id #3. I looked up how to fix it and mysql_fetch_array was supposed to be the answer. What am I doing wrong?

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2条回答 默认 最新

  • dqxmf02844 2014-04-15 03:11
    关注

    I'm looking over your query and I only see you selecting church_id and you want to output shop_id, you should include that in your select like so:

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

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