<?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查詢通過複製/粘貼到phpmyadmin中起作用。我希望腳本只輸出一個shop_id或一系列shop_ids。現在它輸出資源ID#3。我查找了如何修復它,並且mysql_fetch_array應該是答案。我究竟做錯了什麼?PHP語法錯誤?輸出資源ID#3
你應該張貼,你所看到的,看起來錯誤 - 崩潰的結果,輸出等。還請注意** MySQL是過時,新的代碼應寫入的mysqli ** – RobP
...或PDO。請參閱[爲什麼我不應該在PHP中使用mysql_ *函數](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php) – Phil
1) $ day_of_week'從哪裏來? 2)你的'sprintf'字符串只有一個佔位符,但你傳遞了三個值? 3)你'選擇church_id',但試圖回顯'shop_id' 4)上述代碼無法輸出。在任何時候您都不會顯示任何可能成爲資源的信息 – Phil