我有一個PHP陣列介紹如下: -PHP陣列呈現
<?php
$ads = array();
$ads [] = array(
'name' => 'Apple',
'duration' => '3',
'price' => "$5"
);
$ads [] = array(
'name' => 'Orange',
'duration' => '2',
'price' => "$10"
);
$ads [] = array(
'name' => 'Banana',
'duration' => '5',
'price' => "$6"
);
,然後,我想從數據庫替換動態數據的靜態數據: -
$sql = "SELECT * from tb_fruit order by fruit_id ASC";
$result = mysql_query($sql_approve, $conn_fruit);
while($record = mysql_fetch_array($result))
{
$fruit_id = $record['fruit_id'];
$fruit_name = $record['fruit_name '];
$fruit_price= $record['fruit_price'];
$fruit_duration= $record_approve['fruit_duration'];
}
其實,我應該如何將兩個演示文稿結合在一起?謝謝!
這是否應該像一個「默認數據集」,以防萬一您的表沒有您所謂的「動態數據」? – Blake 2012-04-20 04:56:49
**請停止使用古老的'mysql_ *'函數編寫新代碼**不再維護它們並且社區已經開始[棄用過程](http://news.php.net/php.internals/53799) 。相反,您應該瞭解'準備好的語句'並使用[PDO](http://php.net/pdo)或[MySQLi](http://php.net/mysqli)。 – 2012-04-20 05:07:49