好吧,我做這一個,但我有83000字的mysql數據庫,當我執行這個腳本,它會花費太多的時間和一些時間,它不運行。我認爲這個腳本匹配mysql數據庫中的每個標題,因此它在$ row ['full_story']中或者不在。所以如果有任何方法可以使這個過程更快,這使得操作無法使用?或者它只是匹配那些使用的標題$ row ['full_story']代碼是低於在php和mysql中的翻譯
$user_name = "root";
$password = "";
$database = "salar";
$server = "127.0.0.1";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "SET NAMES 'utf8'";
mysql_query($SQL);
$SQL = "SELECT * FROM dle_mylinks ORDER BY LENGTH(title) DESC";
$result = mysql_query($SQL);
while ($db_field = mysql_fetch_assoc($result)) {
$row['full_story'] = str_replace ($db_field['title'],"<a href=\"?newsid=" . $db_field['id'] . "\">" . $db_field['title'] . "</a>" ,$row['full_story']);
$row['short_story'] = str_replace ($db_field['title'],"<a href=\"?newsid=" . $db_field['id'] . "\">" . $db_field['title'] . "</a>" ,$row['short_story']);
}
$mydata =$row['short_story'] . $row['full_story'];
mysql_close($db_handle);
}
else {
print "Database NOT Found ";
mysql_close($db_handle);
}
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