这是我的php文件中的一个函数,用于满足我的android应用程序的请求。
function checkin($DB, $TechID, $ClientID, $SiteID){
$dbConnection = mysql_connect($DB['server'], $DB['loginName'], $DB['password']);
if(!$dbConnection){
die('Error! ' . mysql_error());
}
mysql_select_db($DB['database'], $dbConnection);
$file2 = "C:/wamp/www/file2.txt";
$data2 = "ClientID:".$ClientID." TechID:".$TechID." SiteID:".$SiteID;
file_put_contents($file2, $data2);
$result1 = mysql_query("SELECT COUNT(*) FROM Log") or die('Error! ' . mysql_error());
$query = "SELECT `Type` FROM `Log` WHERE `TechID` = '".$TechID."' ORDER BY LogTime DESC LIMIT 1";
$file5 = "C:/wamp/www/file5.txt";
file_put_contents($file5, $query);
$result2 = mysql_query($query) or die('Error! ' . mysql_error());
while($row1 = mysql_fetch_array($result1)){
$count = $row1['COUNT(*)'];
$file3 = "C:/wamp/www/file3.txt";
$data3 = "ClientID:".$ClientID." TechID:".$TechID." SiteID:".$SiteID." Count:".$count;
file_put_contents($file3, $data3);
while($row2 = mysql_fetch_array($result2)){
$file4 = "C:/wamp/www/file4.txt";
$data3 = "ClientID:".$ClientID." TechID:".$TechID." SiteID:".$SiteID." Count:".$count;
file_put_contents($file4, $data3);
/*if($row2['Type']!="Checkin"){
$count = $count+1;
$Time = date('Y/m/d H:i');
mysql_query("INSERT INTO Log (LogID, TechID, ClientID, SiteID, LogTime, Type)
VALUES (".$count.", ".$TechID.", ".$ClientID.", ".$SiteID.", ".$Time.", Checkin)");
}else{
$query2 = "SELECT TechEmail FROM Tech WHERE TechID=".$TechID;
$result3 = mysql_query($query2) or die('Error! ' . mysql_error());
}
}*/
}
}
}
您会看到我已经隐藏了一些代码,因为我正在调试它,所以创建了一些文件只是为了查看代码的哪一部分无法执行。我发现程序无法进入应创建file4的区域。我已经发现问题可能出在$ query上,当它执行时,mysql会响应“未知表状态:TABLE_TYPE”,这我不明白为什么。
慕神8447489