$class_arr=array();
$sql = "select * from `category` order by sort asc, classid desc";
$query = $mysql -> query($sql);
while($row = $mysql -> fetch_array($query)){
echo ($row['classid'].$row['classname'].$row['parentid'].$row['sort']);
$class_arr[] = array($row['classid'],$row['classname'],$row['parentid'],$row['sort']);
}如上echo ($row['classid'].$row['classname'].$row['parentid'].$row['sort']);为什么没有数出预想的内容?
也没有错误提示输出的内容为空,sql语句放在phpmyadmin里正常执行。php是否有方便点的调试工具得知此错误,刚做php几天对此不熟悉,请把我当做小学生看待吧!
--是不是因为表名两边加了单引号引起的,改为下面的试一下
$sql = "select * from category order by sort asc, classid desc";
resource(5) of type (mysql result)
是什么意思?mysql结果资源5?
$hostname_conn = "localhost"; //host name or ip
$database_conn = "mydb"; //database name
$username_conn = "root"; //mysql root username
$password_conn = ""; //mysql passwordclass mySql_Class
{
function __construct($host, $user, $pass)
{
@mysql_connect($host,$user,$pass) or die("Database connection failed!");
mysql_query("SET NAMES 'utf8'");
}
function select_db($db)//connection table
{
return @mysql_select_db($db);
}
function query($sql)//fecth sql
{
return @mysql_query($sql);
}
function fetch_array($fetch_array)
{
return @mysql_fetch_array($fetch_array, MYSQL_ASSOC);
}
function close() //close db
{
return @mysql_close();
}
function insert($table,$arr) //insert into data
{
$sql = $this -> query("INSERT INTO `$table` (`".implode('`,`', array_keys($arr))."`) VALUES('".implode("','", $arr)."')");
return $sql;
}
}$mysql = new mySql_Class($hostname_conn,$username_conn,$password_conn);
$mysql -> select_db($database_conn);
?>
mysql类文件如上
$mysql = new mySql_Class($hostname_conn,$username_conn,$password_conn);
$mysql -> select_db($database_conn);
把@去掉看报什么错?
你的代码一行一行,还真看不出问题,水平有限
echo ($row['classid'].$row['classname'].$row['parentid'].$row['sort']);
=>
echo ($row['CLASSID'].$row['CLASSNAME'].$row['PARENTID'].$row['SORT']);
字段名换成大写就ok了,为什么?
如何避免这个问题。
你的代码一行一行,还真看不出问题,水平有限
php对变量大小写敏感对函数名不敏感
这样才更能兼容更多的系统
邮箱验证谁能帮我写?
$row['classid']里的classid不是变量名或函数名啊,无论如何问题已经解决了。
感谢大家!结贴