$sqlurl = "211.149.247.151";//
$sqlname = "Suji";//
$sqlusername = "Suji";//
$sqlpassword = "Mars99";//
$sqllit = mysql_connect ($sqlurl, $sqlusername, $sqlpassword);//
if (! $sqllit)
{
Die (' Could not connect: '. Mysql_error ());//
}
mysql_query ("SET NAMES UTF8"); //
mysql_select_db ("Suji", $sqllit);//
$result = mysql_query ("Select *from passlist");//
echo "Passlist:
";//
while ($word = Mysql_fetch_array ($result))//?
{
echo $word [' passname ']; //
}
Mysql_close ($sqllit); //
Reply to discussion (solution)
Online and so on. I hope you can help me, thank you.
Open terminal, input: mysql-u suji-p-H 211.149.247.151
Enter MARS99 after carriage return
The upstairs is!!!!!!!!! the solution.
Open terminal, input: mysql-u suji-p-H 211.149.247.151
Enter MARS99 after carriage return
Where is the terminal? I'm sorry. I'm a little white.
Your code itself is fine, but the remote Access database requires a database authorization
Your code itself is fine, but the remote Access database requires a database authorization
11.149.247.151
Would you like to enter this into the database?
What do you mean? The address of the database is what the database administrator told you, you can't define it yourself.
What do you mean? The address of the database is what the database administrator told you, you can't define it yourself.
http://www.onevisa.com.cn/test.php
To leave a name on this page, I don't know how to get into his database.
The IP of www.onevisa.com.cn is 211.149.247.151
If used $sqlurl = "211.149.247.151"; Can't get in.
Available $sqlurl = "localhost";
Are you sure that MySQL is available?
Weak weak ask, $sqlname parameters not?
Open terminal, input: mysql-u suji-p-H 211.149.247.151
Enter MARS99 after carriage return
Where is the terminal? I'm sorry. I'm a little white.
The terminal is the command line ... Cmd...