Thank you
Search found 6 matches
- Sat Jul 05, 2003 7:43 am
- Forum: PHP - Code
- Topic: Help me
- Replies: 3
- Views: 763
Thank you for your help
Now, It's work
Thank you
Thank you
- Sat Jul 05, 2003 6:38 am
- Forum: PHP - Code
- Topic: Help me!
- Replies: 1
- Views: 300
Help me!
18)if($action==1)
{....
}
40)<FORM METHOD="POST" ACTION="reg1.php?action=1" name="form1" >
after execute the script the error message is,
PHP Notice: Undefined variable:action in line 18
Please help me on this.
Thank you
{....
}
40)<FORM METHOD="POST" ACTION="reg1.php?action=1" name="form1" >
after execute the script the error message is,
PHP Notice: Undefined variable:action in line 18
Please help me on this.
Thank you
- Fri Jun 27, 2003 11:01 am
- Forum: PHP - Code
- Topic: Help me
- Replies: 3
- Views: 763
Help me
<?php $query = "select * from userid where username='" .$_REQUEST['username'] ."'"; $result=mysql_db_query($query); if(mysql_num_rows($result)==1) { header("Location: fail3.php"); exit; } ?> else ........ Execute the above code, no error message, but it seem does not w...
Help me
$query = "select * from userid where username=".$_REQUEST['username'];
$result=mysql_db_query($query);
if(mysql_num_rows($result)==1)
{
header("Location: fail3.php");
exit;
}
Execute the above code, no error message, but it seem does not work.
Thank you
$result=mysql_db_query($query);
if(mysql_num_rows($result)==1)
{
header("Location: fail3.php");
exit;
}
Execute the above code, no error message, but it seem does not work.
Thank you
- Thu Jun 26, 2003 10:44 pm
- Forum: PHP - Code
- Topic: help me!
- Replies: 2
- Views: 377
help me!
Source code: 16)$query = "select * from userid where username=.$_REQUEST[username]"; 17)$result = mysql_query($query); 18)$checknum = mysql_num_rows($result); Error Message : $PHP Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in e:\inetpub\wwwroot\yuan\a...
- Thu Jun 26, 2003 8:20 pm
- Forum: Databases
- Topic: Help me: PHP Parse error
- Replies: 2
- Views: 405
Help me: PHP Parse error
Source code is 1)?php 2)include('connect.php'); 3)mysql_select_db("$table_name1"); 4)$query = "select * from userid where username="$_REQUEST[username]"; The error message is: PHP Parse error: parse error, unexpected T_VARIABLE in e:\inetpub\wwwroot\yuan\adduser1.php on line...