zenag

June 12, 2008

script to validate 10 digit number

function validateNumber(field, msg, min, max) { 
if (!min) { min = 10 } 
if (!max) { max = 10 } 

if ( (parseInt(field.value) != field.value) ||  
       field.value.length < min || field.value.length > max) { 
alert(msg); 
field.focus(); 
field.select(); 
return false; 
} 

return true; 
}

June 12, 2008

open and close braces for if statement

if (string.length !=6)
{
      return false;
}
   return true;

June 12, 2008

(SELECT * FROM table1) UNION (SELECT * FROM table2)  ORDER by date desc

June 12, 2008

left join must have space

SELECT name_details.name,name_details.node,ip_details.ip FROM name_details left join ip_details on ip_details.name=name_details.name

June 12, 2008

SELECT name_details.name,name_details.node,ip_details.ip FROM name_details leftjoin ip_details on ip_details.name=name_details.name

June 12, 2008

SELECT * FROM table1, table2 ORDER by date desc

June 12, 2008

$_SERVER['REQUEST_URI']

June 11, 2008

<?
$con=mysql_connect("localhost","root","");
mysql_select_db("login",$con);
$j=0;
for($i=0;$i<=4;$i++)
{
$z= $j+2;
//echo "select * from user limit($j,$z)";
$sql=mysql_query("select * from user limit $j,$z");
while($row=mysql_fetch_array($sql))
{
?><table><tr><td>heading1</td></tr><tr><td>
<? if($z==2){ echo $row["nickname"];}?></td></tr></table>

<table><tr><td>heading2</td></tr><tr><td>
<? if($z==4){ echo $row["nickname"];}?></td></tr></table>

<table><tr><td>
<? if($z==2){ echo $row["nickname"];}?></td></tr></table>

<?
}
$j=$j+2;
};

?>

April 30, 2008

sorry i cant get u...whether the query displays some error

April 30, 2008

u have not gvn open bracket before..INSERT

$new = ("INSERT INTO travek_user_cols(1,2,3,4,5,6) VALUES('".$result["online"]."','".$result["online"]."','".$result["online"]."','".$result["online"]."','".$result["online"]."','".$result["online"]."'");

April 30, 2008

$server = "79.170.40.51:3306";

April 30, 2008


$server = "localhost:3306"; // this is the server address and port
$username = "web15-ian"; // change this to your username
$password = "1234567 "; 
$link = @mysql_connect ($server, $username, $password)
or die (mysql_error());

April 30, 2008


use mysql_fetch_array() after..
$result = $db->query($sql);

April 30, 2008

CAN U SHOW CODE?

April 30, 2008

return $result["online"];

April 30, 2008

OUT FOR LUNCH BE BACK AT 2.PM

April 30, 2008

while ($row1=mysql_fetch_row($result2)){
$uuid=$row1[0];
echo $uuid;
}
[quote]
i ve assigned this value...echo $uuid;

[/quote]
#$result3 = mysql_query("SELECT users.firstname,users.surname,incidents.owner_id FROM incidents,users where (incidents.owner_id = users.id) OR (incidents.owner_id = NULL)
$result3 = mysql_query("SELECT users.firstname,incidents.description,users.surname,incidents.owner_id FROM incidents,users where (incidents.owner_id = users.id) OR (incidents.owner_id < '$uuid') AND (incidents.owner_id = 0)",$link);

refer this....
this query might help u ...check this query ..it will output results from users table where owner_id in incidents table is null...
SELECT users.firstname,incidents.description,users.surname,incidents.owner_id 
FROM users
LEFT OUTER JOIN incidents
ON incidents.owner_id = users.id
WHERE incidents.owner_id IS NULL

April 30, 2008

sorry pal....ur query works fine ..ur correct...set in ur database for owner_id field as default value as NULL OR 0 in mysql...it will work fine..

$result3 = mysql_query("SELECT users.firstname,incidents.description,users.surname,incidents.owner_id FROM incidents,users where (incidents.owner_id = users.id) OR (incidents.owner_id < '$uuid') OR (incidents.owner_id = NULL)",$link);

April 30, 2008

yeah...just noew i ve came to my seat.....right now hve seen ur mail.....i wl check it out...

April 30, 2008

query hasnt yet accepted .." incidents.owner_id=NULL "..its not working ,we cannot display fields from two tables by satisfying condition from single table (ie.incidents.owner_id satisfies only incidents table but not users).if the condition satisfies both tables only we can able to display fields from both tables..

April 30, 2008

query hasnt yet accepted .." incidents.owner_id=NULL "..its not working ,we cannot display fields from two tables by satisfying condition from single table (ie.incidents.owner_id satisfies only incidents table but not users).if the condition satisfies both tables only we can able to display fields from both tables..

April 30, 2008

since u have selected only four fields here..

$result3 = mysql_query("SELECT users.firstname,incidents.description,users.surname,incidents.owner_id FROM incidents,users where (incidents.owner_id = users.id) OR (incidents.owner_id < '$uuid') OR (incidents.owner_id = NULL)",$link);

it would display that selected fields only....if u need to select all fields from two tables u can write it as...

SELECT * FROM incidents,users where ...

April 30, 2008

ive changed these lines...
$myusername = "administrator";
echo $myusername;
for my convenience ....

change it to 
$myusername = $_SESSION['myusername'];
echo $myusername;

April 30, 2008

this solves all ur probs.....


<?

session_start();

$clicked=$_POST["gender"];

?><?
if($clicked=="open")
{
?>
<html>
<title> Welcome to Project Management Tool</title>
<body bgcolor=#AFC7C7>
<h1> Report</h1>


<?php

$myusername = "administrator";
echo $myusername;

//put the above right at the top of your script. Everything else can follow
?>

<?

//header("Content-Type: text/plain");


/* set's the variables for MySQL connection */

$server = "localhost"; // this is the server address and port
$username = "root"; // change this to your username
$password = ""; // change this to your password

/* Connects to the MySQL server */

$link = @mysql_connect ($server, $username, $password)
or die (mysql_error());

/* Defines the Active Database for the Connection */

if (!@mysql_select_db("helpcore", $link)) {
  echo "<p>There has been an error. This is the error message:</p>";
     echo "<p><strong>" . mysql_error() . "</strong></p>";
     echo "Please Contact Your Systems Administrator with the details";
}

$sql1 = mysql_query("SELECT id from users where loginname ='".$myusername."'",$link);

while($row=mysql_fetch_row($sql1 ))
{

#echo $result[0];
$uid=$row[0];
echo $uid;
#$uid=$result["id"];
#echo "$uid";

}

#$result2 = mysql_query("SELECT incidents.id, incidents.description, users.firstname, users.surname, users.loginname FROM incidents,users WHERE (incidents.owner_id = users.id) AND (incidents.contact_id = '".$result[0]."')", $link);
$result2 = mysql_query("SELECT  id, description  FROM incidents where contact_id = '".$uid."'", $link);
#echo ("SELECT incidents.id, incidents.description, users.firstname, users.surname, users.loginname FROM incidents,users WHERE (incidents.owner_id = users.id) OR (incidents.contact_id = '".$uid."')";
//echo $result2 ;

#$result2 = mysql_query("SELECT incidents.id,incidents.description,users.firstname,users.surname,users.loginname from incidents,users where (incidents.owner_id = users.id) OR (incidents.contact_id =  '".$uid."')", $link);
while ($row1=mysql_fetch_row($result2)){
$uuid=$row1[0];
echo $uuid;
}
#$result3 = mysql_query("SELECT users.firstname,users.surname,incidents.owner_id FROM incidents,users where (incidents.owner_id = users.id) OR (incidents.owner_id = NULL)
$result3 = mysql_query("SELECT users.firstname,incidents.description,users.surname,incidents.owner_id FROM incidents,users where (incidents.owner_id = users.id) OR (incidents.owner_id < '$uuid') OR (incidents.owner_id = NULL)",$link);
//echo "SELECT users.firstname,users.surname,incidents.owner_id FROM incidents,users where (incidents.owner_id = users.id) OR (incidents.owner_id < '$uuid') OR (incidents.owner_id = NULL)";


if (!$result3) {
  echo("<p>Error performing query: " . mysql_error() . "</p>");
  exit();
}
     

/* Starts the table and creates headings */
?>
<table border="1">
<tr>
<td><strong>ID</strong></td>

<td><strong>Description</strong></td>
<td><strong>Owner Name</strong></td>
<td><strong>Notes</strong></td>

<?
/* Retrieves the rows from the query result set
and puts them into a HTML table row */





/* Retrieves the rows from the query result set
and puts them into a HTML table row */
#echo "hello";
#echo "$myusername";
while ($row2 = mysql_fetch_array($result3)) {


echo "\n";
     echo("<tr>\n<td>" . $row2["owner_id"] . "</td>");
    echo("<td>" . $row2["description"] . "</td>");
    echo("<td>" . $row2["firstname"] . $row1["surname"]. "</td>");
    echo("<td>" . $row2["loginname"] . "</td></tr>");

   # echo("<td>" . $row["name"] . "</td>");
   # echo("<td>" . $row["create_time"] . "</td>");
}

/* Closes the table */
?>
</table>
<?

/* Closes Connection to the MySQL server */

/* Closes Connection to the MySQL server */
mysql_close ($link);

}
else
{
if($clicked=="closed")

{
echo "closed";
}
}

?>

April 30, 2008

it cause of loop problem....some braces have to be checked or can u send full code

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zenag

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