PHP+MySQL??

zenag · April 30, 2008

regarding $sql1 we can assign it.variable can be of any type...nothing problem in that...

and...below query is correct it works fine

and regardng..
while ($row1=mysql_fetch_row($result2))(
$uuid=$row1[0];
echo $uuid;

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

tuxbuddy · April 30, 2008

See I have made changes...HEre goes the complete COde:

<?

session_start();

$clicked=$_POST["gender"];

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


<?php

$myusername = $_SESSION['myusername'];
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 = "mysql123"; // 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)
$result3 = mysql_query("SELECT users.firstname,users.surname,incidents.owner_id FROM incidents,users where (incidents.owner_id = users.id) OR (incidents.owner_id <'$uuid') OR (incidents.owner_id = NULL)",$link);




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 "$myusername";
while ($row = mysql_fetch_array($result3, MYSQL_ASSOC)) {


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

   # echo("<td>" . $row["name"] . "</td>");
   # echo("<td>" . $row["create_time"] . "</td>");
}
/* Closes the table */
?>
</table>
<?

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

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

{
echo "closed";
}
}

?>

IS it allright??

tuxbuddy · April 30, 2008

Not working

tuxbuddy · April 30, 2008

Sorry Forgot to change :

row to row1 everywhere:

while ($row1 = mysql_fetch_array($result3, MYSQL_ASSOC)) {

echo "\n";

echo("<tr>\n<td>" . $row1["id"] . "</td>");

echo("<td>" . $row1["description"] . "</td>");

echo("<td>" . $row1["firstname"] . $row["surname"]. "</td>");

echo("<td>" . $row1["loginname"] . "</td></tr>");

But Still its not working...Can yu check code for me?/

zenag · April 30, 2008

have u changed $row1 in every while statement???

like this..

while($row1=mysql_fetch_row($result))

{

while ($row1=mysql_fetch_row($result2)){

if so ...change it...to different variable...$row,$row1

tuxbuddy · April 30, 2008

Sorry I fixed it...missed { } sign near while loop.

Now its showing some o/P like this:

Report
venkat 9Resource id #4383935ID
Description Owner Name         Notes 
               Ad.Administrator  
               Ad.Administrator  
               Ad.Administrator  
               Ad.Administrator  
               PrasanthM  
              PrasanthM  
              PrasanthM  
               PrasanthM  
              SijoJose  
              SijoJose  
              SijoJose  
              SijoJose

I think something going wrong..with query presentation

zenag · April 30, 2008

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

tuxbuddy · April 30, 2008

Here is the complete code:

<?

session_start();

$clicked=$_POST["gender"];

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


<?php

$myusername = $_SESSION['myusername'];
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 = "mysql123"; // 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)
$result3 = mysql_query("SELECT users.firstname,users.surname,incidents.owner_id FROM incidents,users where (incidents.owner_id = users.id) OR (incidents.owner_id < '$uuid') OR (incidents.owner_id = NULL)",$link);




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 "$myusername";
while ($row1 = mysql_fetch_array($result3, MYSQL_ASSOC)) {


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

?>

tuxbuddy · April 30, 2008

What I guess is select statement???

tuxbuddy · April 30, 2008

Heyy...I think we are close to right output.See here it displays:

venkat 9Resource id #4383935ID Description Owner Name Notes

Ad.Administrator

ChandrappaChikkanna

I think ID Description are not being retrieved in the right way But 38 39 and 35 are the right output.Can u put them in ID and Description in the right manner

zenag · 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";
}
}

?>

zenag · April 30, 2008

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

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

tuxbuddy · April 30, 2008

Heyy...One iSse still unsolved.

See 38 39 and 36 which it displays is all I want.But..Under table , it shows:

ID Description Owner Name Notes

17 asdfzsvgfsd Ad.Administrator

26 test1 ChandrappaChikkanna

13 at Ad.Administrator

24 test Ad.Administrator

Thats incorrect.What I want the corresponding entry relative to 38 39 and 36 only.

zenag · 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 ...

tuxbuddy · April 30, 2008

Ok...Leave it ...What I actlly want from you is Just display like this:

ID DEscrition OwnerName Notes

38 ... ..

39 ... ..

36 .. ..

For that the hint I think will work is :

$sql1 = mysql_query("SELECT id from users where loginname ='".$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)

$result2 = mysql_query("SELECT incidents.id,incidents.description,users.firstname,users.surname,incidents.owner_id FROM incidents,users where (incidents.owner_id = users.id) OR (incidents.owner_id = '$uuid') OR (incidents.owner_id = NULL)",$link);



if (!$result2) {
 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 "$myusername";
while ($row1 = mysql_fetch_array($result2, MYSQL_ASSOC)) {


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

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

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

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

WIll it work???I tried but the same o/p as before

zenag · 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..

tuxbuddy · April 30, 2008

Check ur Mail...I want to clearly make u understand what Actlly I want,,,Check ur mail.

tuxbuddy · April 30, 2008

DOne...I have explain you with screenshots.Kindly check.

zenag · 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..

zenag · April 30, 2008

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

zenag · 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);

tuxbuddy · April 30, 2008

What

incidents.owner_id < '$uuid'

means??

zenag · 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

zenag · April 30, 2008

OUT FOR LUNCH BE BACK AT 2.PM

tuxbuddy · April 30, 2008

Th query displays:

SQL-query:

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 LIMIT 0 , 30

[Edit] [Explain SQL] [Create PHP Code] [Refresh]

row(s) starting from record #

in horizontal horizontal (rotated headers) vertical mode and repeat headers after cells

firstname description surname owner_id

Prasanth NULL M NULL

Sijo NULL Jose NULL

Ram NULL Prasad NULL

Josin NULL Jose NULL

Prabhu NULL Hosmani NULL

jose NULL NULL

vinod NULL prasad NULL

venkat NULL NULL NULL

Ajit NULL Raina NULL

Samreen NULL Raina NULL

priya NULL priya NULL

Sumit NULL sumit NULL

vidya NULL NULL

row(s) starting from record #

in horizontal horizontal (rotated headers) vertical mode and repeat

Whats that for?

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PHP+MySQL??

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