not able to get an output pls help

[udaystrad]

Heres the code

dont kno wats the problem

 
<?php
 
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password="india123"; // Mysql password 
$db_name="test"; // Database name 
$tbl_name="members"; // Table name 
 
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");
 
$myusername="maruti"; 
$mypassword="india123";
 

 
$dim="MID"; //  column name 
 
$mid="SELECT $dim FROM $tbl_name WHERE username='$myusername' and password='$mypassword'";
 
echo $mid;
 
 
 
?>
 

Edited November 15, 2013 by udaystrad

[udaystrad]

•  

  phpMyAdmin 2.11.11.3 

• NO errors

out put-- 

SELECT MID FROM members WHERE username='maruti' and password='india123'

•  

the ouput should be -- 

009

(which i m nt able to get)

[mac_gyver]

the code you have posted is not running the sql query statement at all, nor is it fetching and using the data from the query statement, so just exactly how did you expect your code to produce the expected output?

[Barand]

mysql_query

mysql_fetch_assoc

mysql_fetch_row

 

or, better, use the mysqli_ versions as mysql_ functions are deprecated

Edited November 15, 2013 by Barand

[udaystrad]

Hey sir but the same codes worked out for me after some changes.....

or n may be i didnt understand what exactly u r tryin to say bcoz u r an expert and ur chances are more to be right....bt stil i have solved my issue...

 

Here is the code below 

 

 

 
<?php
 
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password="india123"; // Mysql password 
$db_name="test"; // Database name 
$tbl_name="members"; // Table name 
 
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");
 
$myusername="maruti"; 
$mypassword="india123";
 
// If result matched $myusername and $mypassword, table row must be 1 row
 
$dim="MID";
$attendance="attendance";
 // Table name 
$takevalue=mysql_query("SELECT $dim FROM $tbl_name WHERE username='$myusername' and password='$mypassword'");
 
$array = mysql_fetch_array($takevalue); 
 
$mid=$array['MID'];
echo $mid.'<br/>';
 

 
//;
?>

[udaystrad]

Sir pls guide me what exactly u wer tryin to say...or i have got it by myself????

 

 

n i m  also stuck with one more databse problem heres the code

its nt workin????

 

thers no output(not getin inserted in database)

NO error showing. 

 

let me remind you i have tried sme other format also (which didnt work)...although dont have it .... pLs give me the line you think is right coz i have been stuck on this for a while....

 
mysql_query("INSERT INTO $attendance ('MemberId,TimeIn,TimeOut,Day,Month,Year')  VALUES ('$mid','$timein','$timeout','$day','$month','$year') ");
 

Edited November 16, 2013 by udaystrad

[Barand]

put in some checks

$sql = "SELECT $dim FROM $tbl_name WHERE username='$myusername' and password='$mypassword'";
$takevalue=mysql_query($sql);

// check for errors in query
if (!$takevalue) {
    echo "$sql<br>" . mysql_error();
}
else {
    // check if any records found
    if (mysql_num_rows($takevalue)==0) {
        echo "No matching records<br>";
    }
    else {
        $array = mysql_fetch_array($takevalue); 
 
        $mid=$array['MID'];
        echo $mid.'<br/>';    
    }
}
 

[udaystrad]

k...

thank you so much sir....

i did follow this method of checks for some of my codes.

even the last one worked out for me.

My codes are much more easy to understand.