ypkumar

right, I found a solution by myself again PHP - easier than making a tea! LOL

Hi! I don't have any problems with fetching the array, I am having problems in presenting the data fetched. I have solved the problem myself i've used this mysql_fetch_assoc() instead of mysql_fetch_array() ----- now the problem lies in within the next leg of the page. <table> <tr> <td> sender </td> <td> message </td> <td> mark as read </td> </tr> <?php $q = "select customer_fname, customer_message from customer_messages where flag = '0'"; $w = mysql_query($q); while($r = mysql_fetch_assoc($w)) {?> <tr> <td> <?php echo $r['customer_fname'];?> </td> <td> <?php echo $f['customer_message'];?> </td> <td> <input type="checkbox" name="markasread" value="$f['entryid'];" /> </td> </tr> } thats show.php and that gives me, a list of messages from different users. ================================================================ now i need to select the checkbox and click a button to approve them. in the next page 'mesenger.php' i need to fetch the id's of members whose checkbox i have clicked on previous page. i can use $_POST['markasread']; but i need to find out which checkboxes i have checked and which i havent checked. . how do i check that? amny help would be greatly appreciated

hi pikachu () as i mentioned earlier "$t now gives me total number of unread messages in the db." ofcourse it doesnt list all the elements, this is what i thought i must do <table> <tr> <td> sender </td> <td> message </td> <td> mark as read </td> </tr> <?php $q = "select customer_fname, customer_message from customer_messages where flag = '0'"; $w = mysql_query($q); while($r = mysql_fetch_array($w)) {?> <tr> <td> <?php echo $r['customer_fname'];?> </td> <td> <?php echo $f['customer_message'];?> </td> <td> <input type="checkbox" name="read" value="1" /> </td> </tr> } this code retrieves the first entry from the 'db' which has flag set to '0'. now I need to fetch all the entries . what do i do? i'm thinking of a for loop, but still cant get my head around on how to put it here.

hello guys, I'm stuck here: I have a table called customer_messages which has a entryid, custid, fromname, message, flag i have written a script to enter data into this. and another script to count the number of entries in the table. now i want to display entries in the table on to a script what i did was this: flag is by default '0' which mean unread. insertion code: insert into customer_messages (customer_id,customer_fname,customer_message,flag) values('21','mike','hello!','0') fetching number of unread messages select count(*) from customer_messages where flag = '0' $q = "select count(*) from customer_messages where flag = '0'"; $w = mysql_query($q); $r = mysql_fetch_array($w); $t = $r[0]; $t now gives me total number of unread messages in the db. now, I want to display these messages on to a page in the following format: ----------------------------------------------------------------------------------------------- | sender | message | mark as read | ----------------------------------------------------------------------------------------------- | name_from_db | message_from_db | check box | ----------------------------------------------------------------------------------------------- | name_from_db | message_from_db | check box | ----------------------------------------------------------------------------------------------- how do I do this? I'm assuming that I have to use a loop here, I put the code in a while loop, but that didnt work out for me. can I use a for loop here? if yes how do i use it? my brain's froze now so, i cant think atm, can someone help me out please? ================================================================================= in short i need the data from database to be listed as listbox control in asp.. is it possible?

Hello guys, I am a beginner in PHP, so please be gentle. I am building this website for a friend of mine. This website is made up by HTML/CSS. i.e., I have a page made with HTML & CSS. there are a few menus on the page. say Home, About Us, Contact Us, etc., Now I want to separate the page elements (like header/footer) from the page so as its easy to work on. the source is something like this <HTML> <HEAD> <TITLE> My Page </TITLE> <LINK REL="STYLESHEET" HREF="./mystyle.css" TYPE="text/css" /> </HEAD> <BODY> <DIV ID="wrapper"> <DIV ID="header"> <DIV ID="logo"> <UL ID="menu"> <LI> <A HREF="./index.php"><SPAN>Home</SPAN></A> </LI> <LI> <A HREF="./login.php"><SPAN>Login</SPAN></A> </LI> <LI> <A HREF="./contactus.php"><SPAN>Contact Us</SPAN></A> </LI> </UL> <DIV ID="date"> <?php echo Date("d M Y");?> </DIV> </DIV> </DIV> <DIV ID="bodywrapper"> <DIV ID="main"> <DIV ID="content"> THIS IS A TEST! </DIV> </DIV> </DIV> <DIV ID ="clearfooter"></DIV> </DIV> <DIV ID="footer"> © <?php echo date("Y");?> </DIV> </BODY> </HTML> now I will have my content in the DIV called 'content'. now I have separated this one page into three different pages. like this: header.php <HTML> <HEAD> <TITLE> My Page </TITLE> <LINK REL="STYLESHEET" HREF="./mystyle.css" TYPE="text/css" /> </HEAD> <BODY> <DIV ID="wrapper"> <DIV ID="header"> <DIV ID="logo"> <UL ID="menu"> <LI> <A HREF="./index.php"><SPAN>Home</SPAN></A> </LI> <LI> <A HREF="./login.php"><SPAN>Login</SPAN></A> </LI> <LI> <A HREF="./contactus.php"><SPAN>Contact Us</SPAN></A> </LI> </UL> <DIV ID="date"> <?php echo Date("d M Y");?> </DIV> </DIV> </DIV> footer.php <DIV ID ="clearfooter"></DIV> </DIV> <DIV ID="footer"> © <?php echo date("Y");?> </DIV> </BODY> </HTML> and index.php <?php include_once('header.php');?> <DIV ID="bodywrapper"> <DIV ID="main"> <DIV ID="content"> THIS IS A TEST </DIV> </DIV> </DIV> <?php include_once('footer.php');?> Is this the right way to code? I am worried because I have to keep in mind the safety of the website as I have to include a basic login module to this. is there any other style to write this? please do let me know.. thanks

thanks guys, found out that unicode/utf-8 character is the culprit.. couldnt reply faster because i wanted to finish this project.

arrrggh i sit here and write a descriptive information about my problem and tried to upload a file, the write up's all gone! grrr...... ill do it the simple way my system: Windows Vista XAMPP Lite 1.7.3 PHP Version 5.3.1 mysqlnd 5.0.5-dev - 081106 - $Revision: 289630 $ mysql/mysqli 5.1.41 my application: PHP MySQL based banking application, where user can register, login, logout, add remove and edit friends accounts, transfer funds, change password, change personal info, get transaction details etc. etc., my problem: i rewrote the application conforming to OOP concepts, and I separated the functions by its behaviour (authenciation, fetching, dbo etc). when i wrote a function for transferring funds from account a to account b, i recieve an error. Fatal error: Call to undefined function mysql_query() in C:\xampplite\htdocs\mybank\classes\func.inc on line 197 - i did not upgrade/update either PHP nor SQL, and yes, SQL is enabled on my system and works fine with other functions (register,change pasword etc.) i wrote for the same appilcation. FOR THOSE WHO WANT TO HAVE A LOOK, i have uploaded the class file here : http://pastebin.com/PGaXgmkC im not sure whats causing the error, when i remove the code to transfer the funds, it works like charm, but at the same time, the code which is removed is not wrong, it works without errors too! can anyone please look into this issue and let me know? im dying to get this one fixed since 2 days!

ok this is like this.. I have a form in the body as something like this.. <tr> <td> Login </td> <td> <input type="text" name="userid" id ="usertext" onChange="validateUser(this.value);" /> </td> </tr> and in the head I refer to an external js file which has the functions to run... now that particular function validateUser() gets the username from textbox, queries at the db using AJAX and when that query returns a value validateUser changes the color of the textbox (or color of text) based on the validation result. i.e., if validation is successful then the color of plain white text box turns to #CCFF99 and if it was unsuccessful it changes the text box color to #FF6666 and the code (for changing color of font) goes something like this. if(true) { document.getElementById("usertext").style.color = "#CCFF99"; } else if(false) { document.getElementById("usertext").style.color = "#FF6666"; } now lets assume that I have typed in something wrong deliberately in the text box, and I press "TAB" to get to the next text box (control) and to execute the validateUser js function, and as a result of that I get the text box's font color changed to red. now I want to do something that when I click the text box after validateUser has executed once and color of the text box font changes.. the color turn back to normal until I type and press TAB again to execute the function... is there anyway to do that? I have a theory of adding a onFocus or something to change the color.. but i havent tried it out yet.. any suggestions?

hmm.... I've removed all the functions and rewrote everything... and funnily it worked! I've moved the database variables into a file called database.inc and included that file in my php file. and the code is something like this.. .... elseif(isset($_GET['un'])) { include'database.inc'; ..... printf, can you please brief me the disadvantages of declaring the variables global that way? thanks for your input guys, that really made my rusty brain move a bit

its been a long time since i touched php and now i've learnt a bit of sql i decided to come back.. I dont really know where to post this question because this i think is both php and sql related. So please feel free to move it to the right section if i posted it in the wrong section. ta. I need help with this small problem of mine.. I use xampp 1.7 (which i think is a pretty good package for beginners like me) and I've been trying to make a simple login page (which i did), and now I moved some of the code into a function and into a separate file and in a separate directory, and thats when the page started acting wierd :s I have this login.php as .... include 'assets/functions_include.php'; $usrtxt = $_POST['usrtxt']; $pwdtxt = $_POST['pwdtxt']; $query="select count(*) from str_users where usrname='".$usrtxt."'"; doconnectdb($query); ..... and i have this in functions_include.php file in assets directory $dbserver = "localhost"; $dbuser = "root"; $dbpass = "password"; $dbname = "str_db"; function doconnectdb($query) { $connectionattempt = mysql_connect($dbserver,$dbuser,$dbpass); if(!($connectionattempt)) { die("failed to connect because.. ".mysql_error()); } else { $seldbattempt = mysql_select_db($dbname); if(!($seldbattempt)) { die("couldnt find the database because.. ".mysql_error()); } else { querythis($query); } } } function querythis($query) { $query = mysql_query($query); if(!($query)) { die("query failed because.. ".mysql_error()); } else { $count = mysql_num_rows($query); return $count; } } ?> and from sql, i do have a database str_db which has 1 table str_users details for str_users: -------------------------------------------------------------------- | usrid | usrname | usrpassword | usrisauth | usrauthcode | -------------------------------------------------------------------- | 1 | admin | creative | 0 | NULL | -------------------------------------------------------------------- and the username and password provided for the database in PHP are correct. before writing the code in this way, I had the code in 1 single page and everything worked fine!! and when i partitioned the code into separate files and put them in different folders, i started receiving this error.. couldnt find the database because.. No database selected can anyone point out my mistake here? thanks in advance!