Ashish Sood
-
Posts
79 -
Joined
-
Last visited
Content Type
Profiles
Forums
Events
Everything posted by Ashish Sood
-
But i want all the value of option in my dropdown list, Sir , Could you please understand me via example, so it will easy for me to understand . Thanks for a quick response
-
var option=document.createElement("option"); option.text="1"; option.text="2"; option.text="3"; option.text="4"; option.text="5"; btn.appendChild(option); After that i m only getting "5" in my dropdown list, dont know what i did wrong , Please guide .
-
Thanks For the replyI had successfully add a elements on my page. But only one problem i am facing is that how could i add option values in my dropdown list ?<script>function myFunction(){ var text=document.createElement("input"); document.body.appendChild(text); var btn=document.createElement("select"); document.body.appendChild(btn); var area=document.createElement("textarea"); document.body.appendChild(area); };</script>
-
For the security reason i am using ### to hide the information. <?php include_once('validation_turnover.php'); ?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml"><head><meta http-equiv="Content-Type" content="text/html; charset=utf-8" /><title>Daily Turnover</title><style type="text/css">.format {font-family: Verdana, Geneva, sans-serif;font-size: 15px;font-style: normal;text-transform: capitalize;position: relative;text-align: justify;color: #963;letter-spacing: 1px;display: inline;width: auto;visibility: visible;height: 1px;}</style></head><body><form action="turnover.php" method="post"><b> <label for="INCIDENT" class=format>INCIDENT ID</label> <label for="SEVERITY" class=format>SEVERITY</label> <label for="STATUS" class=format>STATUS</label> <label for="description" class=format>TICKET DESCRIPTION</label></b></b><br /><input type="text" name="incident" /> <select name="sev"> <option value="none0"></option> <option value="###">###</option> <option value="###">###</option> <option value="###">###</option> <option value="###">###</option> <option value="###">###</option> </select> <select name="status"><option value="###"></option> <option value="###">###</option> <option value="###">###</option> <option value="###">###</option> <option value="###">###</option> <option value="###">######</option> <option value="###">###</option> </select> <textarea rows="2" cols="30" name="description_tic" ></textarea><br><center><input name="submit" type="submit" value="Submit" /></form><hr></html></body>
-
Hi All, My project is in the last phase, my requirement is i want to dynamic add row on the page in which i want text box, dropdownlist and text area. Detail. Right now i had create a text box ,2 dropdown list and one text area, which is used to stored user information but its not full-fill the requirement because when there is more than one user then we need more element on the page, so i need a button which can call the function. so user click on the button these four element will be create below the earlier element. Hope you guys understand my Req. Please help me guys in my last phase ( i am very new to JS) All helps is appriciated Thanks A lot In Advance
-
Hi All,I am new to java scripts, I had assigned the project which has been completed 90% , only some part is remaining.I want to know how can i create a dropdown and text box on run time, mean when the user click on add new row, textbox and dropdown box create dynamically on the page.What i am thinking is that I can create a function in which text box and dropdown code to mention. and when user click add new row then we can call the function.But dont know how to do..Please help me GuysThanks In Advance
-
No Sir Still not, Could you please give me suitable link for that
-
I am new to CSS, i have created a page in which i had use table, my problem is that the table data is showing different on every system , on my system its show fine , but on the other system(browser)its showing in the improper way.My System (Browser)http://img339.images...75/samplemy.pngOn Other system(Browers)http://img18.imagesh...86/samplequ.pngAs you can see First row Step 1 is in the one line on my systembut when i open same page on another system the Step 1 is not in the one lineI want my website should be open on all the system as same is open on my system.hope your got my point,Thanks In Advance
-
Thanks For A Reply This time i am uploading a snapshot of the page My System (Browser)http://img339.images...75/samplemy.png On Other system(Browers)http://img18.imagesh...86/samplequ.png As you can see First row Step 1 is in the one line on my systembut when i open same page on another system the Step 1 is not in the one line I want my website should be open on all the system as same is open on my system. hope your got my point, Thanks In Advance
-
Hi All, I have created a project in which i am using a table , but i am facing a issues regarding table width and height means to says when i open a website in normal resolution monitor table & rows width and height will be unsettle,but on my system is ok(full HD), is anything i can do to fix the table l x b issue. Hope you understand ThanksAshish Sood
-
@Birbal Could you give me example, so will be more clear to me.
-
Ohh you guys got me wrong. I mean to ask that if i am using "if" condition and the condition get fail (may be user supplied invalid username & password) then i have to show a error message to user like" invalid credentials". which is always show on the top left corner of the page... But i want the error message to show where is want the page. Hope you guys understand my queries
-
Hi , By default in php error shows on the upper left corner of the page, but how could is show error on the specific part of the page. Any IDEA Thanks
-
One more thing after clicking on any hyper link why i am having a error , its working fine on my live site but after import i dont know why i am getting the undefined error messageI know its a variable undifined message but why its not appering on the live site. Notice: Undefined variable: showSpecials in /var/www/html/checklist/project/sample_new/templates/theme162/index.php on line 474
-
Ok I got your point, i gave the permission now my site is working fine . You always help with the good answer , thanks you sir I appriciate your work:)
-
Hi , I just download my whole joomal site to my local system, now i am getting error, i changed my configuration.php with correct username,database and password but still i dont know why i am getting the error, please find the error below. Please find the error below, and please let me know how could i fix this issue.. Warning: require_once(/var/www/html/checklist/project/sample_new/libraries/joomla/config.php): failed to open stream: Permission denied in /var/www/html/checklist/project/sample_new/libraries/joomla/factory.php on line 474 Fatal error: require_once(): Failed opening required '/var/www/html/checklist/project/sample_new/libraries/joomla/config.php' (include_path='.:/usr/share/pear:/usr/share/php') in /var/www/html/checklist/project/sample_new/libraries/joomla/factory.php on line 474 ThanksAshish Sood
-
Thanks for the reply i appriciate your efforts @satishpoul Sir could you please apply the function in my code below, so it will more helpfull to understand.and also please tell me how can i call the function . .<?php if(isset($_POST['submit'])) { $username=$value; //$username=$_POST['username']; $pattern= "/[a-z]/"; if(preg_match($pattern,$username)) { echo "OK"; } else { echo "Not ok "; } } ?> <form action="sample.php" method="post"> <fieldset><legend> USER LOGIN</legend> <p><div><label for="username">*USERNAME:</label><input type="text" name="username" value="<?php if(isset($_POST['username'])){ echo htmlentities($_POST['username']); } ?>"/><br /><p></div> <p><div><label for="username">*PASSWORD:</label><input type="password" name="password" /></div></p> <p><div><label for="username">*FIRSTNAME:</label><input type="text" name="username" value="<?php if(isset($_POST['username'])){ echo htmlentities($_POST['username']); } ?>"/><br /><p></div> <p><div><label for="username">*LASTNAME:</label><input type="text" name="username" value="<?php if(isset($_POST['username'])){ echo htmlentities($_POST['username']); } ?>"/><br /><p></div> <input type="submit" name="submit" /></div></p>
-
Yes sir i find my error and edit my post again please check it.
-
Thanks for your replyi tried below code and its working for me.But the only problem is that if my registration form contain 10 text box then i have to check it again and again, can it be possible to create a function and pass the value this function if yes, please give me a hint..<?php if(isset($_POST['submit'])){ $username=$_POST['username']; $pattern= "/[a-z]/"; if(preg_match($pattern,$username)) { echo "OK"; } else { echo "Not ok "; }}?> Thanks In Advance
-
yes you are right Sir, But i dont want user to enter any no.[0-9],if some try to enter any no and click on submit button, then error should be display like "invalid input type", hope you understand what i want.
-
Hi All, I am creating a form, in this form i want user to enter only string in the text box for that can i create a funtion in php which check the value of the text box Or is there is any other alternative to limit the user to enter only string in the text box. Thanks in advance
-
Thanks For your reply my problem is being solved, Thanks A Lot
-
Hi, I dont know why i am getting a error into my insert query i checked whole query but cant find the error. Parse error: syntax error, unexpected ';' in /var/www/html/checklist/project/main_validation.php on line 159 Please check my below insert query.. $sql = mysql_query("INSERT INTO sample (DATE,JOBNAME,TIME,INITIALS) VALUES ('$date','$amb','$amb_time','$amb_ini'), ('$date','$pro','$pro_time','$pro_ini'), ('$date','$arp','$arp_time','$arp_ini'), ('$date','$cir','$cir_time','$cir_ini'), ('$date','$prenite','$prinite_time','$prinite_ini'), ('$date','$bcl4','$bcl340_time','$bcl340_ini'), ('$date','$bcl5','$bcl350_time','$bcl350_ini'), ('$date','$bcl6','$bcl360_time']','$bcl360_ini''), ('$date','$bcl7','$bcl370_time','$bcl370_ini'), ('$date','$inv','$inv220_time','$inv220_ini'), ('$date','$sto','$sto_time','$sto_ini'), ('$date','$picking','$picking_time','$picking_ini'), ('$date','$arpnight','$arpnight_time','$arpnight_ini'), ('$date','$mstnight','$mstnight_time','$mstnight_ini'), ('$date','$salnight','$salnight_time','$salnight_ini'), ('$date','$arpamb','$arpamb_time','$arpamb_ini'), ('$date','$nreports','$nreports_time','$nreports_ini')" or die (mysql_error()); Thanks In Advance
-
I sorted out my problem, i closely checked my login.php page and found and error and i remove tha line from thereerror line is echo $_SESSION['username']; I removed the above line now every thing is running fine Thanks alot guys for your efforts
-
This line coming from validate_login.php page i think not from login.php page, becuase i included this page at the top of login.php.
