user4fun

I would like to change <img border="1" src="97765/img1.jpg" > to <<img border="1" src=" <? echo $folder ; ?> /img1.jpg" except I need it to work :-) Thank you

I figured it out I am using the readline function, Thank you

This is over my head but I am trying to learn lol the url has a value called mls $fold=$_GET["mls"];$file_array = parse_ini_file("$fold/housefile.ini");print_r $file_array; 1- It is not working for syntax error and I am sure it is in the $fold portion, but not sure how to fix it.2- I need to print the segments of the ini file as I call for them ini file hasName = JohnLocation = 6767 Green RiverPrice = $120,000 I would really like to be able toecho $Name and echo $Location etc etc in different places on the site as I call for them.Thank you so much. I know this is over my head, but I can learn so much from your help.I don'y mind if you just tell me if I am on the right track, or what functions in php should I research. Thx

Excuse my simple question, would that be done using AJAX???

Excellent. Thank you.I started reading about that. I wonder if segment the text fileExample a line for the project titleA paragraph for the project descriptiona a line for the project a manger name for example!Is that possible?Thx

what is the best way to display the content of a .txt on a website example project_Disc.txt hasThis project is designed to develop etc etc etc on the html page, I want to recall the content of the project_Disc.txt file and display it where I want. Thank you Note, the site will eventually be mobile friendly. Thank you

Thank you all for the help with this.1- I am not sure if I should use javascript or something else. The site will "apparently" become mobile friendly as go daddy says. 2- I am trying to make the page loads on big_pic as /img1.jpgon the click of img2 the big_pic would change to /img2.jpgif the user click on img1 the big_pic would become /img1.jpg again. (there are actually 6 pics, but I can duplicate the code) and so forth thank you so much <div class="big_pic"><div class="img"><img src="/img1.jpg" alt="Main Image" width="350" height="291"> <h1 class="pic_caption"> caption note </h1> <div class="img" align="center"><img src="img1.jpg" alt="Home Image1" width="110" height="90"><div class="pic_desc">Image 1 description</div></div><div class="img"><img src="img2.jpg" alt="Home Image2" width="110" height="90"><div class="pic_desc">Image 2 description </div></div>

My name really says it all, i dont design for income, nor a job, or a carreer, I really am a User4fun.From IN, hoping to attend Law Schools at some point.I have been visiting this site often and i think it is just FANTASTIC.

so what would the code look like?

$q2 = mysql_query("SELECT `Site` FROM `checking` WHERE `Site` = '" . strtolower($_POST['site']) ."' LIMIT 1"); IF (mysql_num_rows($q2) == 1) : echo "already in system";//help heredie();ENDIF; help here.I want to add an update section that would pick the column Contacted in table checking where column Site = strtolower($_POST['site']);takes the INT value in the Contacted column and make it ++ ( or i guesss +1 wouldbethe same thing)

okay, and, then , where, so, its, and the variable will....... hmmmmmmI am confusedlolSORRYI may have not explained my self correctlyokay, visitors come to the website, they have a form text box name ="Sname"they hit enterthe system will look in the table 'checking' if column Sname has a value == to user input Sname THEN the column Listed in that same record becomes ++. In which that column is defined as smallint(2) default to 1if Sname does not exsist in table -- just add new recordListed is not a variable within the content of the code, it is a value in an phpmyadmin table called checking.

Not taht i would be ablt to provide much help, i am new to php and i like to read other posts to learn form them??What is the question here?

$q2 = mysql_query("SELECT `Sname` FROM `checking` WHERE `Sname` = '" . strtolower($_POST['name']) ."' LIMIT 1"); IF (mysql_num_rows($q2) == 1) : echo"already in system";die();ENDIF; $str_site = strtolower($name); $sql = "INSERT INTO checking (CheckID, name, Email, Listed) VALUES (' ', '$str_name', 'NA', ' ')"; //email is NA beacuse it gets added throught follow up forms.$query=mysql_query($sql);if($query){echo " "; }else{die('System Failure');} What i am trying to do--The listed column on my table is set to smallint(2) default value 1first of all the default value always ends up being 0 (zero) when a new record is added i am not sure why?and in the section where it saysecho "already in system"; i would like to add function that would make the Listed = Listed+1 that way the table would show how manytimes 'Sname' was listedhow can i do that?

Sounds good, when i get it going i will let you know what happenedthankx for the advise

I have a formwith one text filed calledsitei want to have three buttons instead of the traditional oneeach button will take A DIFFERENT DIRECTIONButton1called check_emwill run action = "check.php method="post"Button 2calledList_emwill run action= "list.php" method = "POST"button 3calledcontact_emwill run action = "contact.php method = "POST"How would i do that?

Thanks for the informationi made my self an easy solutionwhen the erro occured, an error page comes up sayingerror yada yada yadahit your browser back button to try again!.easy fix, for an easy guy

Yes,i am a rookie, and i dont know very much and thatis hwy i got you wonderfull people that like to share their knoweldge.i am trying to have a linktry again!that will take the person to the previouse pagethe previouse page is a form and i want what information already put in it to be retrievedside question?is ther like some kindoff a small command that will show the form again as is before submitting but with the exceptionof a certain fieldexampleretrieve allbutusername value = " "the form page is (.php)the error page is (.php)

The first one works like acharm, i can display the password and all theother infromation that i wantThank you very much for your helpthe first timethe second timeand the third timeand so on

sorryto be so stupid, but the final code would look how?

i tried them both and unfortunatly the email gets send outyour password is: and then there is nothing againi am confused

$get_user = mysql_query("SELECT * FROM associnfo WHERE user_email = '".$_POST['em']."'");$q = mysql_fetch_object($get_user);if(!$q) die("FAILED. The email address does not exsist."); echo "information send to the submitted email";$to = "$em";$subject = "Your Pawword";$message = "<html>Your password is : <? echo '".$q['user_password']."' ?><br>more html code i know that password should be scambled and stuff, but just for nowin the emailed message the results isyour password is : and then it is empty???why is that?

sorry for the confusioni am getting the variables from a form method postbut when i did change the text fields to txt1, txt2, txt3 it works fine?what more efficient methods do i ned to explore

Quick questionall my textfields inthe form should be named txtright

This is great stuff, i think of many other situation that i have that i can use this loop stuff ofr effectively,Thank you both very much