Jump to content

Search the Community

Showing results for tags 'variables'.

  • Search By Tags

    Type tags separated by commas.
  • Search By Author

Content Type


Forums

  • Welcome to PHP Freaks
    • Announcements
    • Introductions
  • PHP Coding
    • PHP Coding Help
    • Regex Help
    • Third Party Scripts
    • FAQ/Code Snippet Repository
  • SQL / Database
    • MySQL Help
    • PostgreSQL
    • Microsoft SQL - MSSQL
    • Other RDBMS and SQL dialects
  • Client Side
    • HTML Help
    • CSS Help
    • Javascript Help
    • Other
  • Applications and Frameworks
    • Applications
    • Frameworks
    • Other Libraries
  • Web Server Administration
    • PHP Installation and Configuration
    • Linux
    • Apache HTTP Server
    • Microsoft IIS
    • Other Web Server Software
  • Other
    • Application Design
    • Other Programming Languages
    • Editor Help (Dreamweaver, Zend, etc)
    • Website Critique
    • Beta Test Your Stuff!
  • Freelance, Contracts, Employment, etc.
    • Services Offered
    • Job Offerings
  • General Discussion
    • PHPFreaks.com Website Feedback
    • Miscellaneous

Find results in...

Find results that contain...


Date Created

  • Start

    End


Last Updated

  • Start

    End


Filter by number of...

Joined

  • Start

    End


Group


AIM


MSN


Website URL


ICQ


Yahoo


Jabber


Skype


Location


Interests


Age


Donation Link

Found 4 results

  1. My login script stores the user's login name as $_SESSION[ 'name'] on login. For some unapparent reason, i'm getting errors stating that $user and $priv are undefined variables, though I've attempted to define $user as being equal to $_SESSION['name'], using $user to look up the the user's privilege level (stored as the su column ) in the SQL table, and then where the result of the sql query is $priv which is then evaluated in an if statement. I can't seem to figure out why this might not be working. The code I'm using: <?php session_start(); function verify() { //verify that t
  2. Hey! I want to create a website that can test your counting pace. I’ll make colored, square div that change color after 10 to 30 seconds when you hover over it, and then you’ll type down how long you think it took in a form to check if you’re right. My solution to do this is to create the PHP variable $time = rand(10,30), and in CSS get the div:hover transition-delay to be $time. Then, after the form, I need the same variable for if($_POST[timeguess] = $time) {echo “Correct!”}. I have no idea how to access the same variable in both the CSS and HTML, so I hope some of you can help me out! Than
  3. I am working on a item helper script. It consists of users selecting from multiple radio buttons for multiple items and updating them all at once. My question is the code below is used for each item the user has (used in "while ($stmt->fetch()){" statement") . With the way I have it set up, how will I make it so it updates the correct item for the right option? Also. how do I grab the bracket variable use_item[$id] ? <tr width=\"100%\"> <td> <p>$name</p> </td> <td> <center><input name=\"use_item[$id]\" type=\"radio\" id=\"use_item[$id]\" va
  4. Help Please! I'm trying to pass variables in a URL and use them in a HTML email. When test1.php calls Test2.php, test2.php should send two HTML emails but nothing happens. I get this error for test2.php: [24-Feb-2016 16:20:42 UTC] PHP Parse error: syntax error, unexpected end of file in /home/my/public_html/test2.php on line 114 HTML - call test1.php <!DOCTYPE html> <html lang="en-US"> <head> <meta charset="utf8" /> <title>Testform.html</title> </head> <body bgcolor="tan" text ="black"> <form action="test1.php" meth
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.