Jump to content

Search the Community

Showing results for tags 'help please'.

  • 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 (PhpStorm, VS Code, 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 1 result

  1. Hey thanks for reading my thread, Currently im working on a Status Page for my Game Hacking Website. What i want todo is have a back Panel that i can Select "Game A" from a drop down box, and select the Detection status from another drop down box. Click the submit button, and it changes the values in the Database. I'm new to coding in PHP. <?PHP // Create connection $con=mysqli_connect("localhost","Username","Password","Database"); // Check connection if (mysqli_connect_errno($con)) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $result = mysqli_query($con,"SELECT * FROM haxstatus"); $Hackname=""; $Status=""; $update=""; while($row = mysqli_fetch_array($result)) { $id=$row["Status"]; $name=$row["GameName"]; $Hackname.="<OPTION VALUE=\"$id\">".$name.'</option>'; } $list = mysqli_query($con,"SELECT * FROM haxstatus"); //dump the sql table in a table echo "<table border='1'> <tr> <th>Game Name:</th> <th>Detection Status:</th> </tr>"; while($row = mysqli_fetch_array($list)) { echo "<tr>"; echo "<td>" . $row['GameName'] . "</td>"; echo "<td>" . $row['Status'] . "</td>"; echo "</tr>"; } echo "</table>"; ?> <p> </p> <p> </p> <p> <SELECT NAME=name> <OPTION VALUE=0>Game Name <? echo $Hackname?> </SELECT> <SELECT Name="Detection"> <OPTION VALUE=0>Edit Detection Status <OPTION VALUE=1>Detected <OPTION VALUE=2>Undetected </SELECT> <form action="update.php" method="post"> <input type="submit"> </form> <?PHP $con=mysqli_connect("localhost","Username","Password","Database"); if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } // i attempted to Update the table haxstatus and set Status to the drop down box named detection $sql="UPDATE haxstatus SET Status=$_POST[Detection] WHERE ('GameName')"; if (!mysqli_query($con,$sql)) { die('Error: ' . mysqli_error()); } echo "Updated"; mysqli_close($con); ?> When i click submit i get this error Warning: mysqli_error() expects exactly 1 parameter, 0 given in /home/gamead/public_html/update.php on line 15 Error: Thank you in advanced if you choose to help.
×
×
  • 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.