Jump to content

Search the Community

Showing results for tags 'mysqli'.

  • 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 20 results

  1. Hi all, I am trying to get data from MySQL to display in a html table in TCPDF but it is only displaying the ID. $accom = '<h3>Accommodation:</h3> <table cellpadding="1" cellspacing="1" border="1" style="text-align:center;"> <tr> <th><strong>Organisation</strong></th> <th><strong>Contact</strong></th> <th><strong>Phone</strong></th> </tr> <tbody> <tr>'. $id = $_GET['id']; $location = $row['location']; $sql = "SELECT * FROM tours WHERE id = $id"; $result = $con->query($
  2. Hi guys, I really hope this will make sense. I am creating a dynamic field on a button click for Pickup Location. That works fine and submitting the form to the database works fine. However, instead of one entry, each time I submit the form with multiple Pickup Locations, it creates multiple separate database entries. Here is the PHP for submitting: if(isset($_POST['new']) && $_POST['new']==1){ $pickups = ''; foreach($_POST['pickups'] as $cnt => $pickups) $pickups .= ',' .$pickups; $locations = count($_POST["pickups"]); if ($locations > 0) { for ($
  3. Hi all, Hope to find you all good. I have the following, which creates a php file. This works fine and without error. However, once created, the content of the page, which is got from the Database, is not showing. <?php include_once('includes/header.php'); if(isset($_POST['new']) && $_POST['new']==1){ if(isset($_POST['submit'])){ $trn_date = mysqli_real_escape_string($con, date("Y-m-d H:i:s")); $name = mysqli_real_escape_string($con, $_POST['name']); $description = mysqli_real_escape_string($con, $_POST['description']); $body = mysqli_real_escape_strin
  4. Hi, this query runs fine when I run it from PHPMyAdmin: UPDATE `tran_term_taxonomy` SET `description` = (SELECT keyword from `good_keywords` ORDER BY RAND() LIMIT 1,1) WHERE `tran_term_taxonomy`.`taxonomy` = 'post_tag' AND `tran_term_taxonomy`.`description` = "" LIMIT 1 However, when I run the same query in a PHP file on my server, the page doesn't load at all. The message I get is: www.somesite.com is currently unable to handle this request. HTTP ERROR 500. This is my PHP code: <?php include("/database/connection/path/db_connect.php"); $result4 = mysqli_query($GLOBALS["__
  5. Hi All, I am adding a button that will delete many things in the system which all have the same $job_id There is going to be 10 or so tables that need to delete * where $job_id = ? Is there a better way to do this rather than just iterating my code 10 times. I was looking at just joining but isnt one benefit of prepared statements that they can be used again and again to increase speed.
  6. HI All Not sure how best to describe what i am trying to do so here goes. People use my system to order food from a menu - this is not for a restaurant where you order one starter main and desert, they will be ordering 00's of meals. The table layout is as follows: ssm_menu menu_id | menu_name | menu_price ssm_menu_connection menu_id | menu_item_id | surrogate_id ssm_menu_items menu_item_id | menu_item_name | menu_item_category ssm_menu_order job_id | menu_id | menu_item_id | menu_item_qty Menu items can appear on more than one menu.
  7. Hi! I'm new at php and I need a help with this code I need to Search an employee from a Database that I already have I have the Search Form and I write the name and the lastname and the details of the employee appear, But I need help. I already have the search, but I tells me when I call the query that I have a fatal error Trying to get property of non-object in in this part of the code $count = $query -> num_rows;and this Fatal error: Call to undefined method EmployeeModel::searchEmployees() in here $employees = $this->model->searchE($_GET['str']); require_once('views
  8. Hi guys, I am trying to display data to a user who submitted it. So, user A submits data to the db and also user B. I want it so that user A cannot see user B's data but only the data they have submitted. Hope that makes sense. Here is the code along with the table of data to be displayed. <div class="panel-body"> <div class="table-responsive"> <table class="table table-striped table-bordered table-hover" id="bookings_table"> <thead>
  9. If anyone could help, I'm not getting an error message and I've tried changing the code around a few times and outputting errors. The database does not update and i'm getting no error message output on submit, I assume i'm missing something? <?php session_start(); include_once 'dbconnect.php'; if (!isset($_SESSION['userSession'])) { header("Location: index.php"); } $query = $DBcon->query("SELECT * FROM tbl_users WHERE user_id=".$_SESSION['userSession']); $userRow=$query->fetch_array(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.o
  10. Hi there I am a new at this website, thanks for helping me... I have this code: <?php $select_active_shopes="SELECT tbl_shopkeepers_info.*, tbl_active_shopkeepers.* FROM tbl_shopkeepers_info INNER JOIN tbl_active_shopkeepers ON(tbl_active_shopkeepers.shopkeeper_id=tbl_shopkeepers_info.shopkeeper_id) ORDER BY tbl_active_shopkeepers.active_shopkeeper_id DESC"; $rs_active_shopes=mysqli_query($link,$select_active_shopes) or die(mysqli_error($link)); ?> <div class="container-fluid text-center my-3"> <h2 class="font-weight-light"></h2> <
  11. Hello everyone, I am trying to submit a comment in a comment box and send it to the DB but is not happening. The connection is good as I am logging in and all but no data is sent to the DB when I post the comment. It doesn't show in my comment section either. Form <!--comment section--> <?php if(isset($_SESSION['id'])) { echo "<form method='POST' action='" . setComments($conn) . "'> <input type='hidden' name='uidUsers' value='".$_SESSION['id']."'> <input type='hidden' name='posted' value='" . date('Y-m-d H:i:s') . "'>
  12. Hi guys and Gals, This has my head wrecked to be honest. I am trying to upload an image to a directory, which is working. However, I also want to put the file name into MySQL. This will work if the image upload script is removed. With the script enabled, the file uploads but I get "Undefined index: userPic" from the following line: $userPic = mysqli_real_escape_string($mysqli, $_POST['userPic']); Here is the complete code: if(isset($_POST['Submit'])){//if the submit button is clicked $company_name = mysqli_real_escape_string($mysqli, $_POST['company_name']); $company_abn = mysqli
  13. I created a recipe site that displays various dishes, and I'm tickled pink that I got it to work (first time doing this!) But I'd like to have pages that displays dishes by its category. The column that houses this is called "category" which has 7 total: Appetizers & Beverages, Soups & Salads, Side Items, Main Dishes, Baked Goods, Desserts, and Cookies & Candy. I have this page connected to a "connection.php" page, but here's the code in question: <!-- Page Title--> <div class="row"> <div class="col-lg-12"> <h1 class="page-header">Appet
  14. I'm being given an error by mysql, "Warning: mysqli_connect() expects parameter 1 to be string, object given in /storage/...", the relevant lines are $code = $this->GenerateCode($num); $enCoded = password_hash($code, PASSWORD_BCRYPT); $query = "INSERT INTO `vaults` " . "(`Alive`, `Contents`, `Code`, `digits`, `Winning_Player_ID`, `Won_Date`, `debug`)" . "VALUES " . "(CONV('1', 2, 10) + 0, '0', '{$enCoded}', '$num', NULL, NULL, '{$code}')"; Followed by the mysqli_connect($this->db, $query).
  15. I have 3 staff members that need to pick vacation in a certain order. Each have X weeks vacation and can pick off a calendar their choice 1-3 weeks per round. I'm trying to loop through the DB to find who is next to pick with weeks left ~~~~~~First round of picking~~~~~~ STAFF 1 - PICK ORDER 1 - 3 weeks available STAFF 2 - PICK ORDER 2 - 5 weeks available STAFF 3 - PICK ORDER 3 - 3 weeks available Staff 1 takes 2 Weeks Staff 2 takes 1 Weeks Staff 3 takes 3 Weeks ~~~~~~Second round of picking~~~~~~ STAFF 1 - PICK ORDER 1 - 1 weeks available STAFF 2 - PICK O
  16. Im trying to write some code for a raffle, when someone buys one ticket it works well but if someone buys ten tickets. i would like it to put each one on a new row, the last column is the ticket number which is got by another table called count and i want the new count in the last column of each row. In the actual script there is more than two columns but this is an example just to try to let you know what im trying to do. As you can see i want the ticket number to increment by one every time someone buys tickets. (the ticket number is in a simple table with just id and ticket number
  17. Is there a way to bind parameters in a select SEPARATELY? I know how to do this in one line, but would like some direction on how to do this one line at a time, OR means to stick that array into the bind_param line where it will work with different array lengths. using mysqli to modify some older code. Understanding of course that the ?'s count must match the amount of parameters. End result to make a generic reader function. $sql = "SELECT ID, foo, bar FROM table WHERE ID = ? AND bar = ?"; $vars[] = array('typ' => "i", 'vlu' => $id); $vars[] = array('typ' => "
  18. Hi, I've question regards to the topic title above for, "how to get/post myorder page code to payment page??" which didn't retrieve any data from myorder page or do I need a database for myorder page?. As myorder page uses GET function to collect information from product page that using mysqli SELECT function to get data from the database. Also "how to get/post the Total from the myorder page to payment page??". Below are the following code: - myorder.php page <?php session_start(); include 'db2.php'; if ( !empty($_SESSION["firstname"])) { } else { $_SESSION["firs
  19. Hi all Trying to build a multi level nav for bootstrap from a database below is the code I am using <?php $table = 'tbl_pages_'.$lang; $sql = "SELECT * FROM $table where parent_id = 0 and visible = 1 order by `order` asc"; //echo $sql; $result = mysqli_query($dbConn,$sql) or die(mysqli_error($dbConn)); if(mysqli_num_rows($result)>0){ ?> <nav class="navbar navbar-default navbar-fixed-top"> <div class="container"> <div class="navbar-header"> <button type="button" class="navbar-toggle collapsed" data-toggle="collapse" dat
  20. Hi all, Trying to update two rows in a MySQL table using MySQLi and PHP with values from a bootstrap dropdown. No errors or anything (have error reporting turned on), just nothing is updated in the database. Here is the update code (if anyone has a prepared version of it would love to have it). if(isset($_POST['Update'])){//if the submit button is clicked if(isset($_GET['issue_id']) && isset($_POST['issue_priority']) && isset($_POST['issue_status'])) { $sql="UPDATE maintenance_requests SET issue_status='$issue_status', issue_priority='$issue_priority' WHERE issue
×
×
  • 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.