PHP - Turning A Table Menu Into A List
I got a tab menu, where the user choose his category and a list of sub categories appear. Personally I like the old fashion way, where everything listed like in craiglist.com ...
How can I turn the following tab menu, into a simple 3 column list? <!--CATEGORIES--> <div id="selCategories"> <div class="slidetabsmenu menu_fix"> <ul> <?php if(isset($groups) and $groups->num_rows()>0) { $i=0; foreach($groups->result() as $group) { ?> <li id="gr<?php echo $i;?>" class="<?php if($i == '0') echo "selected"; ?>"><a href="javascript:;" onclick="getCat('<?php echo $i ?>','<?php echo $groups->num_rows ?>',<?php echo $group->id;?>);"><span><?php echo $group->group_name;?></span></a></li> <?php $i++;} } ?> </ul> </div> <div class="clsInfoBox"> <div class="block"> <div class="grey_t"> <div class="grey_r"> <div class="grey_b"> <div class="grey_l"> <div class="grey_tl"> <div class="grey_tr"> <div class="grey_bl"> <div class="grey_br"> <div class="cls100_p"> <h4><span class="clsCategory"><?php echo $this->lang->line('CATEGORIES');?></span></h4> <div class="clsCategoryList clearfix" id="catInner"> </div> </div> </div> </div> </div> </div> </div> </div> </div> </div> </div> </div> </div> <!--END OF CATEGORIES--> Your tips, directions, and support is highly appreciated. Similar TutorialsI know I'm doing it something right, but can someone tell me why only one table is showing up? Can you help me fix the issue? Heres my code: function showcoords() { echo"J3st3r's CoordVision"; $result=dbquery("SELECT alliance, region, coordx, coordy FROM ".DB_COORDFUSION.""); dbarray($result); $fields_num = mysql_num_fields($result); echo "<table border='1'>"; // printing table headers echo "<td>Alliance</td>"; echo "<td>Region</td>"; echo "<td>Coord</td>"; // printing table rows while($row = mysql_fetch_array($result)) { // $row is array... foreach( .. ) puts every element // of $row to $cell variable foreach($row AS $Cell) echo "<tr>"; echo "<td>".$row['alliance']."</td>\n"; echo "<td>".$row['region']."</td>\n"; echo "<td>".$row['coordx'].",".$row['coordy']."</td>\n"; echo "</tr>\n"; } echo "</table>"; mysql_free_result($result); } I have 2 rows inserted into my coords table. Just frustrated and ignorant to php. Hi I am having a problem with the code below I am using to send a simple php email form. I am not that advanced with php but I want to add a drop down menu to a form I created and not sure how to get get the message (warning not completed) in the field - if(empty($visitortitle)) { - to accept or not to accept? This is the code in the form: <td align="left"><select name="visitortitle"> <option>Please Select</option> <option>Mr</option> <option>Mrs</option> <option>Miss</option> <option>Ms</option> <option>Dr</option> </select> This is the code I have got in the sendemail: if(empty($visitortitle)) { echo "<h2><br /><br /><br /><br />Please hit the back button and enter the title drop down box correctly<br />before you try submitting the form again.</h2>\n"; die ( '<a href="forum.html">click here go back and try again</a>' ); } How can I get this code to recognise that a <option></option> has been selected? Any help would be appreciated. Thanks Gary Hi all, i want my list/menu field values to come from my database. how can i accomplish that? thanks i did Code: [Select] <select name="select"> <option value="0">--select below--</option> <option value="1">Me</option> <?php require_once '../konnect/konex.php'; $result = mysql_query("SELECT * FROM is_clients"); while($row = mysql_fetch_array($result)) { echo "<option value ='2'>".$row"['name']</option>"; echo "<br />"; } ?> </select> Im using dreamweaver, and I have a simple list menu that is linked to a dynamic source in my database. I cannot figure out how to create an action when they select something from the list menu. I suppose that it would have to be passed through the browser somehow. Im not very code fluent, but I try. If you know to accomplish this using dreamweaver cs4 or hard coding, I would appreciate it. I want this: When you click on an item it will bring up all the data pertaining to that particular record. My code: Code: [Select] <form id="form1" name="form1" method="post" action=""> <label for="customerlist">Select your customers</label> <p><label for="customers"></label> <select name="customers" id="customers"> <option value="None">None</option> <?php do { ?> <option value="<?php echo $row_Customers['name']?>"><?php echo $row_Customers['name']?></option> <?php } while ($row_Customers = mysql_fetch_assoc($Customers)); $rows = mysql_num_rows($Customers); if($rows > 0) { mysql_data_seek($Customers, 0); $row_Customers = mysql_fetch_assoc($Customers); } ?> </select> <input type="submit" name="Go" id="Go" value="Go" /> Hi there, Im trying to build a drop down menu that has an if statement to determine the results, heres the code below: Code: [Select] <option value="<?php echo $row_Planet['PlanetName']?>"<?php if ($row_Planet['PlanetName'] == 'Mon Calamari'){ echo"Yavin"; echo"Denab";) ?><?php echo $row_Planet['PlanetName']?></option> Im really unsure of the syntax. If you could help that would be ace! Thanks I have a mysql table with the structure of Code: [Select] ID Menu_Name Parent_ID 1 Finance NULL 2 Business NULL 3 Investment 1 4 Trading 2 How can I create a html <ul><li> list based on the parent? hi every one i want to make a list from my database and if user click on one item it open a link... its a list of models of products so if click in one link me to the page of its full description, my database has the fields: "modelo" (model), and "paginal" the link page i make this code.. works fine cause display all the catalog but when i click in an item dont do nothing... hes the code.. what im missing?? tnx in advance <?PHP // Connect to MySQL Server @mysql_connect('localhost','user','password') or DIE("Couldn't Connect to MySQL Database"); // Select Database @mysql_select_db('zerocctv_almacen') or DIE("Couldn't Select Database"); $sql = "SELECT modelo, paginal,precio,foto FROM modelos"; $result=mysql_query($sql); $row_array=mysql_fetch_row($result); $string='<select name=selection><option value=>modelo</option>'; for ($i=0;$i < mysql_num_rows($result);$i++) { if ($row_array[0]=="") { $row_array=mysql_fetch_row($result); } else { $string .='<option value="'.$row_array[1].'">'.$row_array[0]."</option>"; $row_array=mysql_fetch_row($result); } } $string .='</SELECT>'; //echo "</label>"; echo $string; ?> i have made a simple header page for a project.
now i want to create a dropdown menu for a single menu item in the menu bar.
how i will do that ..??
for ex-under COURSES MENU, THE SUBMENU ARE : DEGREE,DIPLOMA,HIGHSCHOOL.
I am very new new php (wrote my first PHP script 5 Days ago) and am trying to give myself a crash course but I have hit a pit-stop which is killing me a little! I hope that title makes sense.... Basically I created PHP script to take data from a database and display in, I then wrote some code to use a drop down menu to order that data. That all worked ok until I tried to utilise some pagination. I can make the pagination work, and I can make the ordering work, but not at the same time! At the moment the code that I have will allow me to order the list and almost paginate it. There are 40 results and I want to display 10 at a time. When not using the ordering code I can paginate it perfectly but when I try to intergrate the two bits of code it will only display the first 10 results and not give me an option to go to the next page to see the rest! The code: if (!isset($_GET['start'])) { $_GET['start'] = 0; } $per_page = 10; $start = $_GET['start']; if (!$start) $start = 0; $sort = @$_POST['order']; if (!empty($sort)) { $get = mysql_query("SELECT bookname, bookauthor, bookpub, bookisbn FROM booktable ORDER BY ".mysql_real_escape_string($_POST['order'])." ASC LIMIT $start, $per_page"); } else { $get = mysql_query("SELECT bookname, bookauthor, bookpub, bookisbn FROM booktable ORDER BY bookname ASC LIMIT $start, $per_page"); } $record_count = mysql_num_rows($get); ?> <?php if (isset($_GET['showerror'])) $errorcode = $_GET['showerror']; else $errorcode = 0; ?> wont include all the html rubbish and the ordering menu! <div id="mid"> <?php echo "<table>"; echo "<tr>"; echo "<th>"; echo "</th>"; echo "<th>"; echo "Book Title"; echo "</th>"; echo "<th>"; echo "Book Author"; echo "</th>"; echo "<th>"; echo "Book Publisher"; echo "</th>"; echo "<th>"; echo "Book ISBN"; echo "</th>"; echo "<th>"; echo "</th>"; echo "</tr>"; while ($row = mysql_fetch_assoc($get)) { // get data $bookname = $row['bookname']; $bookauthor = $row['bookauthor']; $bookpub = $row['bookpub']; $bookisbn = $row['bookisbn']; echo "<tr>"; echo "<td>"; echo "<a href='addtolist.php?bookname=".$bookname."&bookauthor=".$bookauthor."&bookpub=".$bookpub."&bookisbn=".$bookisbn."'>Add to basket</a>"; echo "</td>"; echo "<td>"; echo $bookname; echo "</td>"; echo "<td>"; echo $bookauthor; echo "</td>"; echo "<td>"; echo $bookpub; echo "</td>"; echo "<td>"; echo $bookisbn; echo "</td>"; echo "</tr>"; } echo "</table>"; $prev = $start - $per_page; $next = $start + $per_page; if (!($start<=0)) echo "<a href='products.php?start=$prev'>Prev</a> "; //set variable for first page number $i=1; //show page numbers for ($x = 0; $x < $record_count; $x = $x + $per_page) { if ($start != $x) echo "<a class='pagin' href='products.php?start=$x'> $i </a>"; else echo "<a class='pagin' href='products.php?start=$x'><b> $i </b></a>"; $i++; } //show next button if (!($start >= $record_count - $per_page)) echo "<a class='pagin' href='products.php?start=$next'> Next </a>"; ?> Thank you so much for reading! having a problem turning this sql into a query and I need it to only update rows where id field is empty. Code: [Select] UPDATE systems SET naqahdah = naqahdah + naqahdah_mines * 2 I tried this out with failure Code: [Select] <?php mysql_query("UPDATE systems SET naqahdah= naqahdah + naqahdah_mines * 2 WHERE id <> ' ' "); ?> so i tried getting rid of WHERE and couldnt even get this to work. I tested the sql out it works fine. Code: [Select] <?php mysql_query("UPDATE systems SET naqahdah= naqahdah + naqahdah_mines * 2 "); ?> Hey guys, I made in php an sql droplist. But I need some help. I want it to work like this. When I select something from the list and click Add To Cart, that also the product gets send to the cart and not only the quantity. you can check it out here : http://fhcs.be/cart-demo2/index.php I also added my index.php as an attachment Greets Now that i have my code that shows my list on the localhost I want to try and put this into a table and beable to add a delete from the table. Here's my code any help would be much appreciated. <?PHP // please add login and pass here// $host = "localhost"; $login = "root" ; $pass = ""; mysql_connect("$host","$login","$pass") OR DIE ("There is a problem with the system. Please notify your system administrator." .mysql_error()); //Seems in this case we can use a general call $connection = mysql_connect("$host","$login","$pass") or die(mysql_error()); $dbs = @mysql_list_dbs($connection)or die(mysql_error()); $db_list="<ul>"; $i =0; while ($i < mysql_num_rows($dbs)){ $db_names[$i] = mysql_tablename($dbs,$i); $db_list .="<li>$db_names[$i]"; $i++; } //Start Create DB// IF (isset($_POST['result'])){ $database=$_POST['database']; $sql="CREATE DATABASE $database "; $result = mysql_query($sql,$connection) or die(mysql_error()); echo "Database $database has been added"; } IF (isset($_POST['delete'])){ $db=$_POST['db']; $query=mysql_query("DROP DATABASE $db"); echo "Database $db has been deleted"; } ?> <html> <head> <title>MySQL Databases</title> </head> <body> <p><strong>Databases on localhost</strong>:</p> <? echo "$db_list"; ?> <?PHP //print_r($_POST); ?> <form action="pretask.php" method="post"> <select name="db"> <?PHP $db_list = mysql_list_dbs($connection); while ($row = mysql_fetch_object($db_list)) { //Here you are listing anything that should not be included if ($row->Database!="information_schema" && $row->Database!="mysql" && $row->Database!="phpmyadmin"){ echo "<option value=\"".$row->Database."\">".$row->Database."</option>"; } } ?> </select> <input type="submit" name="delete" value="Delete"/> </form> <form action="pretask.php" method="post"> Create Database <input type="text" name="database" /> <input type="submit" name="result" value="Create" /> </form> </body> I have some code where i input data and upload images to a database (working 100% correctly) However I want to expand on it - what I want is when a user uploads an image of their choice, to assign it to a directory of their choosing. Then, when input is successful, write the directory of where the image is stored (not just the image) to the database. Here is the code; Code: [Select] <?php ini_set('display_errors',1); error_reporting(-1); require_once ('./includes/config.inc.php'); require_once (MYSQL); $add_cat_errors = array(); if ($_SERVER['REQUEST_METHOD'] == 'POST') { // Check for a name: if (empty($_POST['product'])) { $add_cat_errors['product'] = 'Please enter the name!'; } // Check for a description: if (empty($_POST['prod_descr'])) { $add_cat_errors['prod_descr'] = 'Please enter the description!'; } // Check for a category: if (!isset($_POST['cat']) || !filter_var($_POST['cat'], FILTER_VALIDATE_INT, array('min_range' => 1))) { $add_product_errors['cat'] = 'Please select a category!'; } // Check for a price: if (empty($_POST['price']) || !filter_var($_POST['price'], FILTER_VALIDATE_FLOAT) || ($_POST['price'] <= 0)) { $add_cat_errors['price'] = 'Please enter a valid price!'; } // Check for an image: if (is_uploaded_file ($_FILES['image']['tmp_name']) && ($_FILES['image']['error'] == UPLOAD_ERR_OK)) { $file = $_FILES['image']; $size = ROUND($file['size']/1024); // Validate the file size: if ($size > 512) { $add_cat_errors['image'] = 'The uploaded file was too large.'; } // Validate the file type: $allowed_mime = array ('image/jpeg', 'image/JPG', 'image/jpg'); $allowed_extensions = array ('.jpg', 'jpeg'); $image_info = getimagesize($file['tmp_name']); $ext = substr($file['name'], -4); if ( (!in_array($file['type'], $allowed_mime)) || (!in_array($image_info['mime'], $allowed_mime) ) || (!in_array($ext, $allowed_extensions) ) ) { $add_cat_errors['image'] = 'The uploaded file was not of the proper type.'; } // Move the file over, if no problems: if (!array_key_exists('image', $add_cat_errors)) { // Create a new name for the file: $new_name = (string) sha1($file['name'] . uniqid('',true)); // Add the extension: $new_name .= ((substr($ext, 0, 1) != '.') ? ".{$ext}" : $ext); // Move the file to its proper folder but add _tmp, just in case: $dest = "../db/images/$new_name"; if (move_uploaded_file($file['tmp_name'], $dest)) { // Store the data in the session for later use: $_SESSION['image']['new_name'] = $new_name; $_SESSION['image']['file_name'] = $file['name']; // Print a message: echo '<h4>The file has been uploaded!</h4>'; } else { trigger_error('The file could not be moved.'); unlink ($file['tmp_name']); } } // End of array_key_exists() IF. } elseif (!isset($_SESSION['image'])) { // No current or previous uploaded file. switch ($_FILES['image']['error']) { case 1: case 2: $add_cat_errors['image'] = 'The uploaded file was too large.'; break; case 3: $add_cat_errors['image'] = 'The file was only partially uploaded.'; break; case 6: case 7: case 8: $add_cat_errors['image'] = 'The file could not be uploaded due to a system error.'; break; case 4: default: $add_cat_errors['image'] = 'No file was uploaded.'; break; } // End of SWITCH. } // End of $_FILES IF-ELSEIF-ELSE. // Check for a stock: if (empty($_POST['stock']) || !filter_var($_POST['stock'], FILTER_VALIDATE_INT, array('min_range' => 1))) { $add_cat_errors['stock'] = 'Please enter the quantity in stock!'; } if (empty($add_cat_errors)) { $query = "INSERT INTO product (product, prod_descr, catID, price, image, stock) VALUES (?, ?, ?, ?, ?, ?)"; // Prepare the statement: $stmt = mysqli_prepare($dbc, $query); // For debugging purposes: // if (!$stmt) echo mysqli_stmt_error($stmt); // Bind the variables: mysqli_stmt_bind_param($stmt, 'sssssi', $name, $desc, $_POST['cat'], $_POST['price'], $_SESSION['image']['new_name'], $_POST['stock']); // Make the extra variable associations: $name = strip_tags($_POST['product']); $desc = strip_tags($_POST['prod_descr']); // Execute the query: mysqli_stmt_execute($stmt); if (mysqli_stmt_affected_rows($stmt) == 1) { // If it ran OK. // Print a message: echo '<h4>The product has been added!</h4>'; // Clear $_POST: $_POST = array(); // Clear $_FILES: $_FILES = array(); // Clear $file and $_SESSION['image']: unset($file, $_SESSION['image']); } else { // If it did not run OK. trigger_error('The product could not be added due to a system error. We apologize for any inconvenience.'); unlink ($dest); } } // End of $errors IF. } else { // Clear out the session on a GET request: unset($_SESSION['image']); } // End of the submission IF. require_once ('./includes/form_functions.inc.php'); ?> <form enctype="multipart/form-data" action="add_product.php" method="post" accept-charset="utf-8"> <input type="hidden" name="MAX_FILE_SIZE" value="524288" /> Product<br /><?php create_form_input('product', 'text', $add_cat_errors); ?> Description<br /><?php create_form_input('prod_descr', 'textarea', $add_cat_errors); ?> Category<br /><select name="cat"<?php if (array_key_exists('cat', $add_cat_errors)); ?>> <option>Select One</option> <?php // Retrieve all the categories and add to the pull-down menu: $q = 'SELECT catID, cat FROM category ORDER BY cat ASC'; $r = mysqli_query ($dbc, $q); while ($row = mysqli_fetch_array ($r, MYSQLI_NUM)) { echo "<option value=\"$row[0]\""; // Check for stickyness: if (isset($_POST['cat']) && ($_POST['cat'] == $row[0]) ) echo ' selected="selected"'; echo ">$row[1]</option>\n"; } ?> </select><?php if (array_key_exists('cat', $add_cat_errors)) echo $add_product_errors['cat']; ?> Price<br /><?php create_form_input('price', 'text', $add_cat_errors); ?> Image<br /><?php // Check for an error: if (array_key_exists('image', $add_cat_errors)) { echo $add_cat_errors['image'] . '<br /><input type="file" name="image"/>'; } else { // No error. echo '<input type="file" name="image" />'; // If the file exists (from a previous form submission but there were other errors), // store the file info in a session and note its existence: if (isset($_SESSION['image'])) { echo "<br />Currently '{$_SESSION['image']['file_name']}'"; } } // end of errors IF-ELSE. ?> <br /> <select name="select"> <option value="full_heads">db/images/full_heads</option> <option value="human_hair">db/images/human_hair</option> </select> Stock<br /><?php create_form_input('stock', 'text', $add_cat_errors); ?> <input type="submit" value="Add This Product" class="button" /> </fieldset> </form> How do I achieve this? I am trying to convert and old website from mysql database to sqlite. One of the chores it must do is collect information from the database and put it in a list/menu select box on a page so the user can choose which item to pursue.
In the following (incomplete) snippit, I am doing something incorrectly because the sql query does get the proper information (I can put it in a table on the page just fine). But I'm having trouble getting the information into the select options on a list/menu. It appears to be putting them all, one after the other in the first option spot. The last one seems to be the only one of 6 or 7 that shows up.
It's been 10 or 12 years since I've messed with php, so I think I'm way behind... any help would be appreciated.
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<form action="sqlPropDisplay.php" method="post" id="Residential"> I am trying to turn a .txt file into an array, count the items in the array, and display a random quote from it.
<p class = "center"> <?php $Proverb = file('../proverbs.txt'); $ProverbCount = count($Proverb); echo rand(0, $ProverbCount-1); ?> <br> © 2014 </p>The issue I am having is that it states the file does not exist: FULL PATH: G:\EasyPHP-5.3.2i\www\PHP_Projects\ChineseZodiacs\Includes\inc_footer.php TEXT FILE: G:\EasyPHP-5.3.2i\www\PHP_Projects\ChineseZodiacs\proverbs.txt Please help, thank you. Hey i'm having trouble figuring out how to do this. This is the code grabbing the users who are online and listing them one by one. Code: [Select] $qt=mysql_query("SELECT username FROM users WHERE lastvisit > '$tm' and online='ON'"); echo mysql_error(); while($nt=mysql_fetch_array($qt)){ echo "<tr><td> $nt[username] </td></tr>"; } ?> what i want to do is have each username be a hyperlink that links to their profile how would i go about this? cause i'm completely clueless on this. is there some way i can do it in the php or will i need to make a varchar column for each player that holds their profile link and grab that with the query too? Hi, I've got a basic sign up form but I want a drop down list which will list different catergories that relate to different tables which when selected will input the sign up information into that table which was selected from the catergory drop down. This is the signup form <html><head><title>Birthdays Insert Form</title> <style type="text/css"> td {font-family: tahoma, arial, verdana; font-size: 10pt } </style> </head> <body> <table width="300" cellpadding="5" cellspacing="0" border="2"> <tr align="center" valign="top"> <td align="left" colspan="1" rowspan="1" bgcolor="64b1ff"> <h3>Insert Record</h3> <form method="POST" action="test.php"> <? print "Enter Company Name: <input type=text name=company_name size=30><br>"; print "Enter Contact Name: <input type=text name=contact_name size=30><br>"; print "Enter Telephone: <input type=text name=telephone size=20><br>"; print "Enter Fax: <input type=text name=fax size=30><br>"; print "Enter Email: <input type=text name=email size=30><br>"; print "Enter Address: <input type=text name=address1 size=20><br>"; print "Enter Address: <input type=text name=address2 size=30><br>"; print "Enter Postcode: <input type=text name=postcode size=30><br>"; print "Enter Town / City: <input type=text name=town_city size=20><br>"; print "Enter Website: <input type=text name=website size=30><br>"; print "Enter Company Type: <select name='table'> <option>stationary</option><option>reception</option></select><br>"; print "<br>"; print "<input type=submit value=Submit><input type=reset>"; ?> </form> </td></tr></table> </body> </html> This is the part which I can't figure out and is probably totally wrong! Im trying to use this script to sort the drop down list to then run the correct script to insert the form data. <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Hello!</title> </head> <body> <?php if($_POST['table']=='stationary' 'birthdays_insert_record.php') else if($_POST['table']=='reception' 'insert_reception.php') ?> </body> </html> This is the script which works! that inserts the form data into a specific table <html><head><title>Birthdays Insert Record</title></head> <body> <? /* Change db and connect values if using online */ $company_name=$_POST['company_name']; $contact_name=$_POST['contact_name']; $telephone=$_POST['telephone']; $fax=$_POST['fax']; $email=$_POST['email']; $address1=$_POST['address1']; $address2=$_POST['address2']; $postcode=$_POST['postcode']; $town_city=$_POST['town_city']; $website=$_POST['website']; $db="myflawlesswedding"; $link = mysql_connect('localhost', 'root' , ''); if (! $link) die(mysql_error()); mysql_select_db($db , $link) or die("Select Error: ".mysql_error()); $result=mysql_query("INSERT INTO reception (company_name, contact_name, telephone, fax, email, address1, address2, postcode, town_city, website) VALUES ( '$company_name', '$contact_name', '$telephone', '$fax', '$email', '$address1', '$address2', '$postcode', '$town_city', '$website')") or die("Insert Error: ".mysql_error()); mysql_close($link); print "Record added"; ?> <form method="POST" action="birthdays_insert_form.php"> <input type="submit" value="Insert Another Record"> </form> <br> <form method="POST" action="birthdays_dbase_interface.php"> <input type="submit" value="Dbase Interface"> </form> </body> </html> I hope somebody can help me out here! or can point me in a better way to sort this problem! Thanks for any advice! right now i have my date system to say "X amount of seconds ago, X hours ago, X days ago, X years ago, How could i turn mktime() (1290874161) into a mm/dd/yyyy format? so it would say like, "Nov 27, 2010 at Hour:Minute" (whatever the hour/minute is) |