PHP - Duplication List Values
<?php // Daniel URL duplicator. // Input links list. $linksList = "links.txt"; // How many times to duplicate the url? $manyTimes = 1000; // Read in the list. for ($x = 0; $x <= $manyTimes; $x++) { $handle = fopen($linksList, "r"); $line = fgets($handle); echo $line; fclose($handle); } ?>Hey Guys, I'm stuck on this simple bit of code lol what I'm trying to do is load in a list of urls: site1.com site2.com site3.com etc For each site that is read, I'm trying to duplicate it X times, above would print to screen the same url 1000 times, then move onto the next print it 1000 times etc until the list is done (or how ever many times I select) I can't think of the best way to do it! any help would be appreciated guys! Graham Similar Tutorials
Table Issue - Multiple Location Values For User Pushes Values Out Of Row Instead Of Wrapping In Cell
Hi, In my mysql database i have a text input option, in the registration form and edit my details form i have a multiple select dropdown list, which user selects options to populate the text input box, which ultimately populates the text field in the mysql database. All works perfectly. The dropdownlist consists of 3 parts <optgroups> first is current selection (what is the usesr current selection)works fine, The second <optgroup> is existing words, what words we(the site) have given as options, and the third <optgroup> is the words that others have used. This is where im having a small problem. Because its a text field when i call the data from the database, it calls the entire text box as a single option in my select list.. I want to break the words in the text field (at the comma) and have them listed each one as an option in the select list. Example what i need: Words in text box:(my input allows the "comma") word1, word2, word3, word4, word5, word6, How i want them called/displayed: <option value=\"word1\">word1</option> <option value=\"word2\">word2</option> <option value=\"word3\">word3</option> <option value=\"word4\">word4</option> <option value=\"word5\">word5</option> <option value=\"word6\">word6</option> here's my code: $query = "SELECT allwords FROM #__functions_experience WHERE profile_id = '".(int)$profileId."' LIMIT 1"; $original_functionsexperience =doSelectSql($query,1); $query = "SELECT allwords FROM #__functions_experience WHERE profile_id = '".(int)$profileId."' LIMIT 1"; $functionsexperiencelist=doSelectSql($query); $funcexpList ="<select multiple=\"multiple\" onchange=\"setFunctionsexperience(this.options)\">"; foreach ($functionsexperiencelist as $functionsexperienceal) { $selected=""; if ($functionsexperienceals->allwords == $original_functionsexperience) $selected=' selected="selected"'; $allwords=$functionsexperienceal->allwords; $funcexpList .= "<optgroup label=\"Current selection\"> <option value=\"".$allwords."\" ".$selected." >".$allwords."</option> </optgroup> <optgroup label=\"Existing Words\"> <option value=\"existing1,\">existing1</option> <option value=\"existing2,\">existing2</option> <option value=\"existing3,\">existing3</option> <option value=\"existing4,\">existing4</option> <option value=\"existing5,\">existing5</option> <option value=\"existing6,\">existing6</option> </optgroup> <optgroup label=\"Others added\"> //heres problem <option value=\"".$allwordsgeneral."\">".$allwordsgeneral."</option> </optgroup>"; } $funcexpList.="</select>"; $output['FUNCEXPLIST']=$funcexpList; The result im getting for optgroup others added: word1, word2, word3, word4, word5, how can i get it like this: <option value=\"word1\">word1</option> <option value=\"word2\">word2</option> <option value=\"word3\">word3</option> <option value=\"word4\">word4</option> <option value=\"word5\">word5</option> <option value=\"word6\">word6</option> Hi i'm new to php. I want to get all the values of dropdown list on another page wether selected or not. As part of an image gallery I am trying to INSERT data into a table to make accessible the relevant info of individual images - the problem I have encountered is that each query seems to be being inserted twice, and I can't figure out why. The code = Code: [Select] <?php /*============================================== Connect to MySQL and select the correct database ==============================================*/ mysql_connect("localhost", "uname", "pwd") or die(mysql_error()); echo "Connected to MySQL<br />"; mysql_select_db("gallery_db") or die(mysql_error()); echo "Connected to Database gallery_db.<br />"; /*============================================= Include the files needed to undertake this task =============================================*/ include("functions.php"); /*========================= Set up arrays and variables =========================*/ $store = array(); $exif = array(); $folder = "./images/"; $insrt = 'INSERT INTO images VALUES (img_id, '; $file_name = ''; $file_location =''; $copyright =''; $caption = ''; $file_date = ''; $file_time = ''; $camera = ''; $speed = ''; $fNo = ''; $ISO = ''; /*================================= Open folder and read relevant files =================================*/ if ($handle = opendir('./images')) { while (false !== ($file = readdir($handle))) { if ($file != "." && $file != "..") { $file_name = $file; //>>> File Name <<< $file_location = $folder . $file; //>>> File Location <<< $exif = exif_read_data($file_location, 'EXIF'); //>>> Read exif data <<< /*============================================================== Check if the 'Copyright' info existe in the exif data, and if it doesn't set the $copyright variable to the relevant value. ==============================================================*/ if (array_key_exists('Copyright', $exif)) { $copyright = $exif['Copyright']; } else { $copyright = "copy name"; } $caption = $exif['COMMENT'][0]; //>>> Get the Caption <<< //>>> Re-format DateTime into separate date and time var's $file_date & $file_time <<< $temp = substr($exif['DateTime'], 0, 10); $file_date = substr($temp, 8, 2) . substr($temp, 4, 4) . substr($temp, 0, 4); $file_time = substr($exif['DateTime'], 11, 8); //>>> Set the rest of the variables <<< $camera = $exif['Model']; //>>> Get the camera make and model <<< $speed = $exif['ExposureTime']; //>>> Get the shutter speed <<< $fNo = $exif['COMPUTED']['ApertureFNumber']; //>>> Get the f No. <<< $ISO = $exif['ISOSpeedRatings']; //>>> Get the ISO rating <<< //>>> Fill the $store[] array with the relevant bits for each seperate query <<< $store[] = $insrt . "'" . $file_name . "', " . "'" . $file_location . "', " . "'" . $copyright . "', " . "'" . $caption . "', " . "'" . $file_date . "', " . "'" . $file_time . "', " . "'" . $camera . "', " . "'" . $speed . "', " . "'" . $fNo . "', " . "'" . $ISO . "')"; } } closedir($handle); } /*============================================================================= Get individual queries from $store[] and INSERT INTO images the relevant VALUES =============================================================================*/ $num_queries = count($store); //>>> Check the number of queries in the array $store $loop = 0; //>>> Starting point for query No. bearing in mind that array keys are numbered from zero while ($loop <= $num_queries - 1) { $query = $store[$loop]; if (!mysql_query($query)){ die('Error: ' . mysql_error()); } mysql_query($query); $loop++; } ?> I know it's not pretty, but that can be attended to once working properly. I am quite sure that it will be a very simple solution, probably so simple that I haven't even thought of it. Hi
I am reading in data from a csv file
if (($handle = fopen("data.csv", "r")) !== FALSE) { while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) { echo "User Name: $data[0]"; echo "Booking IDs: $data[1]"; } echo "<hr>";The problem is some of the usernames are duplicated so get outputted the same user name 20 times but with different IDs, obviously but I am unsure how to group them together. I can't seem to figure out how to write if the next username is the same as current then just add another ID not go through the entire loop. Any ideas? Thanks I've recently had a problem with my admin site where PHP sessions, which are stored in a MySQL database, are duplicating and causing the page to return to the login screen. Usually the site works fine in Firefox, but stops working in IE and Safari. This problem suddenly appeared over night with no change to the files. Due to the amount of code, I'm reluctant to post it all, but does anyone have a suggestion on why this could be happening and how to solve it? Hello there, I'm looking for a PHP method to duplicate a line, for example if I wanted to duplicate this line: mail($to,$subject,$message,$headers); How could I specify a value of how many are displayed instead of doing this to send it 3 times: mail($to,$subject,$message,$headers); mail($to,$subject,$message,$headers); mail($to,$subject,$message,$headers); Many thanks Hi all.
how can i make the values show like a list. I tried html line break "<br>" and php \n but all to no avail. It just show all the values in one straigth line.
example of what i want is for the values to appear like this:
1234567890
0987654345
4567890675
instead of :
1234567890 0987654345 4567890675
Thanks
<form data-abide method="post" action=""> <div> <select name=""> <option value="name"> <?php $stmt = $pdo->query("SELECT acct_num FROM table order by id desc"); while ( $row = $stmt->fetch(PDO::FETCH_ASSOC) ) { echo $row['acct_num']; } ?> </option> </select> </div> <div> <label>New Password <small>required</small></label> <input type="password" name="password" id="password" required> <small class="error">New password is required and must be a string.</small> </div> <div> <label>Confirm New Password <small>required</small></label> <input type="password" name="password2" id="password2" required> <small class="error">Password must match.</small> </div> <input name="submit" type="submit" class="button small" value="Change Password"> </form> Edited by Mr-Chidi, 13 November 2014 - 01:34 AM. 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> Hi everyone, I don't know whether this is a PHP or MySql problem, but I think it is the former. The following code queries the database correctly, (and before you ask, there are no duplicate database entries), but the output duplicates every row. e.g., hammer (jpg image) hammer hammer (jpg image) hammer saw (jpg image) saw saw (jpg image) saw screwdriver (jpg image) screwdriver screwdriver (jpg image) screwdriver and so on. I cannot see why the code causes the row to repeat. Code: [Select] <?php session_start(); if (isset($_SESSION['id'])) { // Put stored session variables into local php variable $id = $_SESSION['id']; $userId = $_SESSION['userId']; $userGroup = $_SESSION['userGroup']; $managerId = $_SESSION['managerId']; } include_once("demo_conn.php"); $sql = mysql_query("SELECT * FROM users WHERE id='$id'"); while($row = mysql_fetch_array($sql)) { // Get member data into a session variable $egroup = $row["egroup"]; session_register('egroup'); $_SESSION['egroup'] = $egroup; } $query = mysql_query("SELECT topics.url_big, topics.url_small, topics.title, topics.$egroup, quiz.passState, quiz.userDate FROM topics INNER JOIN quiz ON (topics.managerId = quiz.managerId) WHERE topics.$egroup = 1 ORDER BY title ASC"); while ($row1 = mysql_fetch_array($query)) { echo "<a href='../../wood/wood_tool_images/{$row1['url_big']}' target='_blank'><img src='../../wood/wood_tool_images/{$row1['url_small']}' /><br />\n"; echo "{$row1['title']} <br />\n"; } ?> Hi I have this simple code $one = rand(1,9); $two = rand(1,9); $three = rand(1,9); but I need to change it so that they are all unique and the same number is not used more than once can anyone help? thanks Hi everyone, I am having trouble passing/displaying the values inside of a selected list. I created a add/remove list using Jquery and I tried to display the values passed using foreach and for loops but it is still not working. The values I am trying to get are $existing_mID[$j], which is inside of the option value attribute. Please kindly let me know what should I do in order to get the values and I really appreciate your help. <?php $selected = $_POST['selectto']; if(isset($selected)) { echo "something in selected<br />"; for ($i=0;$i<count($selected);$i++) echo "selected #1 : $selected[$i]"; foreach ($selected as $item) echo "selected: item: $item"; } ?>This is the form Hi i have this drop down list for date which contain 3 selects DAY MONTH YEAR hwo can i make so that when update form select keeps the same value has before help please <?php $months = array('','January','February','March','April','May','June','July','August','September','October','November','December'); echo '<select name="month_of_birth">'; for ($i=1;$i<13;++$i) { echo '<option value="' . sprintf("%02d",$i) . '">' . $months[$i] . '</option>'; } echo '</select>'; echo '<select name="day_of_birth">'; for ($i=1;$i<32;++$i) { echo '<option value="' . sprintf("%02d",$i) . '">' . $i . '</option>'; } echo '</select>'; echo '<select name="year_of_birth">'; $year = date("Y"); for ($i = $year;$i > $year-50;$i--) { $s = ($i == $year)?' selected':''; echo '<option value="' . $i . '" ' . $s . '>' . $i . '</option>'; } echo '</select>'; ?> I have a form where I have inserted 7 pre-populated relational lists. All of the information is pulling correctly from the databases, but when it posts, it's posting the value "ids" instead of the chosen text. The files a www.kcwell.com/gcc_form.php and www.kcwell.com/gccsuccess_form.php How do I set up a query to obtain the data that I need? Help! Please help and amend the code Save without repetition required $read[0] I hope that my example is clear to you I apologize for bothering you. I am not good in English <?php $fileLocation = getenv("DOCUMENT_ROOT") . "/rt.txt"; $file = fopen($fileLocation,"a"); $n1 = $_POST['n1']; $n2 = $_POST['n2']; for($i = 0; $i < count($file); $i++){ $read = explode("|", $file[$i]); if($n1 == $read[0]){ echo 'Repetition'; } else { fwrite($file,$n1.'|'.$n2."\r\n"); }
PHP script return 20 UL LIST values like, < ul >
A < /ul > How to display UL LIST into row wise 5 columns like
A B C D Dear All Members here is my table data.. (4 Columns/1row in mysql table)
id order_no order_date miles How to split(miles) single column into (state, miles) two columns and output like following 5 columns /4rows in mysql using php code.
(5 Columns in mysql table) id order_no order_date state miles 310 001 02-15-2020 MI 108.53 310 001 02-15-2020 Oh 194.57 310 001 02-15-2020 PA 182.22
310 001 02-15-2020 WA 238.57 ------------------my php code -----------
<?php
if(isset($_POST["add"]))
$miles = explode("\r\n", $_POST["miles"]);
$query = $dbh->prepare($sql);
$lastInsertId = $dbh->lastInsertId(); if($query->execute()) {
$sql = "update tis_invoice set flag='1' where order_no=:order_no"; $query->execute();
} ----------------- my form code ------------------
<?php -- Can any one help how to correct my code..present nothing inserted on table
Thank You Edited February 8, 2020 by karthicbabuHi, My company has 240+ locations and as such some users (general managers) cover multiple sites. When I run a query to pull user information, when the user has multiple sites to his or her name, its adds the second / third sites to the next columns, rather than wrapping it inside the same table cell. It also works the opposite way, if a piece of data is missing in the database and is blank, its pull the following columns in. Both cases mess up the table and formatting. I'm extremely new to any kind of programming and maybe this isn't the forum for this question but figured I'd give it a chance since I'm stuck. The HTML/PHP code is below: <table id="datatables-column-search-select-inputs" class="table table-striped" style="width:100%"> <thead> <tr> <th>ID</th> <th>FirstName</th> <th>LastName</th> <th>Username</th> <th>Phone #</th> <th>Location</th> <th>Title</th> <th>Role</th> <th>Actions</th> </tr> </thead> <tbody> <?php //QUERY TO SELECT ALL USERS FROM DATABASE $query = "SELECT * FROM users"; $select_users = mysqli_query($connection,$query);
// SET VARIABLE TO ARRAY FROM QUERY while($row = mysqli_fetch_assoc($select_users)) { $user_id = $row['user_id']; $user_firstname = $row['user_firstname']; $user_lastname = $row['user_lastname']; $username = $row['username']; $user_phone = $row['user_phone']; $user_image = $row['user_image']; $user_title_id = $row['user_title_id']; $user_role_id = $row['user_role_id'];
// POPULATES DATA INTO THE TABLE echo "<tr>"; echo "<td>{$user_id}</td>"; echo "<td>{$user_firstname}</td>"; echo "<td>{$user_lastname}</td>"; echo "<td>{$username}</td>"; echo "<td>{$user_phone}</td>";
//PULL SITE STATUS BASED ON SITE STATUS ID $query = "SELECT * FROM sites WHERE site_manager_id = {$user_id} "; $select_site = mysqli_query($connection, $query); while($row = mysqli_fetch_assoc($select_site)) { $site_name = $row['site_name']; echo "<td>{$site_name}</td>"; } echo "<td>{$user_title_id}</td>"; echo "<td>{$user_role_id}</td>"; echo "<td class='table-action'> <a href='#'><i class='align-middle' data-feather='edit-2'></i></a> <a href='#'><i class='align-middle' data-feather='trash'></i></a> </td>"; //echo "<td><a href='users.php?source=edit_user&p_id={$user_id}'>Edit</a></td>"; echo "</tr>"; } ?>
<tr> <td>ID</td> <td>FirstName</td> <td>LastName</td> <td>Username</td> <td>Phone #</td> <td>Location</td> <td>Title</td> <td>Role</td> <td class="table-action"> <a href="#"><i class="align-middle" data-feather="edit-2"></i></a> <a href="#"><i class="align-middle" data-feather="trash"></i></a> </td> </tr> </tbody> <tfoot> <tr> <th>ID</th> <th>FirstName</th> <th>LastName</th> <th>Username</th> <th>Phone #</th> <th>Location</th> <th>Title</th> <th>Role</th> </tr> </tfoot> </table>
Hi all, I'm a first time poster here and I would really appreciate some guidance with my latest php challenge! I've spent the entire day googling and reading and to be honest I think I'm really over my head and need the assistance of someone experienced to advise the best way to go! I have a multi dimensional array that looks like (see below); the array is created by CodeIgniter's database library (the rows returned from a select query) but I think this is a generic PHP question as opposed to having anything to do with CI because it related to working with arrays. I'm wondering how I might go about searching the array below for the key problem_id and a value equal to a variable which I would provide. Then, when it finds an array with a the matching key and variable, it outputs the other values in that part of the array too. For example, using the sample data below. How would you recommend that I search the array for all the arrays that have the key problem_id and the value 3 and then have it output the value of the key problem_update_date and the value of the key problem_update_text. Then keep searching to find the next occurrence? Thanks in advance, as above, I've been searching really hard for the answer and believe i'm over my head! Output of print_r($updates); CI_DB_mysql_result Object ( [conn_id] => Resource id #30 [result_id] => Resource id #35 [result_array] => Array ( ) [result_object] => Array ( ) [current_row] => 0 [num_rows] => 5 [row_data] => ) Output of print_r($updates->result_array()); Array ( [0] => Array ( [problem_update_id] => 1 [problem_id] => 3 [problem_update_date] => 2010-10-01 [problem_update_text] => Some details about a paricular issue [problem_update_active] => 1 ) [1] => Array ( [problem_update_id] => 4 [problem_id] => 3 [problem_update_date] => 2010-10-01 [problem_update_text] => Another update about the problem with an ID of 3 [problem_update_active] => 1 ) [2] => Array ( [problem_update_id] => 5 [problem_id] => 4 [problem_update_date] => 2010-10-12 [problem_update_text] => An update about the problem with an ID of four [problem_update_active] => 1 ) [3] => Array ( [problem_update_id] => 6 [problem_id] => 4 [problem_update_date] => 2010-10-12 [problem_update_text] => An update about the problem with an ID of 6 [problem_update_active] => 1 ) [4] => Array ( [problem_update_id] => 7 [problem_id] => 3 [problem_update_date] => 2010-10-12 [problem_update_text] => Some new update about the problem with the ID of 3 [problem_update_active] => 1 ) ) Hi all, Just curious why this works: Code: [Select] while (($data = fgetcsv($handle, 1000, ",")) !== FALSE){ $import="INSERT into $prodtblname ($csvheaders1) values('$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]')"; } And this does not: $headdata_1 = "'$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]'"; while (($data = fgetcsv($handle, 1000, ",")) !== FALSE){ $import="INSERT into $prodtblname ($csvheaders1) values($headdata_1)"; }it puts $data[#'s] in the database fields instead of the actual data that '$data[0]','$data[1]'... relates to. I wrote a script to create the values in $headdata_1 based on the number of headers in $csvheaders1 but can't seem to get it working in the sql statement. Thanks Hi , I have one question .. Can I split showing of content of dynamic list in 2 parts , when I echo list in code .. Code: [Select] <?php // Run a select query to get my letest 8 items // Connect to the MySQL database include "../connect_to_mysql.php"; $dynamicList = ""; $sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC LIMIT 8"); $productCount = mysql_num_rows($sql); // count the output amount if ($productCount > 0) { while($row = mysql_fetch_array($sql)){ $id = $row["id"]; $product_name = $row["product_name"]; $price = $row["price"]; $date_added = strftime("%b %d, %Y", strtotime($row["date_added"])); $dynamicList .= '<table width="100%" border="2" cellspacing="2" cellpadding="2"> <tr> <td width="17%" valign="top"><a href="product.php?id=' . $id . '"><img style="border:#666 1px solid;" src="inventory_images/' . $id . '.jpg" alt="' . $product_name . '" width="77" height="102" border="2" /></a></td> <td width="83%" valign="top">' . $product_name . '<br /> $' . $price . '<br /> <a href="product.php?id=' . $id . '">View Product Details</a></td> </tr> </table>'; } } else { $dynamicList = "We have no products listed in our store yet"; } mysql_close(); ?> Code: [Select] <p><?php echo $dynamicList; ?><br /> </p> It works ok, and putting my files, everything works, but when I put 8 pictures with price and other details, it just show one image with details and another image below with details, and the third image below and so on .. Can I split dynamic list to show 4 images with details on the left side and 4 on the right side? Thank you in advance for help , if is possible |