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PHP - Making A List Into A Table
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."\">".$row->Database."</option>"; } } ?> </select> <input type="submit" name="delete" value="Delete"/> </form> <form action="pretask.php" method="post"> Create Database <input type="text" name="database" /> <input type="submit" name="result" value="Create" /> </form> </body> Similar TutorialsI have a form created with code already written. I am in need of a push in the right direction or a potential tutorial on this issue i have. I am running a fanatsy golf website where the user will pick one golfer each week and the cannot select them again. Is there a way I can remove that data from the list for the next week when the user makes his selection or can I have that data another color and unselectable. If you want code, please let me know and i can provide it. Thanks in advance for your help. p.s. the list data is stored in a MySQL database. Okay, really newbie question, but for this code... Code: [Select] <!-- Gender --> <label for="gender">Gender:</label> <select id="gender" name="gender"> <option value="">--</option> <option value="F">Female</option> <option value="M">Male</option> </select> 1.) How do I assign a variable to this? 2.) How do I make this "sticky"? Here is how I have usually done other form types... Code: [Select] <!-- First Name --> <label for="firstName">First Name:</label> <input id="firstName" name="firstName" type="text" maxlength="30" value="<?php if(isset($firstName)){echo htmlentities($firstName, ENT_QUOTES);} ?>" /><!-- Sticky Field --> <?php if (!empty($errors['firstName'])){ echo '<span class="error">' . $errors['firstName'] . '</span>'; } ?> Oh, by the way, at the top of my PHP file I have this code... Code: [Select] if ($_SERVER['REQUEST_METHOD']=='POST'){ // Form was Submitted (Post). // Initialize Errors Array. $errors = array(); // Trim all form data. $trimmed = array_map('trim', $_POST); Thanks, Debbie Ive gotten some results user selcts check box on first page The php page will say which brackets it falls between example price is between 100-300 say I dont seem to be able to populate a table with the data in the database:S Code: [Select] $row_number= 0; while ($row = mysql_fetch_array($result3)) { If (($row["price"] = $price_low) && ($row["price"] <= $price_high)) //price If (($row["storage"] = $storage_low) && ($row["storage"] <= $storage_high)) //storage if (($row["Processor "] = $processor_low) && ($row["Processor"] <= $processor_high)) //Processor { $row_number++; ?> <tr> <td align="center"><?php print $row["Computer_Price"]; ?> </td> <td align="center"><?php print $row["Computer_Storage"]; ?> </td> <td align="center"><?php print $row["Computer_ProcessorSpeed"]; ?> </td> </tr> </table> Thats wat im usign atm the price_low and price_high are what sets the low and high price for the search Hey all! In the code in question I echo out individual records of data from MySQL successfully. For each record there is a number which is used as a var in the javascript that does the count-down-timer part. However when I view the resulting page the timer works dynamically only with the first record. With the rest, the timer is static. Code: [Select] <? $result0 = mysql_query("SELECT * FROM table WHERE field='$value'"); while ($riw0 = mysql_fetch_assoc($result0)) { $seconds1 = $riw0['seconds'] ; //// echo out data and set variable for the number of seconds to count down ?> <script language="JavaScript"> var countDownInterval=<?=$seconds1?>; var c_reloadwidth=200 </script> <ilayer id="c_reload" width=&{c_reloadwidth}; ><layer id="c_reload2" width=&{c_reloadwidth}; left=0 top=0></layer></ilayer> <script> var countDownTime=countDownInterval+1; function countDown(){ countDownTime--; if (countDownTime <=0){ countDownTime=countDownInterval; clearTimeout(counter) window.location.href="military3.php" //Redirection URL return } var mins = Math.floor(countDownTime/60) var secs = countDownTime-(mins*60) if (document.all) //if IE 4+ document.all.countDownText.innerText = mins+" minutes "+secs+ " "; else if (document.getElementById) //else if NS6+ document.getElementById("countDownText").innerHTML=mins+" minutes "+secs+ " " else if (document.layers){ document.c_reload.document.c_reload2.document.write('Soldiers will be ready in... <span id="countDownText">'+countDownTime+' </span> seconds') document.c_reload.document.c_reload2.document.close() } counter=setTimeout("countDown()", 1000); } function startit(){ if (document.all||document.getElementById) document.write('Soldiers will be ready in <span id="countDownText">'+countDownTime+' </span> seconds') countDown() } if (document.all||document.getElementById) startit() else window.onload=startit </script> <? } ?> I tried replacing the javascript vars with PHP echoes for unique variables, but then no timer shows up, even static. So could anyone advice me on how I could use this code to apply for all MySQL records? Thanks in advance, Thauwa P.S. If I am unclear with my quandary, do let me know. Thank you. 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 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! 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. Hello everyone !!! I am sorry if my vocabulary is not exact because english is not my first language. Also i am a newbie at PHP. I am doing this project for myself and if it work might be able to use it at work. But i am doing this to learn. I have been stuck on this problem for 2 weeks and i can t figure it out on my own. Ihave spend many hours searching forums but no success. Oh yeah i almost forgot some part of code are from me, some are scripts from internet i adapted. I have a form with a dropdown menu and when i submit the form the value selected in the dropdown would be inserted in a table. The problem i have is that whatever the value i select it always inserts the last value of the dropdown in the table??? The form is made with the dropdown as an include. It is populated with values from an another table. Here is the code for the dropdown list: Code: [Select] <form> <select name="nom_pcu_form" method="post"> <?php $SQL = "SELECT * FROM pcu ORDER BY nom_pcu"; $res = mysql_query($SQL); while($val=mysql_fetch_array($res)) { $nom=$val["nom_pcu"]; $prenom=$val["prenom_pcu"]; $nom_complet = $prenom . $nom; echo "<option>".$val["nom_pcu"].", ".$val["prenom_pcu"]."</option>\n"; $nom_pcu_form="".$val["nom_pcu"].", ".$val["prenom_pcu"].""; } ?> </select> </form> Here is the part of the form wich calls the dropdown: Code: [Select] <form name="form2" method="post" action="Grille ecoute Permanent.php"> <p>Nom : <?php include 'liste_deroulante_pcu.php' ;?> <p>no carte appel <input name="no_carte_appel" type="text" id="no_carte_appel"> </p> <?php echo date("Y/m/d"); ?> </form> And this is the part where it is inserted in the table: Code: [Select] <?php if($_POST['doSubmit'] == 'Create') mysql_query("INSERT INTO grille_ecoute_pcu_permanent (`user_name`,`nom_pcu`,`no_carte_appel`,`question_1`,`question_2`,`question_3`,`question_4`, `question_5`,`question_6`,`question_7`,`question_8`,`question_9`,`question_10`,`question_11`,`question_12`,`question_13`,`question_14`,`question_15`, `question_16`,`question_17`,`question_18`,`question_19`,`question_20`,`question_21`,`question_22`,`question_23`,`question_24`,`question_25`,`question_26`, `question_27`,`question_28`,`question_29`) VALUES ('$user_name','$nom_pcu_form','$no_carte_appel','$question_1','$question_2','$question_3','$question_4','$question_5','$question_6', '$question_7','$question_8','$question_9','$question_10','$question_11','$question_12','$question_13','$question_14','$question_15','$question_16','$question_17', '$question_18','$question_19','$question_20','$question_21','$question_22','$question_23','$question_24','$question_25','$question_26','$question_27','$question_28', '$question_29') ") or die(mysql_error()); Note:$user_name and all $question are inserted correctly in the table. $nom_pcu_form is the dropdown and it only records the last value of the dropdown even if it s not the value selected. $no_carte_appel are not recorded at all thanks for your time Hi? im just a beginner in php i just want to ask how to insert a data into a table from a dropdown list. I have concatenate the itemid and description to form the dropdown list. But when i viewed my item_table the itemid and description columns are null. can you help me with this.. this is my php code for the dropdown list... <?php $query = "SELECT CONCAT(itemid,' ', '-',' ', description) AS Item FROM item_table"; $result = mysql_query($query) or die(mysql_error()); $dropdown = "<SELECT CONCAT(itemid,' ' '-',' ', description) AS Item FROM item_table>"; while($row = mysql_fetch_assoc($result)) { $dropdown .= "\r\n<option value='{$row['Item']}'>{$row['Item']}</option>"; } $dropdown .= "\r\n</select>"; echo $dropdown; ?> this my code for inserting data into the item_table... <?php if(isset($_POST ['submit'])) { $itemid = $_POST['itemid']; $description = $_POST['description']; $datein = $_POST['datein']; $qtyin = $_POST['qtyin']; $unitprice = $_POST['unitprice']; $unit = $_POST['unit']; $category = $_POST['category']; $empid = $_POST['empid']; $message =''; if(($itemid && $description == "")||($itemid && $description == null)) { header("location:IncomingEntry.php?msg=Incorrect"); exit(); } else { $link = mysql_connect('localhost', 'root', '') or die(mysql_error()); $db_selected = mysql_select_db('inventory', $link); $message=''; $query = "INSERT INTO incoming_table (itemid , description, datein, qtyin, unitprice, unit, category, empid) VALUES ('".$itemid."', '".$description."', '".$datein."', '".$qtyin."', '".$unitprice."', '".$unit."', '".$category."', '".$empid."')"; if (!mysql_query($query,$link)) { die('Error: ' . mysql_error()); } header("location: IncomingEntry.php?msg=1 record added"); } } ?> I am working on a project that uses a drop down list to chose the category when inserting new data into the database. What I want to do now is make the drop down list default to the chosen category on the list records page and the update page. I have read several tutorials, but they all say that I have to list the options and then select the default. But since it is possible to add and remove categories, this approch won't work. I need the code to chose the correct category on the fly. There are two tables, one that has the category ID and category name. The second table has the data and the catid which is referenced to the category id in the first table. Code: [Select] -- -- Table structure for table `categories` -- DROP TABLE IF EXISTS `categories`; CREATE TABLE IF NOT EXISTS `categories` ( `id` int(11) NOT NULL AUTO_INCREMENT, `categories` varchar(37) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=40 ; -- -------------------------------------------------------- -- -- Table structure for table `links` -- DROP TABLE IF EXISTS `links`; CREATE TABLE IF NOT EXISTS `links` ( `id` int(4) NOT NULL AUTO_INCREMENT, `catid` int(11) DEFAULT NULL, `name` varchar(255) NOT NULL DEFAULT '', `url` varchar(255) NOT NULL DEFAULT '', `content` varchar(255) NOT NULL DEFAULT '', PRIMARY KEY (`id`), KEY `catid` (`catid`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=35 ; Then is the list records file, I have Code: [Select] <?php include ("db.php"); include ("menu.php"); $result = mysql_query("SELECT categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $categories=$row["categories"]; $options.= '<option value="'.$row['categories'].'">'.$row['categories'].'</option>'; }; $id = $_GET['id']; $query="SELECT * FROM links ORDER BY catid ASC"; $result=mysql_query($query); ?> <table width="65%" align="center" border="0" cellspacing="1" cellpadding="0"> <tr> <td> <table width="100%" border="1" cellspacing="0" cellpadding="3"> <tr> <td colspan="7"><strong>List data from mysql </strong> </td> </tr> <tr> <td align="center"><strong>Category ID</strong></td> <td align="center"><strong>Category ID</strong></td> <td align="center"><strong>Name</strong></td> <td align="center"><strong>URL</strong></td> <td align="center"><strong>Content</strong></td> <td align="center"><strong>Update</strong></td> <td align="center"><strong>Delete</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td> <SELECT NAME=catid> <OPTION>Categories</OPTION> <?php echo $options; ?> </SELECT> </td> <td><? echo $rows['catid']; ?></td> <td><? echo $rows['name']; ?></td> <td><a href="<? echo $rows['url']; ?>"><? echo $rows['url']; ?></a></td> <td><? echo $rows['content']; ?></td> <td align="center"><a href="update.php?id=<? echo $rows['id']; ?>">update</a></td> <td align="center"><a href="delete.php?id=<? echo $rows['id']; ?>">delete</a></td> </tr> <?php } ?> </table> </td> </tr> </table> <?php mysql_close(); ?> So, how do I get this code Code: [Select] $result = mysql_query("SELECT categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $categories=$row["categories"]; $options.= '<option value="'.$row['categories'].'">'.$row['categories'].'</option>'; }; <SELECT NAME=catid> <OPTION>Categories</OPTION> <?php echo $options; ?> </SELECT> to give me an output that will be something like if catid exactly matches categories.id echo categories.categorie ??? so far everything I have done produces either a default category of the last category, the catid (which is a number), all of the categories (logical since catid will always be = id, or nothing. How do I get just the category name? I will keep reading and try to figure this out, but any help would be greatly appreciated. Thanks in advance Im looking to update the way my merchandise page lists items. Im trying to make it so the query will list items in a table and have it query 6 items in one row before it makes a second table row. heres my code i have now, but it lists them all horizontally down the page cuz of how i have it. but i'd like to change it to my explanation above i know i will have to change it all into a table query but how to get it to list 6 items before it makes next row is what i dont know how to do. Code: [Select] <?php if('/merch.php'==$_SERVER['PHP_SELF']) { $catquery = mysql_query('SELECT * FROM merch GROUP BY merch_cat') or die('mySQL Query Failed: '.mysql_error()); $num_rows = mysql_num_rows($catquery); if($num_rows==0){ echo 'Nothing Here'; break; } echo '<center><font size="6">'; while($row = mysql_fetch_array($catquery)){ list($id, $merch_name, $merch_image, $merch_cat, $merch_info, $buy, $price) = $row; echo ': <font class="myFont"><a href="merch.php?list=false&merch_cat='.$merch_cat.'">'.$merch_cat.'</a></font> '; } echo ':</font><br />'; echo '<center>Select a Category</center>'; echo '<br /><br />'; echo '<div><font class="myFont" size="6">New Items</font></div>'; $newquery = mysql_query('SELECT * FROM merch ORDER BY id DESC LIMIT 4') //Obviously im going to change the limit when i get this working or die('New Query Failed: '.mysql_error()); while($rows = mysql_fetch_array($newquery)){ list($id, $merch_name, $merch_image, $merch_cat, $merch_info, $buy, $price) = $rows; echo '<a href="merch.php?list=true&merch_cat='.$merch_cat.'&id='.$id.'">'.$merch_name.'<br />'; echo '<img src="media/merch/'.$merch_image.'-thumb.png" border="0"></a>'; echo '<br /><br />'; } } break; ?> I would appreciate your assistance, there are tons of login scripts and they work just fine. However I need my operators to login and then list their activities for the other operators who are logged in to see and if desired send their clients on the desired activity. I have the login working like a charm and the activities are listed just beautifully. How do I combine the two tables in the MySQL with PHP so the operator Logged in can only make changes to his listing but see the others. FIRST THE ONE script the member logges in here to the one table in MSQL: <?php session_start(); require_once('config.php'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } $login = clean($_POST['login']); $password = clean($_POST['password']); if($login == '') { $errmsg_arr[] = 'Login ID missing'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'Password missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: login-form.php"); exit(); } $qry="SELECT * FROM members WHERE login='$login' AND passwd='".md5($_POST['password'])."'"; $result=mysql_query($qry); if($result) { if(mysql_num_rows($result) == 1) { session_regenerate_id(); $member = mysql_fetch_assoc($result); $_SESSION['SESS_MEMBER_ID'] = $member['member_id']; $_SESSION['SESS_FIRST_NAME'] = $member['firstname']; $_SESSION['SESS_LAST_NAME'] = $member['lastname']; session_write_close(); header("location: member-index.php"); exit(); }else { header("location: login-failed.php"); exit(); } }else { die("Query failed"); } ?> ................................................. ................................ Now I need the person who logged in to the table above to be able to make multiple entries to the table below <? $ID=$_POST['ID']; $title=$_POST['title']; $cost=$_POST['cost']; $activity=$_POST['activity']; $ayear=$_POST['aday']; $aday=$_POST['ayear']; $seats=$_POST['special']; $special=$_POST['seats']; mysql_connect("xxxxxx", "xxx350234427", "========") or die(mysql_error()); mysql_select_db("xxxx") or die(mysql_error()); mysql_query("INSERT INTO `activity` VALUES ('ID','$title', '$cost','$activity', '$aday', '$ayear', '$special', '$seats')"); Print "Your information has been successfully added to the database!" ?> Click <a href="member-profile.php">HERE</a> to return to the main menu <?php ?> I am trying to print the list of a table which I requested with "SELECT DISTINCT" as below Code: [Select] $db_connect = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); $sql_get = "SELECT DISTINCT category FROM con"; $sql_run = mysqli_query($db_connect, $sql_get) or mysqli_error($db_connect); $sql_assoc = mysqli_fetch_assoc($sql_run); What is now needed to print the list of the table data by this conditions? I tried the while loop, but I seem to approach wrong, and get endless loops or errors. I'm setting up a newsletter thing for my website. I have a newsletter table in MySQL: Quote +----------------+------+--------+ | email | mens | womens | +----------------+------+--------+ | test2@test.com | 1 | 1 | | test1@test.com | 1 | 0 | +----------------+------+--------+ I am using a HTML form and this PHP code I learn't from the manual, which sends out e-mail's. PHP: if ($_POST['newsletter'] == 'Mens') { $to = ''; $subject = $_POST['subject']; $body = $_POST['body']; $header = 'From: Me Someone <me@someone.com>'; mail($to, $subject, $body, $header); } What I want to do with the above code is send out an e-mail to all the e-mails in my MySQL database that are tagged '1' under mens. How would I go about doing this? I'm guessing I will have to use a MySQL query in the $to = ''; that goes something like this: $to = '$query (select from `newsletters` where `email` = 1'); ? 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 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 $dynamicList; ?><br /> </p> It works ok, and putting my files, everything works, but when I put 8 pictures with price and other details, it just show one image with details and another image below with details, and the third image below and so on .. Can I split dynamic list to show 4 images with details on the left side and 4 on the right side? Thank you in advance for help , if is possible Hi, I use the following code to create a select menu from an array of options stored in LISTS.php: include 'LISTS.php'; print('<select id="from" name="from">'); foreach ($langList as $lang) {printf('<option %s>%s</option>', ($from1 == $lang ? 'selected="selected"' : ''), $lang); } echo '</select>'; where LISTS.php includes the following: $langList = array(' ','English', 'French', 'German', 'Dutch', 'Spanish'); This works great, but now I want to do something similar with a checkbox list, where each checkbox has an associated 'onchange' javascript function and I'm getting pretty stuck. My checkbox list is of the following form: Code: [Select] <html> <ul style="height: 95px; overflow: auto; width: 200px; border: 1px solid #480091; list-style-type: none; margin: 0; padding: 0;"> <li id="li1b"><label for="chk1b"><input name="chk1b" id="chk1b" type="checkbox" onchange="function1('chk1b','li1b')">Option1</label></li> <li id="li2b"><label for="chk2b"><input name="chk2b" id="chk2b" type="checkbox" onchange="function1('chk2b','li2b')">Option2</label></li> //etc. </ul> </html> What I want to do is have 'Option1', 'Option2', etc. stored in an array in LISTS.php and have a PHP script that populates the checkbox list accordingly, in a similar manner to my select menu above. I can't work out how to get the ID of the next <li> and the next <input> in the list to go up by one each time, e.g. 'li1b' then 'li2b', 'li3b', etc. Could someone pls help me out? Thanks! Is there a way to take data that is listed like a,b,c,d,e and turn that into an unordered list? Basically, I'm going to have a list of schools that will be listed with comma separators. I'll print that data out in two places, one will be printed as it's in the database. The other I would like to echo as an unordered list. How hard would something like this be? I have a database with all my users. I need an empty list and when a user clicks a button "Add me to the list" they are added to the first position in the list Slots 1. Tom 2. 3. 4. 5. Later, Jerry wants to be added to the list. So this happens Slots 1. Jerry 2. Tom 3. 4. 5. More people join the list 1. Bob 2. John 3. Jerry 4. Tom 5. Tom gets upset clicks the same button to join the list again, and he's bumped back up to first place. Slots 1. Tom 2. Bob 3. John 4. Jerry 5. Is this possible? And is so, how would I go about getting started on such a thing? I've been using PHP for about a year and a half. Is it possible to get bittorrent peer list (IP list) via php, udp protocol ? txh! |
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