PHP - Mysql List Contents Of A Feild
Hi i want to make a form LIST that automaticly gets the content "orderid" to choose for editing.
can i please get a pointing pin on which function(s) to use ? i have tried with: Code: [Select] $result = mysql_query("SELECT * FROM {$table} WHERE 'ordreid'"); if (!$result) { die("Query to show fields from table failed"); } echo($result); $i = 0; while($i<mysql_num_fields($result)) { $meta=mysql_fetch_field($result,$i); echo $i.".".$meta->name."<br />"; $i++; } like i want to get this part: <option>first item in field orderid</option> <option>2'nd item in field orderid</option> and so on...i know its easy but i have struggled wit this enough now. Similar TutorialsHi all, Having a (probably basic) problem here, and as much as I have read about it, I still don't understand. I have a basic form in a page called book.php which upon submit calls sendmail.php to email me the details. Everything works as it should, including my current error handler, which is as follows. Code: [Select] elseif (empty($name) || empty($number) || empty($profexp) || empty($nonprofexp)) { header( "Expires: Mon, 20 Dec 1998 01:00:00 GMT" ); header( "Last-Modified: " . gmdate("D, d M Y H:i:s") . " GMT" ); header( "Cache-Control: no-cache, must-revalidate" ); header( "Pragma: no-cache" ); ?> <html> <head><title>Error</title></head> <body> Error <p> Oops, it appears you forgot to fill in all of your details. Please press the BACK button in your browser and try again. </p> </body> </html> <?php } This is fine, but what I would prefer is a seperate error for each user input section. So, for example, when the user does not enter anything in the $name feild and presses submit, the book.php page is displayed with the error message to the right of the text box. Any help would be appreciated. Thank you I want to display my pictures I stored in Mysql in a 4 column, 2 row table WITH pagination. Here's the code I use to display the data currently: Code: [Select] //your username $username = "username"; //your password $password = "password"; //your mySQL server $host = "host"; //The name of the database your table is in $database = "database"; //connect, but if there is a problem, display an error message telling why there is a problem $conn = mysql_connect($host,$username,$password) or die("Error connecting to Database!<br>" . mysql_error()); //Choose the database from your mySQL server, but if there is a problem, display an error telling why $db = mysql_select_db($database) or die("Cannot select database!<br>" . mysql_error()); // find out how many rows are in the table $sql = "SELECT COUNT(*) FROM myphotos"; $result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR); $r = mysql_fetch_row($result); $numrows = $r[0]; // number of rows to show per page $rowsperpage = 4; // find out total pages $totalpages = ceil($numrows / $rowsperpage); // get the current page or set a default if (isset($_GET['currentpage']) && is_numeric($_GET['currentpage'])) { // cast var as int $currentpage = (int) $_GET['currentpage']; } else { // default page num $currentpage = 1; } // end if // if current page is greater than total pages... if ($currentpage > $totalpages) { // set current page to last page $currentpage = $totalpages; } // end if // if current page is less than first page... if ($currentpage < 1) { // set current page to first page $currentpage = 1; } // end if // the offset of the list, based on current page $offset = ($currentpage - 1) * $rowsperpage; // get the info from the db $sql = "SELECT * FROM myphotos ORDER BY id ASC LIMIT $offset, $rowsperpage"; $result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR); /****** build the pagination links ******/ // range of num links to show $range = 3; // if not on page 1, don't show back links if ($currentpage > 1) { // show << link to go back to page 1 echo "<span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=1'>First</a></span> "; // get previous page num $prevpage = $currentpage - 1; // show < link to go back to 1 page echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$prevpage'>Previous</a></span> "; } // end if // loop to show links to range of pages around current page for ($x = ($currentpage - $range); $x < (($currentpage + $range) + 1); $x++) { // if it's a valid page number... if (($x > 0) && ($x <= $totalpages)) { // if we're on current page... if ($x == $currentpage) { // 'highlight' it but don't make a link echo " <span class=\"paginationDown\"><b>$x</b></span> "; // if not current page... } else { // make it a link echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>$x</a></span> "; } // end else } // end if } // end for // if not on last page, show forward and last page links if ($currentpage != $totalpages) { // get next page $nextpage = $currentpage + 1; // echo forward link for next page echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$nextpage'>Next</a></span> "; // echo forward link for lastpage echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$totalpages'>Last</a></span><br> "; } // end if /****** end build pagination links ******/ // while there are rows to be fetched... while ($list = mysql_fetch_assoc($result)) { extract ($list); // echo data $url = $list['url'];; $title = $list['title']; $description = $list['description']; echo("$title<br><a rel=\"example_group\" title=\"$description\" href=\"$url\"><img src=\"$url\" alt=\"\" width=\"\" height=\"\" class=\"gallery_images\" /></a>"); } // end while /****** build the pagination links ******/ // range of num links to show $range = 3; // if not on page 1, don't show back links if ($currentpage > 1) { // show << link to go back to page 1 echo "<span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=1'>First</a></span> "; // get previous page num $prevpage = $currentpage - 1; // show < link to go back to 1 page echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$prevpage'>Previous</a></span> "; } // end if // loop to show links to range of pages around current page for ($x = ($currentpage - $range); $x < (($currentpage + $range) + 1); $x++) { // if it's a valid page number... if (($x > 0) && ($x <= $totalpages)) { // if we're on current page... if ($x == $currentpage) { // 'highlight' it but don't make a link echo " <span class=\"paginationDown\"><b>$x</b></span> "; // if not current page... } else { // make it a link echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>$x</a></span> "; } // end else } // end if } // end for // if not on last page, show forward and last page links if ($currentpage != $totalpages) { // get next page $nextpage = $currentpage + 1; // echo forward link for next page echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$nextpage'>Next</a></span> "; // echo forward link for lastpage echo " <span class=\"pagination\"><a href='{$_SERVER['PHP_SELF']}?currentpage=$totalpages'>Last</a></span><br>"; } // end if /****** end build pagination links ******/ The above code just displays the pictures vertically. Here's this code live: http://www.djsmiley.net/gallery/albums/my_photos.php (you can't view the page in IE) Now how would I make it so that the first 8 images in the database display in a 4x2 table, etc.? I am in the process of creating a waiting list that I want to work automatically. When a new member registers for membership, the member must be on a waiting list for a while before the member gets permanent membership. When the membership is activated by me, I want this to work like when I delete the person who is number 1 in the list, and number 2 will be number 1. the remaining numbers in the list will automatically move a number up, just now . 5 will be number 4 and so on. PErson number one is the person that allways will be deleted from the waiting list. An idea would be that this persons name ande rest of data connected to hes membership number will be moved automaticaly to the permanent members list when hi is deleted from the waiting list. Only the name will be displayed in the member list, but the form that the new member submits will have several fields. There are t12 fields to be filled in in the MySQL database, but only the name and surname to be displayed in the list. The numbers will be generated automatically in front of the name. I have tried to do this by myself. I have tried to do this myself, but I do not succeed. I am very new when it comes to PHP and programming in general. Hope someone here can help me with this. I can not find out. Edited February 26 by LeonLatexi have been trying to get this code to get a list of usernames from a database and i have now got that to work but when i try and save it it saves all the usernames from the drop down list and not just the one i have selected how can i get it to just use the one i have selected Code: [Select] <?php include "connect.php"; //connection string include("include/session.php"); print "<link rel='stylesheet' href='style.css' type='text/css'>"; print "<table class='maintables'>"; print "<tr class='headline'><td>Post a message</td></tr>"; print "<tr class='maintables'><td>"; // Write out our query. $query = "SELECT username FROM users"; // Execute it, or return the error message if there's a problem. $result = mysql_query($query) or die(mysql_error()); $dropdown = "<select name='username'>"; while($row = mysql_fetch_assoc($result)) { $dropdown .= "\r\n<option value='{$row['username']}'>{$row['username']}</option>"; } $dropdown .= "\r\n</select>"; if(isset($_POST['submit'])) { $name=$session->username; $yourpost=$_POST['yourpost']; $subject=$_POST['subject']; $to=$dropdown; if(strlen($name)<1) { print "You did not type in a name."; //no name entered } else if(strlen($yourpost)<1) { print "You did not type in a post."; //no post entered } else if(strlen($subject)<1) { print "You did not enter a subject."; //no subject entered } else { $thedate=date("U"); //get unix timestamp $displaytime=date("F j, Y, g:i a"); //we now strip HTML injections $subject=strip_tags($subject); $name=strip_tags($name); $yourpost=strip_tags($yourpost); $to=strip_tags($to); $insertpost="INSERT INTO forumtutorial_posts(author,title,post,showtime,realtime,lastposter,name) values('$name','$subject','$yourpost','$displaytime','$thedate','$name','$to')"; mysql_query($insertpost) or die("Could not insert post"); //insert post print "Message posted, go back to <A href='forum.php'>Forum</a>."; } } else { print "<form action='newtopic.php' method='post'>"; print "Your name:<br>"; print "$session->username<br>"; print "User to send to:<br>"; print "$dropdown"; print "Subject:<br>"; print "<input type='text' name='subject' size='20'><br>"; print "Your message:<br>"; print "<textarea name='yourpost' rows='5' cols='40'></textarea><br>"; print "<input type='submit' name='submit' value='submit'></form>"; } print "</td></tr></table>"; ?> MOD EDIT: Changed PHP manual link [m] . . . [/m] tags to [code] . . . [/code] tags. I am trying to send a list of every name in a database via email but everything i have tried just sends lots of emails with a different name in each email. How can i send every name in one email? Hi. I am using this script to populate a dropdown list box from sql, it works but does anyone know how to sort the list in alphabetical order? $sql="SELECT * FROM Fish WHERE ***** = '".$_GET['stocktype']."'"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $ID=$row["ID"]; $Stock=$row["Commonn"]; $Options.="<OPTION VALUE=\"$ID\">".$Stock; } <SELECT NAME='stock1'> <OPTION VALUE='$Options'>$Options</option> </SELECT> Say I have a column in a MySQL database, that contains the following data (each piece of data is in its own row, as stored as a string): 16b 166 13A 13a 4 402c A66 A66b Currently the list sorts as follows: A66 A66b 13A 13a 16b 166 4 402c I need it to sort as follows: A66 A66b 4 13a 13A 16b 166 402c So that any strings that start with a letter are first, followed by numbers in numerical order, with lower case letters coming before upper case letters. This is a huge issue on my website, which has a database of over 35k such numbers, split up into lists. I can't get these sorted properly. At least it would be nice to sort as follows: 1 4 100 344 Instead of 1 100 344 4 Know what I mean? I know that this is complicated, because the variables are strings and not numbers, but is there an easy way to do this? Been using a function I found here a while back for listing categories and sub categories and it works perfect. Code: [Select] function listSubcats ($parent, $level=0){ global $abc; $sql = "SELECT id, title FROM Cat WHERE parent = $parent"; $res = mysqli_query($abc, $sql); while (list($id, $title) = $res->fetch_row()) { $indent = str_repeat('-', $level); echo "<OPTION value='$id'>$indent $title</OPTION>\n"; listSubcats ($id, $level+1); // list its subcats } } But this is the first time I need change the way the results are displayed. And that leads me to realizing I am a bit confused on how list really works. The mysql table has the typical ID, CatName, parent I need to have each Cat with all subcats related to it listed in its own div. But whatever I try I cannot get the placements of the opening and closing divs in the right spots. So can anyone tell if its even possible to do it with this function. I did work it out using multiple queries but then read on alot of forums that queries inside while statements is not good. Just looking for he best (Correct) way of geting the results laid out properly. like this Code: [Select] <div class="one-third column"> Cat1 </div> <div class="one-third column"> Cat 2 </div> <div class="one-third column"> Cat 3 - Subcat 1 - Subcat 2 -- Sub Subcat 1 </div> <div class="one-third column"> Cat 4 </div> Hope that makes sense... Thanks for any guidance. 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, basically, here's the deal: I have a lit of checkboxes that are added by the admin (there's an unlimited amount, just depends on how many are added). Then, those are put in a form, in which the user picks whichever ones need to be chosen and those values get sent to a MySQL table. Here's the code that displays the checkboxes Code: [Select] <?php $sql="SELECT * FROM category ORDER BY categoryID ASC"; $result=mysql_query($sql); while($row=mysql_fetch_array($result)){ echo "<input type='checkbox' id='". $row['categoryName'] ."' name='licensed[]' value='" . $row['categoryID'] ."' /><label for='". $row['categoryName'] ."'><br>" . $row['categoryName'] ."</label><br>"; } ?> What I'm making now, is an edit form where whichever checkboxes were checked, will show up checked in the edit form. But I'm not really sure how to go about this, since there is only one actual input tag in the code, I can't select the different ones. I also have this SQL query which selects whichever boxes were inputted into the DB Code: [Select] $sql2="SELECT * FROM category, categoryInfo WHERE category.categoryID = categoryInfo.categoryID AND categoryInfo.parentID = $parentID"; $result2=mysql_query($sql2); Where $parentID is the ID of whichever form you're editing. But yes, I'm basically not really sure how to go about this and would like some help figuring this out Thanks for your time I'm making a checklist. One table holds the list with IDs. There are about 224 rows, each with its own ideas. Now I have another table to hold user accounts. When you create an account, it shows you a fresh new checklist that you need to start checking off. Could anyone please share techniques so I can have multiple accounts have their own list they need to check off? (ie, when a new person creatures a new account they should have their own list with NOTHING checked) The only way I can think of doing this is making 224 fields for the user account with the IDs of the checklist table to check if I checked it or not. Surely there's an easier way? Thanks I have created a drop down list and it does retrieve information from mysql but now I want to use what is been selected to retrieve information. How Do I do this? <?php MYSQL_CONNECT(localhost,'root','') OR DIE("Unable to connect to database"); @mysql_select_db(Examination) or die( "Unable to select database"); $query=("SELECT * FROM subject"); $result=mysql_query($query) or die ("Unable to Make the Query:" . mysql_error() ); echo "<select name=myselect>"; while($row=mysql_fetch_array($result)){ echo "<OPTION VALUE=".$row['Sub_ID'].">".$row['Sub_Name']."</OPTION>"; } echo "</select>"; ?> 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 $options; ?> </SELECT> to give me an output that will be something like if catid exactly matches categories.id echo categories.categorie ??? so far everything I have done produces either a default category of the last category, the catid (which is a number), all of the categories (logical since catid will always be = id, or nothing. How do I get just the category name? I will keep reading and try to figure this out, but any help would be greatly appreciated. Thanks in advance Hi, I'm a php newbie, with some mysql experience. I have a mysql database as follows: Database=watch, Table=events - fields id, reportno, sdate, comments What I need is: 1. A dropdown list to display reportno from mysql database. 2. Depending on which reportno I choose, I'd like to open a popup(or separate) page to display the stored information. Tks in advance for any help hi i have generated a product list from a mysql table called product_list, once i enter a new product in to the table the product will be shown in the generated list and the list will grow and the table grows. and i want to allow user the edit/delete/save the products from the generated table, i have no idea how to do it and what is the algorithmic idea to do it so. here is the php and the html code. <!--Body container for creating a new product in to the list--> <div class="body_orderviewform"> <form name="form1" method="post" action="upload_file.php" enctype="multipart/form-data"> <p> <label for="user_id">User ID:</label> <input type="text" name="user_id" id="user_id"> <label for="customer_name">Customer Name:</label> <input type="text" name="customer_name" id="customer_name"> <label for="customer_family">Customer_family</label> <input type="text" name="customer_family" id="customer_family"> <label for="freelancer_name">Purchaser:</label> <input name="freelancer_name" type="text" id="freelancer_name"> <? if (isset($_COOKIE['picAdd'])) echo $_COOKIE['picAdd'];?> </p> <!------------------------------------------------------------------------------------------------------- for generating the list --> <div class="div.neworder_list" > <div class="div.neworder_listheader" align="center"> <table width="637" border="1" > <tr> <td width="193"><label for="link">link:</label> <label for="link_new"></label> <input type="text" name="link_new" id="link_new"></td> <td width="202"><label for="unitprice">Unit Price:</label> <label for="Unit_price_new"></label> <input type="text" name="Unit_price_new" id="Unit_price_new"></td> <td width="220"><label for="qty">Quantity:</label> <label for="quantity_new"></label> <input type="text" name="quantity_new" id="quantity_new"></td> </tr> <tr> <td><label for="express">Express Fee:</label> <label for="express_new"></label> <input type="text" name="express_new" id="express_new"></td> <td><label for="commission_new">Commission:</label> <input type="text" name="commission_new" id="commission_new"></td> <td><label for="customer_description">Description</label> <label for="description_new"></label> <textarea name="description_new" id="description_new" cols="45" rows="5"></textarea></td> </tr> <tr> <td> </td> <td colspan="2">Picture Upload: <input type="hidden" name="<?php echo ini_get("session.upload_progress.name");?>" value="123" /> <input name="file" type="file" autofocus="autofocus"/> <br /> </tr> <tr> <td colspan="3"><input type="reset" name="reset" id="reset" value="Reset"> <input type="submit" name="submit" id="submit" value="Submit The Product"></td> </tr> </table> </form> </div> <div class="neworder_listview"> <p> <form action="" method="post" name="list"> <input type="submit" name="del" id="del" value="Save"> <input type="submit" name="save" id="save" value="Del"> </p> <table width="1022" border="1" align="center"> <tr> <th width="24" scope="col"> </th> <th width="24" scope="col">Row#</th> <th width="137" scope="col">Manager</th> <th width="137" scope="col">Purchaser Desc</th> <th width="40" scope="col"><p>Link</p> <p>/Ссылки</p></th> <th width="53" scope="col">ФОТО</th> <th width="50" scope="col">Unit Price/Цена за еденицу товара</th> <th width="46" scope="col">Quantity/ Кол-во</th> <th width="138" scope="col">Total Unit Price/ Общая цена</th> <th width="89" scope="col">Express/Доставка по Китаю</th> <th width="119" scope="col">Description/Описание</th> <th width="89" scope="col">ADDITIONAL LINKS/ЗAMЕНЫ</th> </tr> <?php $username = "my username"; $password = "my pass"; $database = "userinfo"; $link = mysql_connect("localhost", "$username", "$password"); if(!$link) {echo("Failed to establish connection to mysql server"); exit();} $status = mysql_select_db($database); $query = "SELECT * FROM order_list"; $result = mysql_query($query); $num = mysql_num_rows($result); $i=0; while ($i < $num) { $field1_name=mysql_result($result,$i,"admin_st"); $field2_name=mysql_result($result,$i,"freelancer_st"); $field3_name=mysql_result($result,$i,"link"); $field4_name=mysql_result($result,$i,"picture"); $field5_name=mysql_result($result,$i,"unitprice"); $field6_name=mysql_result($result,$i,"qty"); $field7_name=mysql_result($result,$i,"express"); $field8_name=mysql_result($result,$i,"customer_st"); $i++; } ?> <?php $i=0; $row=1; while ($i < $num) { $f1=mysql_result($result,$i,"admin_st"); $f2=mysql_result($result,$i,"freelancer_st"); $f3=mysql_result($result,$i,"link"); $f4=mysql_result($result,$i,"pic_address"); $f5=mysql_result($result,$i,"unitprice"); $f6=mysql_result($result,$i,"qty"); $f7=mysql_result($result,$i,"express"); $f8=mysql_result($result,$i,"customer_st"); $totao_unit_price = $f5*$f6; ?> <tr> <td><input type="checkbox" name="del_chbox" id="del_chbox"> <td><p><font face="Arial"><input name="row_txtbox" type="text" id="row_txtbox" size="2" value="<?php echo $row; ?>"></font></td> <td><p><font face="Arial"> <textarea name="manager_txtbox" cols="10" id="manager_txtbox"><?php echo $f1; ?></textarea></font></td> <td><p><font face="Arial"> <textarea name="purchase_txtbox" cols="10" id="purchase_txtbox"><?php echo $f2; ?></textarea> </font></td> <td><font face="Arial"><a href="<?php $f3 ?>" target="_blank"><?php echo $f3; ?></a></font></td> <td><font face="Arial"><img src="<?php echo $f4;?>" width="100" align="middle"100></font></td> <td><font face="Arial"><input name="unitprice_txtbox" type="text" id="unitprice_txtbox" size="2" value="<?php echo $f5; ?>"></font></td> <td><font face="Arial"> <input name="qty_txtbox" type="text" id="qty_txtbox" size="2" value="<?php echo $f6; ?>"></font></td> <td><p><font face="Arial"><?php echo $totao_unit_price; ?></font></td> <td><p><font face="Arial"><input name="express2" type="text" id="express3" size="2" value="<?php echo $f7; ?>"></font></td> <td><p><font face="Arial"> <textarea name="custdesc_txtbox" cols="20" id="custdesc_txtbox"><?php echo $f8; ?></textarea></font></td> <td><input name="express2" type="text" id="express3" size="2" value="<?php echo "new link" ?>"></td> </tr> <p> <?php $i++; $row++; } ?> </table> </p> <p> </p> </form> Lets start out by saying I'm a nube to sql/php things so I am learning as I go. I try to read all that I can before I post, and only post when I cant figure it out on my own. That being said. What I want to do in simplest terms is be able to assign a variable to each item in an sql table. So say I have an mysql table that has ID, username, fontcolor. I want to be able to pull those out so say.... while($row = mysql_fetch_array($users)) { $username[$i] = $row[username]; $fontcolor[$i] = $row[fontcolor]; } Then on the page I can just call to $username[1] type thing. I have tried mixing this several different ways with for and while and I keep getting errors on the page. I realize the code isn't showing everything but its just there to show you the idea of what I'm trying to do. I just want it to generate a list(array) that will make it easier for me to call back just the items I need on parts of the page with out having to have extra coding everywhere. Thanks in advance. Jim Ok, so I've spent quite a bit of time piecing together this solution from a variety of sources. As such, I may have something in my code below that doesn't make sense or isn't neccessary. Please let me know if that is the case. I'm creating an administrative form that users will you to add/remove items from a MySQL table that lists open positions for a facility. The foreach loop generates all of the possible job specialties from a table called 'specialty_list'. This table is joined to a second table ('open_positions') that lists any positions that have been selected previously. Where I'm stuck is getting the checkbox to be checked if the facility_ID from the open_positions table matches the $id passed in the URL via ?facility_id=''. Here's where I am so far: $query = "SELECT specialty_list.specialty_displayname , specialty_shortname , open_positions.position , facility_ID FROM specialty_list LEFT OUTER JOIN open_positions ON open_positions.position = specialty_list.specialty_shortname ORDER BY specialty_list.specialty_shortname"; $results = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_assoc($results)) { $positions[$row['specialty_shortname']] = $row['specialty_displayname']; } echo "<form method='POST' action='checkbox.php'>"; foreach($positions as $specialty_shortname => $specialty_displayname) { $facility_ID = $row['facility_ID']; $checked = $facility_ID == $row['facility_ID'] ? ' checked' : ''; echo "<input type='checkbox' name='position[]' value=\"{$specialty_shortname}\"{$checked}> {$specialty_displayname}</input><br/>"; } echo "<input type='hidden' name='facility_ID' value='$id'>"; echo "<input type='submit' value='Submit Checkboxes!'>"; echo "</form>"; Any ideas how to get this working? I feel like I'm very close, but I just can't get it. I also tried starting from scratch with a WHILE statement instead of a FOREACH, but haven't tweaked it enough to prevent duplicate checkboxes. With that in mind, here it is, just in case that's a better direction: $query = "SELECT specialty_list.specialty_displayname , specialty_shortname , open_positions.position , facility_ID FROM specialty_list LEFT OUTER JOIN open_positions ON open_positions.position = specialty_list.specialty_shortname ORDER BY specialty_list.specialty_shortname"; $results = mysql_query($query) or die(mysql_error()); echo "<form method='POST' action='checkbox.php'>"; while($row=mysql_fetch_assoc($results)) { $facility_ID = $row['facility_ID']; $specialty_shortname = $row['specialty_shortname']; $specialty_displayname = $row['specialty_displayname']; if ($facililty_ID==$id) { $checked=' checked'; } echo "<input type='checkbox' name='position[]' value=\"$specialty_shortname\"$checked> $specialty_displayname</input><br/>"; } echo "<input type='hidden' name='facility_ID' value='$id'>"; echo "<input type='submit' value='Submit Checkboxes!'>"; echo "</form>"; Hey guys, I can't wrap my head around how to make this work right... I have three tables: Code: [Select] CREATE TABLE `games` ( `g_id` int(11) NOT NULL AUTO_INCREMENT, `name` varchar(150) DEFAULT NULL, PRIMARY KEY (`g_id`)); CREATE TABLE IF NOT EXISTS `game_player` ( `r_id` int(11) NOT NULL AUTO_INCREMENT, `p_id` int(11) DEFAULT NULL, `g_id` int(11) DEFAULT NULL, `bool` int(1) NOT NULL DEFAULT '0', PRIMARY KEY (`r_id`)); CREATE TABLE IF NOT EXISTS `players` ( `p_id` int(11) NOT NULL AUTO_INCREMENT, `playerid` varchar(150) NOT NULL, PRIMARY KEY (`p_id`), UNIQUE KEY `playerid` (`playerid`)); The players table is my list of users, and they're tied to the list of games via the game_player table. So here's my issue... I'm trying to show the full list of games, and then check mark each record where the player does play it. This is what I have so far - it shows all the games, but it's not checking the boxes. Code: [Select] $result = mysql_query("SELECT * FROM games") or die(mysql_error()); while($row = mysql_fetch_array($result)) { $newquery = "SELECT * FROM game_player, players WHERE game_player.p_id = players.p_id AND game_player.g_id = ".$row['g_id']. " AND players.playerid = {$userid}"; $query = mysql_query($newquery) or die(mysql_error()); if($query['bool'] == 1) { $set_checked = " CHECKED"; } else{ $set_checked = ""; } echo "<input type=\"checkbox\" name=\"box1\" value=\"".$query['g_id']."\"" . $set_checked . "/>".$row['name']."<br />\n"; } |