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PHP - Dropdown List To Show The Pubs In The Town
hi,
I would like to show a list of pubs in a town that the user has selected from the drop down list http://www.mypubspace.com/dashtest/order.html it is all there working but in a different way If you type Brighton into the Town search, it shows a list of pubs in that town, GREAT! If you click the 'show towns list' link, a drop down appears, (this is where my problem is) When you click the town, I would like to get the list of pubs in that town basically get rid of the form where you type in the box and click the button here is my code HTML and Javascript Code: [Select] <html> <body> <script language="javascript" type="text/javascript"> <!-- //Browser Support Code function ajaxFunction(){ var townRequest; // The variable that makes Ajax possible! try{ // Opera 8.0+, Firefox, Safari townRequest = new XMLHttpRequest(); } catch (e){ // Internet Explorer Browsers try{ townRequest = new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { try{ townRequest = new ActiveXObject("Microsoft.XMLHTTP"); } catch (e){ // Something went wrong alert("Your browser broke!"); return false; } } } // Create a function that will receive data sent from the server townRequest.onreadystatechange = function(){ if(townRequest.readyState == 4){ var ajaxDisplay = document.getElementById('ajaxDiv'); ajaxDisplay.innerHTML = townRequest.responseText; } } var name = document.getElementById('name').value; var county = document.getElementById('county').value; var town = document.getElementById('town').value; var queryString = "?name=" + name + "&county=" + county + "&town=" + town; townRequest.open("GET", "http://www.mypubspace.com/dashtest/pubs.php" + queryString, true); townRequest.send(null); } function townlistFunction(){ var ajaxRequest; // The variable that makes Ajax possible! try{ // Opera 8.0+, Firefox, Safari ajaxRequest = new XMLHttpRequest(); } catch (e){ // Internet Explorer Browsers try{ ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { try{ ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP"); } catch (e){ // Something went wrong alert("Your browser broke!"); return false; } } } // Create a function that will receive data sent from the server ajaxRequest.onreadystatechange = function(){ if(ajaxRequest.readyState == 4){ var ajaxDisplay = document.getElementById('ajaxTownlist'); ajaxDisplay.innerHTML = ajaxRequest.responseText; } } ajaxRequest.open("GET", "http://www.mypubspace.com/dashtest/town-select.php", true); ajaxRequest.send(null); } function countylistFunction(){ var ajaxRequest; // The variable that makes Ajax possible! try{ // Opera 8.0+, Firefox, Safari ajaxRequest = new XMLHttpRequest(); } catch (e){ // Internet Explorer Browsers try{ ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { try{ ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP"); } catch (e){ // Something went wrong alert("Your browser broke!"); return false; } } } // Create a function that will receive data sent from the server ajaxRequest.onreadystatechange = function(){ if(ajaxRequest.readyState == 4){ var ajaxDisplay = document.getElementById('ajaxCountylist'); ajaxDisplay.innerHTML = ajaxRequest.responseText; } } ajaxRequest.open("GET", "http://www.mypubspace.com/dashtest/county-select.php", true); ajaxRequest.send(null); } function MM_jumpMenu(targ,selObj,restore){ //v3.0 eval(targ+".location='"+selObj.options[selObj.selectedIndex].value+"'"); if (restore) selObj.selectedIndex=0; } //--> </script> <a href="#" onClick="townlistFunction();">show towns list</a> <a href="#" onClick="countylistFunction();">show counties list</a> <div id="ajaxTownlist"></div> <div id="ajaxCountylist"></div> <form name='myForm'> Pub Name: <input type='text' id='name' /> <br /> County: <input type='text' id='county' /> <br /> Town: <input type='text' id='town' /> <input type='button' onclick='ajaxFunction()' value='Query MySQL' /> </form> <div id='ajaxDiv'></div> </body> </html> here is my pubs.php code Code: [Select] <?php $dbhost = "xxx"; $dbuser = "xxx"; $dbpass = "xxx"; $dbname = "xxx"; //Connect to MySQL Server mysql_connect($dbhost, $dbuser, $dbpass); //Select Database mysql_select_db($dbname) or die(mysql_error()); // Retrieve data from Query String $name = $_GET['rsPubName']; $sex = $_GET['rsTown']; $wpm = $_GET['rsCounty']; // Escape User Input to help prevent SQL Injection $name = mysql_real_escape_string($name); $town = mysql_real_escape_string($town); $county = mysql_real_escape_string($county); //build query $query = "SELECT * FROM pubs WHERE rsTown = '$town'"; $qry_result = mysql_query($query) or die(mysql_error()); //Build Result String $display_string = "<table>"; $display_string .= "<tr>"; $display_string .= "<th>Region</th>"; $display_string .= "<th>Pub Name</th>"; $display_string .= "<th>Town</th>"; $display_string .= "<th>County</th>"; $display_string .= "</tr>"; // Insert a new row in the table for each person returned while($row = mysql_fetch_array($qry_result)){ $display_string .= "<tr>"; $display_string .= "<td>$row[Region]</td>"; $display_string .= "<td>$row[rsPubName]</td>"; $display_string .= "<td>$row[rsTown]</td>"; $display_string .= "<td>$row[rsCounty]</td>"; $display_string .= "</tr>"; } $display_string .= "</table>"; echo $display_string; ?> here is my town_select.php code Code: [Select] <?php $dbhost = "xxx"; $dbuser = "xxx"; $dbpass = "xxx"; $dbname = "xxx"; //Connect to MySQL Server mysql_connect($dbhost, $dbuser, $dbpass); //Select Database mysql_select_db($dbname) or die(mysql_error()); // Retrieve data from Query String $town = $_GET['rsTown']; // Escape User Input to help prevent SQL Injection $town = mysql_real_escape_string($town); //build query $query = "SELECT DISTINCT rsTown, COUNT(PUBID) As PubCount FROM pubs GROUP BY rsTown ORDER BY rsTown ASC"; $qry_result = mysql_query($query) or die(mysql_error()); // Insert a new row in the table for each person returned $display_string .= "<form name=\"form3\" method=\"post\" action=\"\">"; $display_string .= "<select name=\"menu2\" onChange=\"MM_jumpMenu('parent',this,0)\" class=\"textbox\">"; $display_string .= "<option value=\"\">Search by Town...</option>"; while($row = mysql_fetch_array($qry_result)){ $display_string .= '<option value="rsTown='.$row['rsTown'].'" onclick=\"ajaxFunction()\" id=\"town\">'.$row['rsTown'].' ('.$row['PubCount'].')</option>'; } $display_string .= "</select>"; $display_string .= "</form>"; echo $display_string; ?> Similar TutorialsCode: [Select] for ($i = 1; $i<$rows+1; $i++) { $Choicequery ="SELECT PersonA, PersonB FROM modules WHERE GroupID = '$UpdateNumber'"; $Choiceresult = mysql_query($Choicequery); $ChoiceRows = mysql_fetch_array($Choiceresult); $PersonQuery = "SELECT FirstName, LastName FROM Persons p, modules m, users e, group t WHERE m.modID = $ChoiceRows[0] AND e.UserID = m.PersonA AND p.PersonID = e.PersonID"; $PersonResult = mysql_query($PersonQuery); $PersonRows = mysql_fetch_array($PersonResult); $Person2Query = "SELECT FirstName, LastName FROM Persons p, modules m, users e, group t WHERE m.modID = $ChoiceRows[0] AND e.UserID = m.PersonB AND p.PersonID = e.PersonID"; $Person2Result = mysql_query($Person2Query); $Person2Rows = mysql_fetch_array($Person2Result); echo "<Option value = \"$i\">$PersonRows[0] $PersonRows[1] and $Person2Rows[0] $Person2Rows[1] </Option>"; } How do I get it to show the next result when $i increases? All the options are the first people A and B $i number of times. Not sure how to use a loop here to go though the results. Whats the best way? ie $PersonRows[0] $PersonRows[1] are the same for every option I tried inserting a town into my database but only the first word gets inserted?! eg: Abbots Langley only Abbots gets inserted?! Code: [Select] if(isset($_POST["register"])) { // Your code here to handle a successful verification $RSTOWN = $_POST['rsTown']; $rsGender = $_POST['rsGender']; $RSUSER = $_POST['RSUSER']; $RSPASS = $_POST['RSPASS']; $rsEmail = $_POST['rsEmail']; $rsMobile = $_POST['rsMobile']; $rsAge = $_POST['rsAge']; $to = 'info@mypubspace.com, '.$rsEmail; //define the subject of the email $subject = 'Welcome '.$RSUSER.' to My Pub Space'; // message $message = ' <html> <head> <title>'.$subject.'</title> </head> <body> <table> <tr> <td>Name:</td> <td>'.$RSUSER.'</td> </tr> <tr> <td>Email:</td> <td>'.$rsEmail.'</td> </tr> <tr> <td>Town:</td> <td>'.$rsTown.'</td> </tr> <tr> <td>Telephone:</td> <td>'.$rsMobile.'</td> </tr> <tr> <td>Age:</td> <td>'.$rsAge.'</td> </tr> <tr> <td>Password:</td> <td>'.$RSPASS.'</td> </tr> </table> </body> </html> '; // To send HTML mail, the Content-type header must be set $headers = 'MIME-Version: 1.0' . "\r\n"; $headers .= 'Content-type: text/html; charset=iso-8859-1' . "\r\n"; // Additional headers $headers .= 'To:' .$to. "\r\n"; $headers .= 'From:' .$rsEmail. "\r\n"; // Mail it mail($to, $subject, $message, $headers); $sql = "INSERT INTO members_copy (RSTOWN, RSGENDER, RSUSER, RSPASS, RSEMAIL, RSMOBILE, RSAGE) VALUES ('$rsTown', '$rsGender', '$RSUSER', '$RSPASS', '$rsEmail', '$rsMobile', '$rsAge');"; //echo $sql; mysql_query($sql); Hi, I am trying to call the data from Mysql but I am getting an empty drop down list, this is the code: mysql: Code: [Select] create table years ( yearID integer auto_increment, year varchar(30), primary key (yearID) ); insert into years (yearID, year) values ('1', '2007-2008'); insert into years (yearID, year) values ('2', '2008-2009'); insert into years (yearID, year) values ('3', '2009-2010'); insert into years (yearID, year) values ('4', '2010-2011'); insert into years (yearID, year) values ('5', '2011-2012'); insert into years (yearID, year) values ('6', '2012-2013'); PHP: Code: [Select] <?php require_once('../Connections/connection.php'); ?> <?php $result = @mysql_query( "select yearID, year, from sss.years"); print "<p>Select a year:\n"; print "<select name=\"yearID\">\n"; while ($row = mysql_fetch_assoc($result)){ $yearID = $row[ 'yearID' ]; $year = $row[ 'year' ]; print "<option value=$yearID>$year\n"; } print "</select>\n"; print "</p>\n"; ?> Thank you! friend the code display all the managers name
<?phpand the variable below displays the result $Manager= $row['Manager'] ; but how do i do it on the above dropdown list the it shows which result is in the $Manager variable? Hey, I have the following coding: Quote <? $dbuser="*******"; $dbpass="*******"; $dbname="virtuda_db"; //the name of the database $chandle = mysql_connect("localhost", $dbuser, $dbpass) or die("Connection Failure to Database"); mysql_select_db($dbname, $chandle) or die ($dbname . " Database not found. " . $dbuser); $mainsection="license"; $query1="select name from license"; $result = mysql_db_query($dbname, $query1) or die("Failed Query of " . $query1); //do the query while($thisrow=mysql_fetch_row($result)) { $i=0; while ($i < mysql_num_fields($result)) { $field_name=mysql_fetch_field($result, $i); echo $thisrow[$i] . " "; //Display all the fields on one line $i++; } echo "<br>"; //put a break after each database entry } ?> How would I set up this so that instead of just "listing" them out on new lines, it would list the results into a drop down list? Thanks! I'm trying to sort this dropdown box. It reads from a directory, and lists the file name in the dropdown box. Here's the tricky part... the filename is listed differently in the dropdown than in the directory by using explode(). I want to sort it though since it's still being sorted by the directory listings... For example: Filename starts out as: 123_abc_567.pdf then gets listed as abc_123_567.pdf in the dropdown, but it's still getting sorted as if it were 123_abc_567.pdf How can I do that? Here's my code: // Define the full path to folder from root $path = "C:/Work_Orders/"; // Open the folder $dir_handle = @opendir($path) or die("Unable to open $path"); echo "<form method=\"POST\" action='".$_SERVER['PHP_SELF']."' name='selectworkorder'><select name='ordernumber2'>"; // Loop through the files while ($file = readdir($dir_handle)) { //Remove file extension $ext = strrchr($file, '.'); if($ext !== false) { $file = substr($file, 0, -strlen($ext)); } if($file == "." || $file == ".." || $file == "index.php" ) continue; //explode file name $changedordernumber = explode("_",$file); //put in new order $changedordernumber = $changedordernumber[1]."_".$changedordernumber[0]."_".$changedordernumber[2]; $changedordernumber=trim($changedordernumber,"_"); //list options echo "<option name='$file' value='$file'>$changedordernumber</option>\n"; } echo "</select><input type='submit' value='Change' name='submit'/></form></div>"; // Close closedir($dir_handle); Alright, so I have an xml file differences.xml that is being parsed in XML. This is what the xml looks like: <item code="lM" name="dog"> <cost>5000</cost> <Start>12/15/2010</Start> <End>01/13/2011</End> </item> <item code="lF" name="cat"> <cost>5000</cost> <Start>04/15/2010</Start> <End>04/23/2011</End> </item>[/ I want to have the item names (dog, cat) show in a dropdown menu so that I can select these items for editing before storing in my mysql database. This is the php code I have so far: <?PHP $xml = simplexml_load_file("differences.xml"); $object = $xml->xpath("//item"); echo '<SELECT name=object>'; foreach ($object['name'] as $key => $value) { echo '<OPTION value='.$value.'> '.$value; } echo '</select>'; ?> I do have a dropdown list but there are no values inside it (it is empty). Can anyone help me figure out why? I do have this code that does work which lists the items in plaintext (not in a dropdown) so hopefull this will help us out: <?PHP $xml = simplexml_load_file("differences.xml"); $object = $xml->xpath("//item"); $count = count($object); $i = 0; while($i < $count) { echo '<h1>'.$object[$i]['name'].'</h1>'; $i++; } ?> 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> If I can get this fixed, I will have completed all but the admin login for this project - my first php/mysql project. Here is what I need. I have a list_records.php that list all the records in the table 'links' and the category each entry is in from the table 'categories'. Here are my table structures. 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 ; On the update.php file, I have a form that lets me make changes to the record. Here is the codes for update.php Code: [Select] <? include "menu.php" ?> <? include "db.php" ?> <?php $id=$_GET['id']; $sql = "select * from links where id =$id"; $query = mysql_query($sql); while ($row = mysql_fetch_array($query)){ $id = $row['id']; $catid = $row['catid']; $name = $row['name']; $url = $row['url']; $content = $row['content']; //we will echo these into the proper fields } mysql_free_result($query); ?> <table width="65%" align="center"> <tr><td align="left"> <form action="updated.php" method="post"> <input type="hidden" value="<?php echo $id; ?>" name="id"/> <br> <b>Website Name:</B><br> Change the name of the website listing.<br> <input type="text" value="<?php echo $name; ?>" name="name"/> <br> <br> <b>URL:</b><br> Change the URL of the website listing.<br> <input type="text" value="<?php echo $url; ?>" name="url"/> <br> <br> <b>Description:</b><br> Change the description of the website listing.<br> Limit 255 characters.<br/> <textarea name="content" cols="45" rows="4" wrap="soft"><?php echo($content);?></textarea> <br> <?php $result = mysql_query("SELECT id, categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $id = $rows["id"]; $categories=$row["categories"]; $options.= '<option value="'.$row['id'].'">'.$row['id'].'-'.$row['categories'].'</option>'; }; ?> <SELECT NAME=catid> <OPTION VALUE=selected><? echo $catid; ?><? echo $options; ?></OPTION> </SELECT> <?php mysql_close(); ?> <div align="center"> <input type="submit" value="submit changes"/> </div> </form> <br> </td></tr></table> The part of he code I need help with is Code: [Select] <?php $result = mysql_query("SELECT id, categories FROM categories") or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $id = $rows["id"]; $categories=$row["categories"]; $options.= '<option value="'.$row['id'].'">'.$row['id'].'-'.$row['categories'].'</option>'; }; ?> <SELECT NAME=catid> <OPTION VALUE=selected><? echo $catid; ?><? echo $options; ?></OPTION> </SELECT> I want it to default to the category that the entry is in. If you look, you will see in the select portion that I I have Code: [Select] <$ echo $catid; ?> which echos the proper category ID, but if I use Code: [Select] <? echo $categories; ?> it echos Writing, which is the last category in the list. Yet, the $options echo the catid and it corresponding category. How can I get the default option to echo BOTH the catid and category name while also listing all the other categories so that the records can be moved to a new category is needed? Any help will be appreciated. Thank you in advance. Hi i'm new to php. I want to get all the values of dropdown list on another page wether selected or not. i 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 "$dropdown"; print "Subject:<br>"; print "<input type='text' name='subject' size='20'><br>"; print "Your message:<br>"; print "<textarea name='yourpost' rows='5' cols='40'></textarea><br>"; print "<input type='submit' name='submit' value='submit'></form>"; } print "</td></tr></table>"; ?> MOD EDIT: Changed PHP manual link [m] . . . [/m] tags to [code] . . . [/code] tags. This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=316599.0 I have the following code currently: Code: [Select] <?php foreach ((array)$node->field_buy_at as $item) { ?> <?php print $item['view'] ?> <?php } ?> I would like to make the list a drop down with a link so that when a user selects, he goes to a new page. I tried the following: Code: [Select] <select name="select"> <?php foreach ((array)$node->field_buy_at as $item) { ?> <?php $url = $node->field_buy_at[0]['url']; $store = $item['view']; ?> <? echo "<option value='$url'>$store</option>";?> <?php } ?> </select> I'm pretty sure it's this "$url = $node->field_buy_at[0]['url'];" that I don't have correct. Hey Guys, I know it may seem pretty simple, but im having trouble populating a drop down list. Here is my code at the moment, but what it's doing is displaying the names all in one value, where it should be in separate select values. *Note that i have only done it to the first one. See attachment. 'AntonMatt' are next to each other, they should be separate select values. Code: [Select] <? $id = $_GET['id']; $selectplayers="SELECT * FROM players WHERE club='$club' AND team='$team'"; $player=mysql_query($selectplayers); ?> <table class='lineups' width="560" cellpadding="5"> <tr> <td colspan="2">Starting Lineup</td> <td colspan="2">On the Bench</td> </tr> <tr> <td width="119"> </td> <td width="160"> </td> <td width="69"> </td> <td width="160"> </td> </tr> <tr> <td>Prop</td> <td><select name="secondary" style="width: 150px"> <option value='' selected="selected"><? while($rowplayer = mysql_fetch_array($player)) { echo $rowplayer['fname']; } ?></option> </select></td> <td>16.</td> <td><select name="secondary16" style="width: 150px"> <option value='' selected="selected">Secondary Position</option> </select></td> </tr> <tr> <td style="padding-top: 8px;">Hooker</td> <td style="padding-top: 8px;"><select name="secondary2" style="width: 150px"> <option value='' selected="selected">Secondary Position</option> </select></td> <td style="padding-top: 8px;">17.</td> <td style="padding-top: 8px;"><select name="secondary17" style="width: 150px"> <option value='' selected="selected">Secondary Position</option> </select></td> </tr> </table> </div> </div> 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 have the following code: <td width='100px'>Suppliers <select name="supplier"> <?php $catcher_id = $service->getCatcherId(); $supplier_names = LpmAdnetworkPeer::getByName($catcher_id); foreach($supplier_names as $row) { ?> <option><?php echo $row->getName(); ?></option> <?php } ?> </select> </td> then on the same form i have a submit button that takes me to the next form..the problem now is how can i access the ID of the item seleted in the dropdown on the NEXT form please? i can get the name from the list by $_POST['supplier'] on the next form thanks Hello everyone I have two dropdown lists. First one you can see the tables of my db and the second one the files inside a folder. The problem is that i can't send my selections from the dropdown list to an other file. My project is to choose a table, then a CSV file and then load the file in db. index.php: Code: [Select] <?php //EMFANISH PINAKWN include("../forma/dbCon.php"); echo "<form action='load_table_test.php' method='POST'>"; $sql="SHOW TABLES FROM misthodosia"; $result=mysql_query($sql); if (!$result) { echo "Υπάρχει πρόβλημα με την βάση\n"; echo 'MySQL Error: ' . mysql_error(); exit; } echo"<select>"; while ($row = mysql_fetch_row($result)) { echo "<option name='{row[0]}' value='{row[0]}'>{$row[0]}</option>"; } echo"</select>"; mysql_free_result($result); //==========================EMFANISH ARXEIWN========================== function populateDropdown(){ $path = glob("C:\\xampp\\htdocs\\dbadmin\\upload_files\\*"); //the double \\ is not a typo you //have to escape it //if the path cannot be found if(!($path)){ echo("Δεν υπάρχει ο φάκελος"); return; }//if //the path is valid else{ foreach($path as $k){ $i = basename($k); echo "<option name='$i'>".$i."</option>"; }//foreach }//else }//populateDropdown echo"<select>"; populateDropdown(); echo"</select> "; echo"<input type='submit' name='submit' value='Φόρτωσε1111111' />"; echo "</form>"; $hello="hello"; ?> load_table_test.php: Code: [Select] <?php include("../forma/dbCon.php"); echo $_POST['hello']; echo $_POST['{row[0]}']; echo $i; //$pinakas=$_POST['pinakas'] //$arxeio=$_POST['arxeio'] //$pinakas=$_POST['row'] //$arxeio=$_POST['diadromi'] //mysql_query("LOAD DATA LOCAL INFILE "$arxeio" INTO TABLE $pinakas //FIELDS //TERMINATED BY ',' //ENCLOSED BY '\"' //LINES TERMINATED BY '\n') echo "$arxeio"; echo "$pinakas"; echo "<br/><br/><br/><br/><br/><a href='index.php'>Επιστροφή στην προηγούμενη σελίδα.</a>"; ?> errors: Code: [Select] Notice: Undefined index: hello in C:\xampp\htdocs\dbadmin\load_table_test.php on line 6 Notice: Undefined index: {row[0]} in C:\xampp\htdocs\dbadmin\load_table_test.php on line 8 Notice: Undefined variable: i in C:\xampp\htdocs\dbadmin\load_table_test.php on line 9 Notice: Undefined variable: arxeio in C:\xampp\htdocs\dbadmin\load_table_test.php on line 27 Notice: Undefined variable: pinakas in C:\xampp\htdocs\dbadmin\load_table_test.php on line 28 i skipped solving this the other day with an easier way but now i am stuck with this stumble again in another area and i have no way out.... so here it goes, i have an ajax dropdown box...i need to get the value that is selected by the user when it is clicked and then pass this value to a new pop window by appending to its url....any suggestions? how come the lines: PHP Code: echo $form['catcher_id']; $catcher_names = $form['catcher_id'] foreach($catcher_names as $val) { echo "testing loop"; } produce nothing?? its not doing the foreach. $form['catcher_id']; is a dropdown list containing catcher names please help? thanks |
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