PHP - Simple Xml Problem
so I have an xml string
the following expression $xpath = $response->xpath('/atom:feed/atom:entry/atom:content'); returns the following array SimpleXMLElement Object ( [@attributes] => Array ( [type] => application/vnd.google-earth.kml+xml ) [Placemark] => SimpleXMLElement Object ( [name] => M1 [description] => SimpleXMLElement Object ( ) [Style] => SimpleXMLElement Object ( [IconStyle] => SimpleXMLElement Object ( [Icon] => SimpleXMLElement Object ( [href] => http://maps.gstatic.com/intl/en_ALL/mapfiles/ms/micons/blue-dot.png ) ) ) [Point] => SimpleXMLElement Object ( [coordinates] => -79.395018,43.645423,0.0 ) ) ) However what I'm trying to get are the coordinates, but neither $xpath = $response->xpath('/atom:feed/atom:entry/atom:content/Placemark'); OR $xpath = $response->xpath('/atom:feed/atom:entry/atom:content/Placemark/child::*'); work I even tried just loading atom:content into an array and going from there no luck. Any ideas. Oh. BTW this might help http://code.google.com/apis/maps/documentation/mapsdata/developers_guide_protocol.html#RetrievingFeatures Similar TutorialsSOLVED, sorry. ... but i can't work it out. all i want to do is; if ( $this_item exists within the table ) { do a } else { do b } for example, $this_item = 'john', and i want the script to do (a) if he's in the table, or (b) if he isn't. thank you! This is my website: www.wearenip.com/home.html This is the menu that should appear on the left hand side, but doesn't: www.wearenip.com/menu.php This is my server's output saying that php is running just fine: http://wearenip.com/phpinfo.php I'm using this code in the left side bar div: Code: [Select] <?php include("menu.php"); ?> The PHP file will not display from the HTML file no matter what I do. I've tried using every possible combination of file linking including /menu.php ./menu.php and even the full http://www.wearenip.com/menu.php etc etc etc I'm genuinely upset about this. Any help would be greatly appreciated. PS: I can't get any PHP file to properly include in any HTML file on any computer I have. Nothing displays whatsoever regardless of what is in the PHP file, even if it's just a text word. Doesn't matter if it's hosted on a PHP compatible internet server or just on a local server or just on the local machine itself. Nothing. It shouldn't be that difficult I just don't know how to do it myself. I'm using the following code for a simple navigation, i has been placed in the content of my home page, it works fine however i don't know how to set the home page. <div id="content"> <?php $page = $_GET['page']; $file = $page.".php"; if(file_exists($file)) { include($file); } else { include "error.php"; } ?> </div> Sorry this page is not available or has been accessed in error An error occurred in script 'C:\xampp\htdocs\public_html\login\plugins\delete_product.php' on line 45: Undefined variable: id Date/Time: 8-24-2010 00:50:10 Array ( [GLOBALS] => Array *RECURSION* [_POST] => Array ( ) [_GET] => Array ( [id] => 2 ) <?php # Script 9.2 - delete_product.php // This page is for deleting a album. require_once ('../includes/config.inc.php'); // Check for a valid product ID, through GET or POST: if ( (isset($id)) && (is_numeric($id)) ) { // From editproducts.php $id = $_GET['id']; } elseif ( (isset($_POST['id'])) && (is_numeric($_POST['id'])) ) { // Form submission. $id = $_POST['id']; } else { echo '<p class="error">Sorry this page is not available or has been accessed in error</p>'; } echo $id; require_once(MYSQL); // Check if the form has been submitted: if (isset($_POST['submitted'])) { if ($_POST['sure'] == 'Yes') { // Delete the record. // Make the query: $q = "DELETE FROM products WHERE product_id='$id' LIMIT 1"; $r = @mysqli_query ($dbc, $q); if (mysqli_affected_rows($dbc) == 1) { // If it ran OK. // Print a message: echo '<p>The Product has been deleted.</p>'; } else { // If the query did not run OK. echo '<p class="error">The Product could not be deleted due to a system error.</p>'; // Public message. echo '<p>' . mysqli_error($dbc) . '<br />Query: ' . $q . '</p>'; // Debugging message. } // Make the query: $q = "DELETE FROM photos WHERE product_id='$id'"; $r = @mysqli_query ($dbc, $q); } else { // No confirmation of deletion. echo '<p>The Product has NOT been deleted.</p>'; } } else { // Show the form. // Retrieve the albums information: $q = "SELECT name FROM products WHERE product_id='$id' Limit 1"; $r = @mysqli_query ($dbc, $q); if (mysqli_num_rows($r) == 1) { // Valid user ID, show the form. // Get the user's information: $row = mysqli_fetch_array ($r, MYSQLI_NUM); $form = $_SERVER['PHP_SELF']; // Create the form: echo '<form action="' . $form . '" method="post"> <h1>Delete ' . $row[0] . '?</h1> <p>Are you sure you want to delete ' . $row[0] . '?<br /> <p><b>Warning if you delete this album all content inside of it will be losed!</b><br /> <input type="radio" name="sure" value="Yes" /> Yes <input type="radio" name="sure" value="No" checked="checked" /> No</p> <p><input type="submit" name="submit" value="Submit" /></p> <input type="hidden" name="submitted" value="TRUE" /> <input type="hidden" name="id" value="' . $id . '" /> </form>'; } } mysqli_close($dbc); ?> Im simply having a problem with structuring a directory structure for mkdir Code: [Select] mkdir("saves/".$worldname."/".$playername, 0777); The middle "/" is causing issues, if I take it out I have no problem in creating a folder name saves/worldnameplayername/ just want to add playername as a sub-dir. I have tried every combination I can think of to get it to accept the divider Any help? Hi -- This query seems to be problematic because the UPDATE is not being performed. Could you please take a gander and let me know what is the problem? BTW, the table "teamy" does contain 630 records. Thanks in advance! $sql = "SELECT COUNT( * ) AS records FROM teamy" ; $result = mysql_query( $sql ) ; if ( ! $result ) { die ( __line__ . "_teamy_" . mysql_error() ) ; } $a_count = mysql_fetch_assoc($result); if ( $a_count['records'] = 0 ) { echo "No records found in teamy." ; } else { /*** Update r_rost_rma ***/ $sql = "UPDATE r_rost_rma JOIN teamy ON r_rost_rma.student_id = teamy.student_id SET r_rost_rma.teachername = teamy.team WHERE RTRIM( UPPER( r_rost_rma.localcourse ) ) = 'HOMEROOM'" ; $result = mysql_query ( $sql ) ; if ( ! $result ) { die ( __line__ . "_update_r_rost_rma_" . mysql_error() ) ; } } Ok I decided to brush up on some functions and this one was giving me a hard time. Basically the if condition is printing the statement even though its not what the expression was asking for. I'm positive I am just missing something small, but I can't figure it out. Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xm1ns="http://www.w3.org/1999/xhtml" xml:lang="en"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8"/> <title>Counting Words...</title> </head> <body> <?php function count_words() { $totalwords = str_word_count($_POST['total']); $amountneeded = $_POST['amount']; $wordsneeded = $amountneeded - $totalwords; $extrawords = $totalwords - $amountneeded; if (isset($_POST['submitted'])) { if(str_word_count($_POST['totalwords']) <= $_POST['amountneeded']) { print "\nYou need $wordsneeded more words"; } else { print "\nYou wrote an extra $extrawords words"; } if(str_word_count($_POST['totalwords']) == $_POST['amountneeded']) { print "You hit the jackpot and got just as much words as you needed"; } } } ?> <form action="wordcount.php" method="post"> <p>Words Needed: <input type="text" name="amount" value="" /></p> <p>Copy and Paste your words:</p> <textarea name="total" rows="5" cols="30"></textarea> <input type="submit" name="submit" value="Count My Words" /> <input type="hidden" name="submitted" value="true" /> </form> <?php count_words(); ?> </body> </html> Hi! I am not great at using php but do use it to process forms on websites. I always use the same script and just change the relevant sections for each website. This usually works with no problems. I am writing an mailing list subscribe/unsubscribe form my website and it works fine as in I get the email sent through with all the information on it and the automated reply gets sent out etc. The problem is that once the user hits Submit instead of going to the redirect page it stays on the php page. if($from == '') {print "You have not entered an email, please go back and try again";} else { if($name == '') {print "You have not entered a company name, please go back and try again";} else { $send = mail($to, $subject, $body, $headers); $send2 = mail($from, $subject2, $autoreply, $headers2); if($send) {header( "Location: http://www.hillsideweb.co.uk/unsubthanks.html" );} Like I say this exact script works on another site of mine perfectly well, can someone please help me solve this. The form is found on http://www.hillsideweb.co.uk/unsubscribe.html I hope this makes sense! Many thanks Swenglish Hello, i am trying to create a simple funtion that i can all upon to "turn off a website" i am calling the function by <?php siteonline(n); ?> and the function is function siteonline($msg){ $offlinecheck = mysql_query("SELECT * FROM acp") or die(mysql_error()); $siteoffline = mysql_fetch_array($offlinecheck); $ifsiteoffline = $siteoffline['site_offline']; $offline = $siteoffline['offline_msg']; if ($ifsiteoffline == "y") { echo("$offline"); die();} } now this will turn off the site but it shows no message what so ever and i can not figure out why can anyone help? Cheers, Hi guys.I've got a problem.I'm building my php browsergame and i'm stuck into the inventory page.I have a code that seems ok to me but it won't work.Here it is: $inventory = array(); $query = sprintf("SELECT id, item_id, quantity FROM user_items WHERE user_id = '%s'", mysql_real_escape_string($userID)); $result = mysql_query($query); while($row = mysql_fetch_assoc($result)) { $item_query = sprintf("SELECT name FROM items WHERE id = '%s'", mysql_real_escape_string($row['item_id'])); $item_result = mysql_query($item_query); list($row['name']) = mysql_fetch_row($item_result); array_push($inventory,$row); } I've spotted the problem.The $inventory array is empty.I checked it with if(empty($inventory)) .Thanks in advance and keep in mind that I'm a noob yet. I have a script.. it works fine on my localhost.. now i am trying to work it with on my remote server.. the script reads the folder in my c drive and shows me the file on display here is the code <?php $main_dir ="C:\\xampp\\htdocs\\xtrajam"; $main_open = opendir($main_dir); while(($main_file = readdir($main_open)) != FALSE) { echo $main_file."<br />"; } closedir($main_open); ?> Now the thing is this works fine on local host as it can easy access the path C:\\xampp\\htdocs\\xtrajam Now i want this path to be accessed sumhow wen i try it on my remote server.. how should i do that like when i run the script online it reads the contents from my c drive directory..? is it sumhow possible? I am thining this will be pretty simple for someone to help me with, I have been hassling with it for several hours and really feel dumb that I can't get it to work. Here is the problem - I have one script wherein I have a form that passes a variable using Post method. I have previously assigned a value "A" to a variable $rtn the input is: <input name="Rtn" type="hidden" value="<?php echo $rtn; ?> /> In the script that I am calling I first get the value of the variable: $rtn = (_POST['Rtn']); then there is other stuff goin on and I come to an if stmt: if ($rtn == "A" { do some stuff } else { echo $rtn; } It will never get into the do some stuff - it always falls through to the else the output is \'A\' I have tried to use double quotes, single quotes, used stripslasches and addslashes funtions, trim, and various combination of all of these. Nothing I do seem to get me into the first part of the if stmt, to do some stuff. Can someone explain to me whats wrong with this? <?php session_start(); // store session data $_SESSION['views']=1; ?> <html> <body> <?php //retrieve session data echo "Pageviews=". $_SESSION['views']; ?> </body> </html> The output it gives me in the browser is this: Quote Warning: session_start() [function.session-start]: Cannot send session cookie - headers already sent by (output started at D:\Program Files\xampp\xampp\htdocs\sessions.php:1) in D:\Program Files\xampp\xampp\htdocs\sessions.php on line 2 Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at D:\Program Files\xampp\xampp\htdocs\sessions.php:1) in D:\Program Files\xampp\xampp\htdocs\sessions.php on line 2 Pageviews=1 Explanation? Hi, This is a image tooltip from database. but I have got the problem, Not Working! How can i fix it. <a href="index.php?MD=urundetay&resimid=<?php echo $row_uruns['resimid']; ?>"><img src="urun_resim/<?php echo $row_image['image']; ?>" width="150" border="0" /></a> Haven't posted here for quite a while but I'm working on my first php driven website and am having a few problems with some includes that I'm sure you'll find to be fairly basic and obvious. What I'm trying to do is get my "home" page to load into a division when you initially visit my site. The header and footer had an include that contains various links which include this home page. The navigation links to the start page work perfectly and as expected, however when the site is visited initially all I get is a blank space in the main content division. The site in question is linked here... http://valvetech.byethost14.com/ The code that I'm using is as follows and I'm at a complete loss as to what I'm doing wrong. Code: [Select] <div id="center_column"> <?php $url = ''; if (!empty($_GET['category'])) { $url .= $_GET['category'] . '/'; } if (!empty($_GET['page'])) { $url .= $_GET['page'] . '.php'; } include $url; ?> </div> Thanks in advance for any assistance anybody can give me because I'm at the end of my rope on what should be a fairly easy problem to fix. -Carl heres the code... <?php mysql_connect("localhost","peter","tizer") or die("Error: ".mysqlerror()); mysql_select_db("matquant"); //replace TestTable with the name of your table //also in a real app you would get the id dynamically $idtochange=$_POST['idtochange']; $sql="select * from `quantites` where id = $idtochange"; $query=mysql_query($sql); while ($row=mysql_fetch_array($query)){ $id = $row['id']; $material = $row['material']; $quantity = $row['quantity']; $unit = $row['unit']; //we will echo these into the proper fields } mysql_free_result($query); ?> <html> <head> <title>Edit material list</title> <link href="css/quant1.css" rel="stylesheet" type="text/css"> </head> <body> <div id="leftcol"> <form "method="post"> <p><br/> <input type="text"value="1" name="idtochange" > <p> <input type="submit"value="submit changes"/> </form> <form action="changequan.php"method="post"> <p>Material Id currently active<br/> <input type="text"value="<?php echo $id;?>" name="id" disabled/> </p> <p> Enter new material name: </p> <input type="text"value="<?php echo $material;?>" name="material"/> <br/> change quantity: <input type="text"value="<?php echo $quantity;?>" name="quantity"/> <br/> unit of mesurement for quantity<br/> <input type="text"value="<?php echo $unit;?>" name="unit"/> <br/> </p> <input type="submit"value="submit changes"/> </form> </div> <div id="rightcol"><?php // make connection mysql_connect ("localhost", "peter", "tizer") or die ('I cannot connect to the database becuase: ' . mysql_error()); mysql_select_db ("matquant"); // build query $query = mysql_query("SELECT * FROM `quantites` ORDER BY `id` ASC"); // display results while ($row = mysql_fetch_array($query)) { echo "<br /> ..... " .$row ['id']. " ....... " .$row ['material']. " ...... " .$row['quantity']. " ...... " .$row['unit']. "<br />";} echo mysql_error(); ?> <a href="index.html">Home </a> <p> <a href="quanform.php">Insert new material and quantity form</a></p></div> </body> </html> I'm struggling with php. Any help anyone can give would be appreciated. So, I'm going through a tutorial in my PHP book and we're using the code below. The page shows up and connects to the database (no error reported), but no data comes up. I'm wondering if I'm just missing something small or what? Let me know if you can help <html> <head> <title>Search Results</title> </head> <body> <h1>Book-0-Rama Search Results</h1> <?php //create short variable names $searchType=$_POST['searchType']; $searchTerm=trim($_POST['searchTerm']); if(!$searchType || !$searchTerm){ echo 'You have not entered enough information to process the search'; exit; } if(!get_magic_quotes_gpc()){ $searchType = addslashes($searchType); $searchTerm = addslashes($searchTerm); } echo $searchType; echo $searchTerm; @ $db = new mysqli('localhost','root','root'); if(mysqli_connect_errno()){ echo 'Error: Could not connect to the database man...'; exit; } $query = "SELECT * FROM books WHERE ".$searchType." LIKE '%".searchTerm."%'"; $result = mysqli_query($query); $num_results = mysqli_num_rows($result); echo "<p>Number of books found: ".$num_results."</p>"; for ($i=0;$i<$num_results;$i++){ $row = mysqli_fetch_assoc($result); echo "<p><strong>".($i+1).". Title: "; echo htmlspecialchars(stripslashes($row['title'])); echo "</strong><br/>Author: "; echo stripslashes($row['author']); echo "<br/>ISBN: "; echo stripslashes($row['isbn']); echo "<br/>Price: "; echo stripslashes($row['price']); echo "</p>"; } $result->free(); $db->close(); ?> </body> </html> Can anybody see how to fix the table below. My results are being stacked horizontally rather than vertically. I have attached a picture for you to see what I mean. Thanks //SKILL - query the database table $sql="SELECT Skill_Name, Rating FROM resource_skill ln inner join skill n on ln.Skill_ID = n.Skill_ID WHERE ln.Resource_ID=" . $contactid; //SKILL - run the query against the mysql query function $result=mysql_query($sql) or die(mysql_error() . '<br />' . $sql); echo "<table id=skill>"; echo "<tr>"; echo "<th>Skill</th>"; echo "<th>Rating</th>"; echo "</tr>"; //create while loop and loop through result set while($row=mysql_fetch_array($result)){ $Skill_Name=$row['Skill_Name']; $Rating=$row['Rating']; //display the result of the array echo "<ul>\n"; echo "<td>" . "" .$Skill_Name . " <td>" . $Rating . "</td></td>\n"; echo "</ul>"; } } echo "</tr>"; echo "</table>"; ?> Hi all I am grabbing the _SERVER['REQUEST_URI'] and trying to compare it but this doesnt work $currpage = $_SERVER['REQUEST_URI'];//when i echo this i get ' / ' $home = '/'; //but this doesnt work if(strpos($currpage, $home)){echo 'test works';} PS It will work if i use if($currpage == $home){echo 'test2';} Any ideas? Thanks, Magnetica I'm trying to learn arrays and I have this problem: I have a variable $a which is a result from a from a database query If I echo it I get: Code: [Select] forenamesurnamedate_of_birthgenderpaypal_addresspayment_planaddress_line_1address_line_2address_cityaddress_zip_post_codeaddress_countrytelephonemobilewhere_did_you_hearreceive_newsletter If I var_dump($a) I get: Code: [Select] string(8) "forename" string(7) "surname" string(13) "date_of_birth" string(6) "gender" string(14) "paypal_address" string(12) "payment_plan" string(14) "address_line_1" string(14) "address_line_2" string(12) "address_city" string(21) "address_zip_post_code" string(15) "address_country" string(9) "telephone" string(6) "mobile" string(18) "where_did_you_hear" string(18) "receive_newsletter" My question is, how do I get each value from a? It looks like an array to me because it holds multiple strings. I need to get the first, second and third elements printed on the page and used in other areas of my program. Thanks in advance |