|
PHP - Accessing Resource
If $result contains the result of mysql_query, a select count distinct query, how do I access the count? I've tried $result[0] to no avail and $result only returns the resource #.
Similar TutorialsHi guys, I have xml that looks like: <item> <title></title> <link></link> <description></description> <dc:date>2011-02-27T16:42:31-05:00</dc:date> </item> I can access title, link, and description just fine. But how do I access a tag that has a colon? I tried this: Code: [Select] $x = new SimpleXmlElement($content); $entry = $x->item; $entry->children('dc', true)->date; Thanks for looking. LR HEy Guys Here is a snippet of an array: SimpleXMLElement Object ( [report1] => SimpleXMLElement Object ( [ReportHost] => Array ( => SimpleXMLElement Object ( [HostName] => 192.168.2.34 [startTime] => Thu Dec 17 13:58:49 2009 [stopTime] => Thu Dec 17 15:13:13 2009 [netbios_name] => (unknown) [mac_addr] => (unknown) [dns_name] => (unknown) [os_name] => (unknown) [num_ports] => 2 [num_lo] => 0 [num_med] => 0 [num_hi] => 0 [ReportItem] => Array which is printed using print_r($a) now how can i access [HostName] => 192.168.2.34 in the $a array?? Thanks in advance I am a part of an institue which has internal web pages. It also provides me with my web space with php servers. Now I know those internal web pages are available to the server. I want a code that can fetch those web pages and display it in my web space. And whatever forms i fill or links i click it should do those on internal page and again return the new web page. Hello everyone, I am writing some code for a login script. I keep getting the error Resource ID #13. What does this mean and how can I fix it? Here is the code that I am having trouble with Code: [Select] function user_id_from_username($username) { $username = sanitize($username); $query = mysql_query("SELECT `user_id` FROM `users` WHERE `username` = '$username'"); return mysql_result($query, 0, 'user_id'); } function login($username, $password) { $user_id = user_id_from_username($username); $username = sanitize($username); $password = md5($password); $query = "SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username' AND `password` = '$password'"; return (mysql_result(mysql_query($query), 0) == 1) ? $user_id : false; } Hello everyone, I am writing some code for a login script. I keep getting the error Resource ID #13. What does this mean and how can I fix it? Here is the code that I am having trouble with Code: [Select] function user_id_from_username($username) { $username = sanitize($username); $query = mysql_query("SELECT `user_id` FROM `users` WHERE `username` = '$username'"); return mysql_result($query, 0, 'user_id'); } function login($username, $password) { $user_id = user_id_from_username($username); $username = sanitize($username); $password = md5($password); $query = "SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username' AND `password` = '$password'"; return (mysql_result(mysql_query($query), 0) == 1) ? $user_id : false; } I was expecting a return string, but got Resource id #2 instead. How do I have a string returned instead of that? heres my table user code Bob One Ted Two I dont get it Code: [Select] <html> <body> <?php $con = mysql_connect("localhost","user","PassWord"); if (!$con) { echo 'Could not connect to MySQL server. <br />Error # ', mysql_errno(), ' Error msg: ', mysql_error(); exit; } $db = mysql_select_db("userdb") or die("Unable to select database"); if (!$db) { echo 'Could not select db. <br />Error # ', mysql_errno(), ' Error msg: ', mysql_error(); exit; } $query = "SELECT code from usertbl WHERE user = 'Ted' LIMIT 0 , 30"; $result = mysql_query($query, $con); if (!$result) { echo 'Could not query server. <br />Error # ', mysql_errno(), ' Error msg: ', mysql_error(); exit; } echo $result; ?> </body> </html> When I use that query in phpmyadmin it works Any pointers much appreciated Hey PHPFreaks. I made a php code, thats only needed to be showed for admin accounts only. I tryed to echo the mysql_num_rows($result); and it gave me this: Resource id #51 Heres a piece of my code where the problem is: $result = mysql_query("SELECT adminlevel FROM accounts WHERE name = '".$_SESSION['auth_username']."'") or die(mysql_error()); echo $result; echo mysql_num_rows($result); if(mysql_num_rows($result) == 1) { echo '<br /><br /><a href="home.php?admin">Admin Area</a>'; } Hope you can help I keep getting a resource #6 at the top of my script and not sure why. <?php if (isset($_REQUEST['option'])) { switch ($_REQUEST['option']) { case 0: ?> <h1 class="backstage">Biographies Management</h1><br /> <h2 class=backstage>Bio Types</h2><br /> <?php $query = "SELECT * FROM efed_list_styles AS styles"; $result = mysql_query ( $query ); $rows = mysql_num_rows($result); if ($rows > 0) { print'<table width="100%" class="table1"> <tr class="rowheading"> <td> </td> <td width="40" align="center">ID</td> <td>Name</td> </tr>'; $i = 0; while ( $row = mysql_fetch_array ( $result, MYSQL_ASSOC ) ) { $sClass = 'row2'; if ($i++ % 2) $sClass = 'row1'; printf ( "<tr class=\"%s\">", $sClass ); print "<td valign=\"top\" align=\"center\" width=\"30\"><a href=\"#\" onclick=\"ajaxpage('backstage_libs/biolayout.php?option=1&id=$row[id].', 'content'); return false;\">Edit</a></td>"; printf ( "<td align=\"center\" valign=\"top\" width=\"40\">%s</td>", $row ['id'] ); printf ( "<td valign=\"top\">%s</td>", $row ['name'] ); echo '</tr>'; } echo '</table><br>'; } else { echo '<span>There are no bio types.</span><br /><br />'; } returnmain(); footercode(); break; case 1: require_once('../backstageconfig.php'); require_once('../backstagefunctions.php'); $id = $_GET['id']; $query = mysql_query("SELECT * FROM `efed_list_styles` WHERE `id` = '" . $id . "'"); $row = mysql_fetch_array($query); echo $query; ?> <h1 class="backstage">Bio Layouts Management</h1><br /> <h2 class="backstage"><?php echo $row['name']; ?> Biography Layout</h2><br /> <?php } } ?> This works...
$result = mysql_query("SELECT * FROM login_attempts");
while($row = mysql_fetch_array($result)) {
$diff = strtotime($row['login_date'])-time();
if ($diff > -300){
$count = $count + 1;
}
if ($count > 4) {
$result = "locked";
}
}
This returns Resource ID 5...
$result = mysql_query("SELECT * FROM login_attempts WHERE login_username='$username'");
while($row = mysql_fetch_array($result)) {
$diff = strtotime($row['login_date'])-time();
if ($diff > -300){
$count = $count + 1;
}
if ($count > 4) {
$result = "locked";
}
}
What's the problem?Many thanks, Hi all, I am still trying to figure what went wrong. It doesn't work over here. The error shown was the {else} command else { echo '<p class="error">There was a problem accessing your profile.</p>'; } I have attached my 4 sql tables for reference. Appreciate any advices.... Code: [Select] <?php $query3 = "SELECT ols.subject_level_id, sl.level_id, sl.subject_id, tl.level_name AS level_name, ts.subject_name AS subject_name " . "FROM tutor_overall_level_subject AS ols " . "INNER JOIN tutor_subject_level AS sl USING (subject_level_id) " . "INNER JOIN tutor_level AS tl USING (level_id) " . "INNER JOIN tutor_subject AS ts USING (subject_id) " . "WHERE ols.tutor_id = '" . $_GET['tutor_id'] . "'"; $data3 = mysqli_query($dbc, $query3) or die(mysqli_error($dbc)); if (mysqli_num_rows($data3) == 1) { echo '<div id="panel4">'; echo'<table><tr>'; // Start your table outside the loop... and your first row $count = 0; // Start your counter while($row3 = mysqli_fetch_array($data3)) { /* Check to see whether or not this is a *new* row If it is, then end the previous and start the next and restart the counter. */ if ($count % 5 == 0) { echo "</tr><tr>"; $count = 0; } echo '<td class="label">' . $row3['level_name'] . '</td><td>' . $row3['subject_name'] . '</td>'; $count++; //Increment the count } echo '</tr></table><br/>'; //Close your last row and your table, outside the loop echo '</div>'; //End of panel 4 } //End of if (mysqli_num_rows($data3) == 1) else { echo '<p class="error">There was a problem accessing your profile.</p>'; } ?> I have the following code in zend: Code: [Select] $arrErrors=array(); if (!empty($this->post['submit'])) { // Each time theres an error, add an error message to the error array // using the field name as the key. if (empty($this->post['client_name'])) $arrErrors['client_name'] = "Please Enter Client's name as it appears in the carrier software"; } if i set $this->view->arrErrors=$arrErrors in the controller, Can I access it as $this->arrErrors['client_name'] in the view? I use a CURL script API with the header "$headers[] = 'Content-Type: application/json';" The script ends with "$result = curl_exec($ch);" If I simply echo the $result to my browser, I get an array that looks like this:
stdClass Object
[toAddress] => stdClass Object Do I really need to "json_decode" this array? I've spent 2 1/2 days trying every which way, with every permutation to access that one value:
echo $result["fromAddress"]["phone"]; Unfortunately I must have fallen asleep during the array lesson in PHP class ? Thank you. Can somebody help me what i'm doing wrong? The value from database is not visible when using: <?php $artikel = new Artikel(); $content= $artikel->printArtikel(); echo $content; ?> <?php class Artikel { // db connection protected $mydb; // database value protected $land; public function __construct($artikelnummer = false) { // databaseconnection: $this->mydb = new MyDB(); if($artikelnummer) { $this->load($artikelnummer); } else // load everything from database. { $sql = "SELECT * FROM artikel"; $this->mydb->doQuery($sql); $this->mydb->close(); } } public function load($artikelnummer) { $sql = "SELECT * FROM artikel"; $this->mydb->doQuery($sql); if($artikel = $this->mydb->fetch()) { // i got a feeling there is something missing here, like query from database?? } $this->mydb->close(); } public function printArtikel() { // return database value as table. $html = "<table border='1'>"; $html .= "<tr> <td>".$this->land."</td> </tr>"; $html .= "</table>"; return $html; } } ?> Other class are accessing extern. For example database connection with class MyDB. please, can somebody correct my script?? Hi all, I've tried to access the admin.php of the xcart installation he http://www.cleaningshopuk.com but it doesn't allow me to access it, saying that it's a 404 page not found error. That page does in fact exist, but I think the issue might be something to do with a redirection. The site owner apparently clicked on something in the admin section (possibly https or something similar) but what can I do to avoid this issue without having to re install everything? Thanks, Neil How do I access an object on different pages? I created object "A" on pageone.php and I need to access it on pagetwo.php. How do I do that? TomTees Hey everyone, I'm creating a website which requires users to login. I didn't want to have to do my own crypto, so I basically created it with Wordpress. I'm at the point where I don't want to use Wordpress for the whole site--just the login system. So I was wondering how I can piggyback off the user management system of Wordpress while using my own script for the rest of the site. I know that Wordpress ends up storing passwords in the database as HMACs, for what that's worth. Thank you, Ryan Is it possible to have a class access another class it its parent directory without using a fully qualified name? While the following script is not valid, it is how I would have thought it would be done should it be possible. If it is not possible, I expect there is a good reason why one would not want to do so, and would appreciate your thoughts why that is so. Thanks
// src/PrimaryTopic/SubTopic/SubSubTopic/ParentClass.php
namespace PrimaryTopic\SubTopic\SubSubTopic;
class ParentClass {}
// src/PrimaryTopic/SubTopic/SubSubTopic/Child/ChildClass1.php
namespace PrimaryTopic\SubTopic\SubSubTopic\Child;
class ChildClass1 extends ..\ParentClass {}
// src/PrimaryTopic/SubTopic/SubSubTopic/Child/ChildClass2.php
namespace PrimaryTopic\SubTopic\SubSubTopic\Child;
class ChildClass2 extends ..\ParentClass {}
Hello All: I am trying to access a value within an object. The value name has whitespace in it and is called "zip code". When I try to access the value via regular object call such as: Code: [Select] $sub = $sub->zip code; PHP goes crazy on me. I've tried to encapsulate the value name within a variable like this: Code: [Select] $zip_code = "Zip Code"; $sub = $sub->zip_code; and that doesn't work either. Does anyone have any ideas about how I should go about yanking the data out? So basically I'm using PHP to solve a math problem for me. The user puts in a few parameters and the program runs those numbers through the algorithm and spits out a bunch of xy coordinates. I'm using the PHP SELF method to retrieve the user input. This all works fine, but it's afterwords that I run into problems. I create a drop down menu of all the x values and I want the user to be able to choose any x value they wish and to have the corresponding y value be displayed. The problem is that I have to use another POST command for this and when I do, it wipes out all the computed data. Of course the y-value is never displayed, and there lies the problem. I'm sure there are many ways to get around this, but I could not find one out myself. Anyone have any ideas? I am currently trying create a sales system where it checks the user's username against the database to check whether they are in the list of buyers. The mysql query returns "Resource id #35", I need it to return the actual username (which I manually inserted into the database to test). PHP code that fetches from database: <?php $con = mysql_connect("x","x","x"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("x", $con); $check_buyer = "SELECT * FROM Buyers WHERE Buyer='x'"; $buyer = mysql_query($check_buyer); ?> Product page: <?php include("/home/x/public_html/scripts/buyer.php"); if ($user->data['user_id'] == ANONYMOUS) { echo 'To use ' . $product . 'you must be logged in!'; echo '<br /><a href="http://x/forum/ucp.php?mode=register">Register</a>'; echo ' or '; echo '<a href="http://x/forum/ucp.php?mode=login">Sign In</a>'; } elseif ($user->data['username_clean'] == $buyer) { echo "<h3>Welcome to x</h3>"; } else { echo "You need to buy this product to use it!"; echo $user->data['username_clean']; //test whether username is outputted correctly - which it did echo $buyer; //Fetched from mysql - returned "Resource id #35", not the desired username } ?> Any help would be great! Thanks, otester |
|||||||