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PHP - Simple Array Not Working ! Very Strange
Hi,
I am performing some whois query using phpwois. http://www.phpwhois.org/ So after doing its object output i.e. Code: [Select] $whois = new Whois(); $result = $whois->Lookup($domain); $winfo = ''; $utils = new utils; $winfo = $utils->showHTML($result); , I am getting following data for for phpfreaks.com> Code: [Select] regrinfo->Array domain->Array name->phpfreaks.com nserver->Array ns1.serverpowered.net->204.13.168.4 ns2.serverpowered.net->216.12.222.4 ns3.serverpowered.net->66.97.171.74 ns4.serverpowered.net->66.97.171.254 status->Locked changed->2011-08-29 created->2001-10-11 expires->2016-10-11 registered->yes owner->Array name->Web Freaks address->Array 0->Contact Administrative () admin->Array email->domains@thewebfreaks.com name->Contact Administrative phone->4072751117 fax->+1.4072757706 address->Array 0->Web Freaks 4->3700 Commerce Blvd 5->Suite 154 6->Kissimmee, FL 34741 7->US tech->Array email->domains@thewebfreaks.com name->Contact Administrative phone->4072751117 fax->+1.4072757706 address->Array 0->Web Freaks 4->3700 Commerce Blvd 5->Suite 154 6->Kissimmee, FL 34741 7->US regyinfo->Array registrar->ENOM, INC. referrer->http://www.enom.com servers->Array 0->Array server->com.whois-servers.net args->domain =phpfreaks.com port->43 1->Array server->whois.enom.com args->phpfreaks.com port->43 type->domain Now I want to show Domain registrar (Enom, INC.) So I did following Code: [Select] echo $winfo['regyinfo']; Now above code shows following >> Code: [Select] registrar->ENOM, INC. referrer->http://www.enom.com servers->Array 0->Array server->com.whois-servers.net args->domain =phpfreaks.com port->43 1->Array server->whois.enom.com args->phpfreaks.com port->43 type->domain To obtain I mean to show ENOM, INC, I used following > Code: [Select] echo $winfo['regyinfo']['registrar']; But it gave error, so I used following >> Code: [Select] $temp = $winfo['regyinfo']; echo $temp['registrar']; But it give output as follows : < I dont know what wrong I am doing ? If it is showing regyinfo, then it should also show registrar. Thanks Similar TutorialsHi there i have an array that catches all appropriate records from my database and selects a random row which is then outputted as a hyperlink. Code: [Select] $colname_Class3 = "3"; $colname_Ships3 = "-1"; if (isset($_SESSION['MM_Username'])) { $colname_Ships3 = (get_magic_quotes_gpc()) ? $_SESSION['MM_Username'] : addslashes($_SESSION['MM_Username']); } mysql_select_db($database_swb, $swb); $query_Ships3 = sprintf("SELECT * FROM ships WHERE PlayerName = %s AND Class=%s AND Template='1'", GetSQLValueString($colname_Ships3, "text"), GetSQLValueString($colname_Class3, "int")); $Ships3 = mysql_query($query_Ships3, $swb) or die(mysql_error()); $totalRows_Ships3 = mysql_num_rows($Ships3); $count = 0; while ($row = mysql_fetch_assoc($Ships3)){ $array[$count]['ShipID'] = $row['ShipID']; $array[$count]['ShipName'] = $row['ShipName']; $array[$count]['Class'] = $row['Class']; $count++; } $total3 = count($array3); $random3 = rand(0,$total3 - 1); $random_ship3 = $array3[$random3]; $ShipID3 = $random_ship3['ShipID']; $url3 = '<a href="add_ship.php?recordID='.$ShipID3.'">Test3</a>'; echo '<span style="color: #009900">'; echo $url3;} This all worked fine until i added another similar query: Code: [Select] $colname_Class4 = "4"; $colname_Ships4 = "-1"; if (isset($_SESSION['MM_Username'])) { $colname_Ships4 = (get_magic_quotes_gpc()) ? $_SESSION['MM_Username'] : addslashes($_SESSION['MM_Username']); } mysql_select_db($database_swb, $swb); $query_Ships4 = sprintf("SELECT * FROM ships WHERE PlayerName = %s AND Class=%s AND Template='1'", GetSQLValueString($colname_Ships4, "text"), GetSQLValueString($colname_Class4, "int")); $Ships4 = mysql_query($query_Ships4, $swb) or die(mysql_error()); $totalRows_Ships4 = mysql_num_rows($Ships4); $count = 0; while ($row = mysql_fetch_assoc($Ships4)){ $array[$count]['ShipID'] = $row['ShipID']; $array[$count]['ShipName'] = $row['ShipName']; $array[$count]['Class'] = $row['Class']; $count++; } $total4 = count($array4); $random4 = rand(0,$total4 - 1); $random_ship4 = $array4[$random4]; $ShipID4 = $random_ship4['ShipID']; $url4 = '<a href="add_ship.php?recordID='.$ShipID4.'">Test4</a>'; echo $url4;} On first look it seems fine, however once the page is refreshed the parsed recordID's both become the same number, e.g: Test3 recordID 8 Test4 recordID 24 then after refreshed: Test3 recordID 24 Test4 recordID 24 Is there anything wrong with my code?? Thanks I am working with the Amazon API, and I am trying to display a default image if a product does not have an image associated with it. The query is coming back as an array. Typically, each product image array looks like this: $d = SimpleXMLElement Object ( http:// => [url]http://ecx.images-amazon.com/images/I/51aUIul6XjL._SL160_.jpg [Height] => 160 [Width] => 112 ) The code goes like this: Code: [Select] if ($d=$E->MediumImage) { $iu=$d->URL; $ih=$d->Height; $iw=$d->Width; echo count($d); if (strlen($iu) > 0) {echo "<center><a href='$url' target='_blank'><img src='images/amazon_noimage.jpg' width='175' height='175' border='0'></a></center>";} else {echo "<center><a href='$url' target='_blank'><img src='$iu' width='$iw' height='$ih' border='0'></a></center>";} } However, images/amazon_noimage.jpg never shows up (even though it is linked correctly, as I've tested this link). I have tried the following: if (strlen($iu) > 0) if (count($iu) > 0) if (strlen($d->URL) > 0) if (count($d->URL) > 0) if (isset($d)) if (!isset($d)) etc ... If I display the following, where there is no image, I get nothing displayed: echo $iu; print_r($iu); echo $d->URL; etc ... However, if there is an image, I get a link, such as the following: http://ecx.images-amazon.com/images/I/51c2BFpDN0L._SL160_.jpg There seems to be NOTHING that I can do to trigger the 'ELSE' part of the if statement. This is a total enigma to me ... any ideas?? I've got to be missing something pretty basic here.. considering the query is pretty basic. I'm trying to figure out how to pull a query as an array so I can compare it against another array (array_diff) I'm doing a mysql_fetch_array, and I'm getting an error ( mysql_fetch_array(): supplied argument is not a valid MySQL result resource): Quote $checker = "SELECT ID FROM edible_uses"; $result2 = mysql_fetch_array($checker) or die(mysql_error()); //echoing to see if I'm getting what I need. echo $row['ID']; I've done a mysql_query and I get results. The table name and all that is correct. I'm stumped. This seems like a pretty simple query? I tried mysql_fetch_assoc as well. Same result? I tried it with an extra set of parenthesis around it. nope. Hi, I am new to php, and I have run into a problem. The tutorial I am using has provided me with this exact code. But it does not work for me. Its very simple. Here is the HTML page: Code: [Select] <body> <FORM ACTION="welcome.php" METHOD=POST> First Name: <INPUT TYPE=TEXT NAME="name"> <INPUT TYPE=SUBMIT VALUE="GO"> </FORM> </body> And here is the php page: <body> <?php echo( "Welcome to our Web site, $name!" ); ?> </body> You can see the problem live at <http://www.freewaycreative.com/test> (dont mind the digits below) The name just does not show. Anyone know why? Thanks! I'm trying my first PHP code :
<!DOCTYPE html>
<?php
</body>
Result : nothing, just a blank screen ... Am I missing something ?
Regards, Martin Hey all,
This code was given to me by my client who swears it works, yet I can't seem to get it to function.
<?php
$now = time(); // or your date as well
$your_date = strtotime("2016-06-01");
$datediff = $now - $your_date;
$referrals = number_format(162250 + (floor($datediff/(60*60*24)) * 527) + (87920 + (floor($datediff/(60*60*24)) * 45)));
?>
//javascript:
<script>
//vars from template
var referrals = "<?php echo $referrals;?>";
var currentdate = new Date();
jQuery("document").ready(function() {
jQuery ("#sp-menu > div > nav > ul > li:nth-child(1) > a").html("Home");
jQuery(".referrals").html(referrals);
jQuery(".current_date").html((currentdate.getMonth()+1) + "/" + currentdate.getDate() + "/" + currentdate.getFullYear());
jQuery(window).on("scroll", function() {
var scrollPos = jQuery(window).scrollTop();
if (scrollPos <= 0) {
jQuery(".counter").fadeIn();
jQuery(".sec-nav").fadeIn();
} else {
jQuery(".counter").fadeOut();
jQuery(".sec-nav").fadeOut();
}
});
});
</script>
The following mysql query is not returning rows like I expect it to. '$update_field' is a variable, matching an actual field name in table 'users'. 'user_task[1]' is an integer value. What am I missing here? Code: [Select] $query_update_user = "UPDATE users SET ".$update_field." = 'Y' WHERE user_no = '".$user_task[1]."'"; I am running PHP Version 5.3.1. The following Code does not write to my database. It is code that I took from the PHP Pocket Reference from O'Reilly but it does not work... What's wrong with this code? Thanks in advance. Guy <?php if($vote && !$already_voted) SetCookie('already_voted',1); ?> <html> <head> <title>Name the Baby</title> </head> <h3>Name the Baby</h3> <form action="baby.php" method="POST"> <p>Suggestion: <input type="text" name="new_name"/> </p> <input type="submit" value="Submit idea and/or vote"/> <?php mysql_pconnect("localhost","root","password"); $db = "babynames"; $table = "baby_names"; if($new_name) { if(!mysql_db_query($db, "insert into $table values ('$new_name',0)")) { echo mysql_errno().': '. mysql_error()."<br />\n"; } } if($vote && $already_voted) { echo '<p><b>Hey, you voted already '; echo "Vote ignored.</b></p>\n"; } else if($vote) { if(!mysql_db_query($db, "update $table set votes=votes+1 where name='$vote'")) { echo mysql_errno().': '. mysql_error()."<br />\n"; } } $result=mysql_db_query($db,"select sum(votes) as sum from $table"); if($result) { $sum = (int) mysql_result($result,0,"sum"); mysql_free_result($result); } $result=mysql_db_query($db, "select * from $table order by votes DESC"); echo <<<EOD <table border="0"><tr><th>Vote</th> <th>Idea</th><th colspan="2">Votes</th></tr> EOD; while($row=mysql_fetch_row($result)) { echo <<<FOO <tr><td align="center"> <input type="radio" name="vote" value="$row[0]"></td> <td>$row[0]</td> <td align="right">$row[1]</td> <td> FOO; if ($sum && (int)$row[1]) { $per = (int)(100 * $row[1]/$sum); echo '<img src="bline.gif" height=12 '; echo "width=$per> $per %</td>"; } echo "</tr>\n"; } echo "</table>\n"; mysql_free_result($result); ?> <input type="submit" value="Submit idea and/or vote" /> <input type="reset" /> </form> </body></html> Hi, I have a php form that I use to try to get matching data from the database that I put into the form. So if I enter date of birth 9-4-80 and first name Dave and lastname Smith. When I submit it the code should pull all of the matching terms out of the database and display. Now I get the following error when I submit the form. Query: Resource id #2 Failed with error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Resource id #2' at line 1 any help Greatly appreciated. thank you! here is the code Code: [Select] <?php ini_set ("display_errors", "1"); error_reporting(E_ALL); $host = " "; $database = " "; $username = " "; $password = " "; $tbl_name = "users"; $conn = mysql_connect($host, $username, $password) or die("Could not connect: " . mysql_error()); if($conn) { mysql_select_db($database); echo "connected to database!!"; } else { echo "failed to select database"; } //include('bouncer.php'); //$currentUser = $_SESSION['myusername']; if(isset($_POST['submit'])) { $first = mysql_real_escape_string( $_POST['first']); $last = mysql_real_escape_string( $_POST['last']); $dob = mysql_real_escape_string( $_POST['dob']); //THE SEARCH FUNCTION $sql = mysql_query ( "SELECT * FROM users WHERE firstname LIKE '%$first%' OR lastname LIKE '%$last%' OR dob LIKE '%$dob%' ") or die(mysql_error()); $result = mysql_query($sql) or die( "<br>Query: $sql<br>Failed with error: " . mysql_error() ); if (!$result) { echo "Could not successfully run query ($sql) from DB: " . mysql_error(); exit; } if (mysql_num_rows($result) == 0) { echo "No rows found, nothing to print so am exiting"; exit; } while ($row = mysql_fetch_assoc($result)) { echo $row["firstname"]; echo $row["lastname"]; echo $row["dob"]; } }//if(isset($_POST['submit'])) ?> <html> <body> <form action="login_success8.php" method="post"> <p> <input type="text" name="first" size="20" /> First name<br /> <input type="text" name="last" size="20" /> Last name<br /> <input name="dob" type="text" size="20" id="dob" /> Date of Birth<br /> <input type="submit" name="submit" value="Search" /> <input type="reset" value="Reset fields" /> </p> </form> </body> </html> Not really sure what to ask because I don't know what the problem is. Maybe just need a fresh set of eyes to find out whats wrong. I am trying selecting content from a database table but its not working. I am not getting any errors just a blank screen where the content should be displayed. html Code: [Select] <table> <tbody> <tr> <th>Topic</th> <th>Name</th> <th>Date</th> </tr> <?php include 'server/forum.php'; ?> </tbody> </table> php Code: [Select] <?php require_once('load_data.php'); $con = mysql_connect($db_host, $db_user, $db_pwd); if (!$con) { die('Could not connect to database: ' . mysql_error()); } $dbcon = mysql_select_db($database); if (!$dbcon) { die('Could not select database: ' . mysql_error()); } $sql = "SELECT * FROM $table ORDER BY id"; $result = mysql_query($sql) or die("Error ". mysql_error(). " with query ". $sql); if (!$result) { die("Query to show fields from table failed:".mysql_error()); } $row = mysql_fetch_array($result) while($row) { echo '<tr>'; echo '<td class="forumtd"><b>'; echo '<a href="topic.php?id='.$row['id'].'">'.stripslashes(htmlspecialchars($row['topic'])).'</a>'; echo "</b></td>"; echo '<td class="forumtd"><em>'; echo stripslashes(htmlspecialchars($row['name'])); echo "</em></td>"; echo '<td class="forumtd">'; echo date("l M dS, Y", $row['date']); echo "</td>"; echo "</tr>"; } mysql_close($con); ?> well I lowered my standards massively. LOL. I decided to google, "php upload and display image", instead of "php gallery". Here is the error I get Warning: copy(/images/sheila.jpg) [function.copy]: failed to open stream: No such file or directory in /hermes/bosweb/web173/b1739/sl.brendansite1/public_html/ealike2/smallgallery/smallgallery.php on line 59 and here is the script that I mostly understand. I thought it was the ..images/, but now I don't know what it is. any help greatly appreciated. thank you. below is the code for the page. Code: [Select] <!-- Start PHP Code For Image Upload --> <?php //define a maxim size for the uploaded images in Kb define ("MAX_SIZE","5060"); //This function reads the extension of the file. It is used to determine if the file is an image by checking the extension. function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } //This variable is used as a flag. The value is initialized with 0 (meaning no error found) //and it will be changed to 1 if an errro occures. //If the error occures the file will not be uploaded. $errors=0; //checks if the form has been submitted if(isset($_POST['Submit'])) { //reads the name of the file the user submitted for uploading $image=$_FILES['image']['name']; //if it is not empty if ($image) { //get the original name of the file from the clients machine $filename = stripslashes($_FILES['image']['name']); //get the extension of the file in a lower case format $extension = getExtension($filename); $extension = strtolower($extension); //if it is not a known extension, we will suppose it is an error and will not upload the file, //otherwise we will do more tests if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { //print error message echo '<h2>Unknown extension!</h2>'; $errors=1; } else { //get the size of the image in bytes //$_FILES['image']['tmp_name'] is the temporary filename of the file //in which the uploaded file was stored on the server $size=filesize($_FILES['image']['tmp_name']); //compare the size with the maxim size we defined and print error if bigger if ($size > MAX_SIZE*1024) { echo '<h2>You have exceeded the file size limit! Please reduce the image size to 100 Kb or less!</h2>'; $errors=1; } //we will give an unique name, for example the time in unix time format $image_name=$filename; //the new name will be containing the full path where will be stored (images folder) $newname="../images/".$image_name; //we verify if the image has been uploaded, and print error instead $copied = copy($_FILES['image']['tmp_name'], $newname); if (!$copied) { echo '<h2>Copy unsuccessful!</h2>'; $errors=1; }}}} //If no errors registred, print the success message if(isset($_POST['Submit']) && !$errors) { echo "<h2>File Uploaded Successfully!</h2><br />"; echo "<img src='http://ealike.com/images/<?php echo $image_name; ?> />"; } ?> <!-- End PHP Code For Image Upload --> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <!-- Start Image Upload Form --> <form name="newad" method="post" enctype="multipart/form-data" action=""> <input type="file" name="image"> <input name="Submit" type="submit" value="Upload image"> </form> <!-- End Image Upload Form --> </body> </html> Hi, I'm trying to make a simple slideshow with forward and back buttons that change the image inside a div in php. When I press the "further" button, it jumps from index[0] to index[1] and never shows the first image. Then when I click more it doesn't go forward. When I click back, it goes back to a black screen div. Any help getting this to work is GREATLY appreciated because I've been trying for 2 days with tutorials and can't get it. Thanks. Derek Here are the php parts that are relevant, my page was too large and confusing to include. first, the $background array where I store my images. Code: [Select] $background = array ( "<img src='sundragon_environments/ocean/ocean1_FRAME.jpg'/>", "<img src='sundragon_environments/ocean/ocean1_FRAME2.jpg'/>" ); then the code to move the images forward and backwards if the buttons are pressed. Code: [Select] if(!empty($_POST['further'])) { $currentBackground=next($background); } elseif(!empty($_POST['back'])) { $currentBackground=prev($background); } and now to echo out the images inside the div. Code: [Select] <div id="background"><?php echo $currentBackground;?></div> I am trying to use JSONReader (along with PHP & XPATH) to parse a very large JSON file, then display search results. A stream parser (such as JSONREader) is recommended over JSON_decode when parsing large files. This simple code below is not displaying any results (in the echo statements). Any advice is greatly appreciated. I have a form on my website which actions login.php. The login.php code is below: <?php include('includes/classes.php.inc'); session_start(); $link = new BaseClass(); $data = $link->query("SELECT * FROM logins"); $pass_accepted = false; if($_REQUEST['username'] && $_REQUEST['password']){ $username = $_REQUEST['username']; $password = $_REQUEST['password']; while($row = mysql_fetch_array($data)){ if(($row['username']==$useranme)&&($row['password']==$password){ echo 'Password correct!'; $_SESSION['loggedin']=true; $pass_accepted = true; } } } else { echo 'You did not enter a username or password!'; } if(!$pass_accepted){ echo 'Your password is incorrect'; } echo '<br>Please <a href="index.php">click here</a> to return to page'; ?> I have checked that my references are all correct however even when I enter the correct password it returns saying the password is incorrect. Any idea on why this could be? I am happy to answer any follow up questions. Regards Hello. i am writing an API for lua... that sends POST requests to a php script. this php script is malfunctioning. below is the code and error. my php array knowledge is still basic and i can't seem to get this code to yield an answer in the form i want. i would like to be able to get at $m and $b separately so I can manipulate them. how, for example do i just print $m? Code: [Select] <?php var_dump( linear_regression(array(1, 2, 3, 4), array(1.5, 1.6, 2.1, 3.0)) ); /** * linear regression function * @param $x array x-coords * @param $y array y-coords * @returns array() m=>slope, b=>intercept */ function linear_regression($x, $y) { // calculate number points $n = count($x); // ensure both arrays of points are the same size if ($n != count($y)) { trigger_error("linear_regression(): Number of elements in coordinate arrays do not match.", E_USER_ERROR); } // calculate sums $x_sum = array_sum($x); $y_sum = array_sum($y); $xx_sum = 0; $xy_sum = 0; for($i = 0; $i < $n; $i++) { $xy_sum+=($x[$i]*$y[$i]); $xx_sum+=($x[$i]*$x[$i]); } // calculate slope $m = (($n * $xy_sum) - ($x_sum * $y_sum)) / (($n * $xx_sum) - ($x_sum * $x_sum)); // calculate intercept $b = ($y_sum - ($m * $x_sum)) / $n; return array("m"=>$m, "b"=>$b); } ?> Hello. I have a newbie question regarding arrays: How can I echo an element's subsequent arrays and elements? For example Code: [Select] $pers = array("Steve" => array("eyes" => "green", "age" => "43", "race" => "caucasian")); and if I echo $pers["Steve"] to make it display: green 43 caucasian. Thanks. 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 |
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