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PHP - Gathering Print Outs Of Url Links In <div>?
Code: [Select]
foreach($html->find('a') as $element) echo $element->href . '<br>'; I understand that the code above results in printing out all links on the page in text. I want the same but only in a <div> I choose. See below: Code: [Select] <div id=web> <A href = myweb.com/url1>home</a> <A href = myweb.com/url2>about</a> <A href = myweb.com/url3>contact</a> </div> I want for example <div id=web> to print out onto page the urls in plain text from within that <div> or put into an array so I can create variables out of each url to do different actions. I'm using simple html dom I've tried different ways but I either get an error, a blank white page, or a print out of the word array? Either direct me to some web sources or write a code out for me, thanks Similar TutorialsHi everyone! I'm on a mission to make a simple (I hope :/ ) script that takes some files from a folder and sends them off to a snippet of javascript. I have a folder creatively named 'files' that is full of Google Earth .kml files. I want to get some PHP code to grab those files, and pass them to the javascript that handles the Google Earth plugin, which looks like this: Code: [Select] var link = ge.createLink(''); var href = '<?php echo($testvar); ?>' link.setHref(href); var networkLink = ge.createNetworkLink(''); networkLink.set(link, true, true); The php in there is me testing out php-to-javascript variables, which seems to be working fine for a single file. My plan for the code was to: 1) Detect all the files in the folder 2) Make the paths for each file into string variables 4) Loop the above javascript once for each file The end result would be the kml files, which are all building models, populating the earth. The files in the folder will be added and removed all the time, so It has to check it every time. Unfortunately this had to be my first experience with php and so this is as far as I got: foreach(glob("files/"."*.kml") as $file){ echo $file." - "; } So, it lists out the files in the directory with a ' - ' between each one, but from there I've been lost for days Does anyone have some pointers on what to do with this? Thanks, Love ya! Hi I am trying to find out the total value of dontations raised per email address but in order to find this information I have to do a few queries in order to find out the email address by using there ID and also need to see what percentage of each donation and which cause this donation went to. Find out donation amount Code: [Select] SELECT wp_supporters_donations.SFMemberNumber, wp_supporters_donations.KickbackAmount FROM supporter_2_cause RIGHT JOIN wp_supporters_donations ON supporter_2_cause.supp_id=wp_supporters_donations.SFMemberNumber GROUP BY wp_supporters_donations.SFMemberNumber Find out percentage here Code: [Select] function getdonationperc($supp_id, $causeid, $ProcessDate){ global $wpdb; $fu=0; $ProcessDate = substr($ProcessDate, 0, 6); //echo $ProcessDate; //echo $supp_id." "; //echo $causeid; $new = $wpdb->get_results("SELECT * FROM dp_log WHERE sid=$supp_id AND cid=$causeid AND date=$ProcessDate AND fp != 'yes' ORDER BY id LIMIT 0, 1"); foreach ($new as $new){ $fu = $new->perc; //echo $fu; } //echo " RESULT11 ".$fu; if($fu=="0" || $fu==NULL){ $newresult = $wpdb->get_results("SELECT * FROM supporter_2_cause WHERE supp_id=$supp_id AND caus_id=$causeid AND deleted!='1' AND fp!='yes' ORDER BY id LIMIT 0, 1"); foreach ($newresult as $newresult){ $fu = $newresult->donation; //echo "wtf"; } //echo $fu; if($newresult == NULL) { $fu="0"; } }else{ //$fu="0"; } //echo $supp_id." - ".$causeid." - ".$ProcessDate." - ".$fu."<br/>"; //echo " RESULT ".$fu; return $fu; } Find out email address attached to each donation Code: [Select] SELECT email FROM supporters WHERE id='SFMemberNumber' Then I need to add all donations up for each email address per each cause and output these / insert into a new table. <html> <?php $id = $_GET['id']; $dbusername="web148-matt"; $dbpassword="matt"; $dbdatabase="web148-matt"; mysql_connect(localhost,$dbusername,$dbpassword); @mysql_select_db($dbdatabase) or die( "Unable to select database"); mysql_query("UPDATE count SET clicks=clicks+1 WHERE id='$id'"); $sql = mysql_query("SELECT link FROM count WHERE id='$id'"); $fetch = mysql_fetch_row($sql); $result = mysql_query("SELECT * FROM count"); while($row = mysql_fetch_array($result)) { echo "<a href=" .$row['link']. ">Link</a>"; } ?> <a href='http://www.google.com'>Google</a> <a href='/index.php?id=2'>link2</a> </html> So I have been working on my website for a while which all is php&mysql based, now working on the social networking part building in similar functions like Facebook has. I encountered a difficulty with getting information back from a link. I've checked several sources how it is possible, with title 'Facebook Like URL data Extract Using jQuery PHP and Ajax' was the most popular answer, I get the scripts but all of these scripts work with html links only. My site all with php extensions and copy&paste my site links into these demos do not return anything . I checked the code and all of them using file_get_contents(), parsing through the html file so if i pass 'filename.php' it returns nothing supposing that php has not processed yet and the function gets the content of the php script with no data of course. So my question is that how it is possible to extract data from a link with php extension (on Facebook it works) or how to get php file executed for file_get_contents() to get back the html?
here is the link with code&demo iamusing: http://www.sanwebe.c...-php-and-jquery
thanks in advance.
Hello.
i am totally new to php and just started to learn now. i just dont understand why the following code is not printing the username that i enter on the page.
Please note that the code itself is saved with the name "basicForm.php".
Thanks.
<html> Hello, How to customize print page in php ? I had a page, but I need to print it like an invoice look page. Thanks in advance Code: [Select] $query ="SELECT oneID FROM table WHERE table.PersonID = 'game.PlayerA'" ; $result = mysql_query($query); $row = mysql_fetch_array($result); $oneID = $row[0]; [code] If I then echo "$oneID" why does it not print anything? $result echos resource7 Hi all this might be a bit of a novice question, but if anyone knows the following i would be pleased and my eyes will too I am looking for a way to print on paper or pdf (en print the pdf in the end) all the stringfunctions on php.net (http://www.php.net/manual/en/ref.strings.php) If anyone knows a way to do so i would love to hear it, because after a few days behind a computer my eyes are about to go on strike. Hi Guys I want to knw if its possible to send job to print with php. example, in my database I have a table called Letters which stores letters, while I loop through the table, I want to print each letter off, any examples available? Thank you This topic has been moved to Installation in Windows. http://www.phpfreaks.com/forums/index.php?topic=351150.0 Hi I have this code to echo th econtents of a table, which contains 100 rows. I want to be able to display all 100 rows. The problem is that it displays 99 (it excludes the first). I've tried to backtrack to reconstruct the code to find out where the error is but no joy. Any ideas? Thanks in advance! Code: [Select] .... // get all entries from table $sql = "SELECT * FROM 100words ORDER BY word_id"; $result = mysqli_query($dbc, $sql); $r = mysqli_fetch_row($result); //print table entries to screen in columns echo '<div id="container">'; // results presented in html table 5 columns, 20 rows per page echo '<div id="outerbox">'; echo '<div class="innerbox">'; echo '<table class="centerresults" border="0">' . "\n"; echo '<td>'; $i = 0; $max_columns = 5; while ($list = mysqli_fetch_assoc($result)) { extract($list); // open row if counter is zero if($i == 0) echo '<tr>' . "\n"; echo '<td width="150px"><a href="word.php?w=' . $list['word'] . '">' . $list['word'] . "</a></td> \n "; // increment counter - if counter = max columns, reset counter and close row if(++$i == $max_columns) { echo '</td>' . "\n"; $i=0; } // END if(++$i == $max_columns) { } // END while (!empty($myArray)) { //} END while ($list = mysql_fetch_array($result)) { //END if($i < $max_columns) { echo '</tr>' . "\n"; echo '</table>' . "\n"; echo '</div>'; echo '</div>'; Is there a way to dynamically print the url of a web page once it loads? If so, how? This is for metadata purposes. Thanks! I currently have a search page on my site that prints the products but it prints the products more than once if its in more than one category I have tried getting distinct item in my SQL. But this doesnt work so im trying an if statement that if there is more than one specific result then to just print this once. I was wondering if anyone had any ideas of how to do this using an if statement I just dont know how to go about just printing the result just once if its greater than 1. The code is below to make it clearer. $searchterm = $_POST['searchterm']; trim ($searchterm); /*check if search term was entered*/ if (!$searchterm){ echo 'Please enter a search term.'; echo $searchterm; } /*add slashes to search term*/ if (!get_magic_quotes_gpc()) { $searchterm = addslashes($searchterm); } /*query the database*/ $query = "SELECT * from (products LEFT JOIN categories_products_link ON products.prod_id = categories_products_link.prod_id) LEFT JOIN categories ON categories_products_link.cat_id = categories.cat_id WHERE prod_title LIKE '%" . $searchterm . "%' ORDER BY cat_title, prod_title"; $result = mysql_query($query); /*number of rows found*/ $num_results = mysql_num_rows($result); echo '<p><h1>Search Results: '.$num_results.'</h1></p><br />'; /*loops through results*/ for ($i=0; $i <$num_results; $i++) { $num_found = $i + 1; $row = mysql_fetch_assoc($result); echo "$num_found. "?><a href="store-<?php echo $row['cat_id'];?>-<?php echo $row['prod_id']; ?>/<?php echo seo_makeSafeURI($row['prod_title']); ?>.html"><strong><?php echo $row['prod_title']; ?></strong></a> <br /> I have the following code but I cannot get it to print the results of the array into my web page. Help would be so appreciated!! Code: [Select] $sql = "SELECT RegName, SireID, DamID FROM pedigrees WHERE ID="; $DoggieIDQ = $sql . $values["ID"]; $DoggieIDR = db_query($DoggieIDQ,$conn); $DoggieID = $_GET['ID']; $Depth = 1; $DepthLimit=8; function getParents($DoggieID, $Depth, $DepthLimit) { $Sire = getSire($DoggieID); $Dam = getDAM($DoggieID); $Depth++; if ($Depth == $DepthLimit) { return array($Sire, $Dam); } else { # Now get grandparents too $SireParents = getParents($Sire, $Depth, $DepthLimit); $DamParents = getParents($Dam, $Depth, $DepthLimit); return array($Sire, $Dam, $SireParents, $DamParents); } } I am trying to get the code at the bottom of the script to print just once during the loop but it either doesn't print at all or repeats with the loop im am using if (!$i++) to print once and i works the first time i use it. foreach($uploadFilename as $key => $myvar) { if (!$i++) print "<!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>Upload Complete....</title> </head> <body> <body onload=\"document.forms.formname.submit);\"> <form id=\"formname\" name=\"form1\" method=\"post\" action=\"reg5.php\">\n"; echo "<input type=\"hidden\" name=\"image$key\" value=\""; echo end(explode('/',$myvar)); echo "\">\n"; if (!$i++) print "</form> </body> </html>\n"; } This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=346728.0 Hello guys and gals, I am pretty green to PHP! I have an empty array that that I am trying to put images into. The thing is I have a certain file name in the folder, I want to exclude that file. This is what I have tried, any advice would be appreciated! Code: [Select] $thumbImg[] = array(); foreach (glob($DImg) as $PImg) { if (!is_file("thumbnail.jpg")) { $thumbImg[] = "<img src=\"pathtoimage\">"; } } Later on the page I am printing it out with this. It is still including the thumbnail.jpg image. Thank you in advance!! Code: [Select] for ($i=0; $i<count($thumbImg); $i++) print $thumbImg[$i]; can someone please give me some guidance on how to do this please I am wanting to create a status updating type application on my site and i have this idea in my head i want it to retrive and print the last 3 posts (max ids) made by the user if someone could give me some example code please and i can hopefully work from that. Thanks James hihi, so I have the following, except it goes all the way up to server 400. Is there a way to make it print after each echo? As it is right now it will not print the entire list until all 400 servers are done echo "Server 01: " . count($server01->listaccts()) . " / 130" . "<br />" ; echo "Server 02: " . count($server02->listaccts()) . " / 130" . "<br />" ; echo "Server 03: " . count($server03->listaccts()) . " / 130" . "<br />" ; echo "Server 04: " . count($server04->listaccts()) . " / 130" . "<br />" ; echo "Server 05: " . count($server05->listaccts()) . " / 130" . "<br />" ; echo "Server 06: " . count($server06->listaccts()) . " / 130" . "<br />" ; echo "Server 07: " . count($server07->listaccts()) . " / 130" . "<br />" ; echo "Server 08: " . count($server08->listaccts()) . " / 130" . "<br />" ; thanks! Im trying to print several image url's and names from an sql database into 3 columns, Aiming to get it to go 1,2,3 4,5,6 7,8,9 etc etc. but for some reason myne is going 1,3,5 2,4,6 etc. Dont bother mentioning that my loops do nothing, i realised that about 10 minutes ago, Any help would be appreciated. <?php include 'config.php'; mysql_connect($host, $user, $pass) or die(mysql_error()); mysql_select_db($database) or die(mysql_error()); $result = mysql_query("SELECT * FROM tracks"); echo '<div id="left_wrapper">'; for ($i=0;$i<mysql_num_rows($result);$i+=3) { $row = mysql_fetch_array($result); $id = $row['id'] + 1; echo "<img src='Thumbnails/" . $id .".gif'></img><br>"; echo $row['name']. "<br>"; } echo "</div>"; echo '<div id="middle_wrapper">'; for ($i=1;$i<mysql_num_rows($result);$i+=3) { $row = mysql_fetch_array($result); $id = $row['id'] + 1; echo "<img src='Thumbnails/" . $id .".gif'></img><br>"; echo $row['name']. "<br>"; } echo "</div>"; echo '<div id="right_wrapper">'; for ($i=2;$i<mysql_num_rows($result);$i+=3) { $row = mysql_fetch_array($result); $id = $row['id'] + 1; echo "<img src='Thumbnails/" . $id .".gif'></img><br>"; echo $row['name']. "<br>"; } echo "</div>"; ?> i know each of those loops does nothing, But you can see where they are meant to do, Each div is aligned to different positions, first loop is left, 2nd is center, 3rd is right. |
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