Search Results

Search found 8646 results on 346 pages for 'echo flow'.

Page 52/346 | < Previous Page | 48 49 50 51 52 53 54 55 56 57 58 59  | Next Page >

  • How to allow multiple filters to be selected - pagination?

    - by NewSOuser
    Hey, I am still trying to allow multiple filters to be selected for my pagination script but not sure how to do it being very new to php and programing in general. So in my pagination, when a user clicks the 'marketing' button(link) it queries the database just for the category that = marketing. The same goes for the other 2 filter buttons as seen in the script below. (automotive, sports). The problem is, I want to be able to select multiple filters like only marketing and auomotive or automotive and sports, for example if I click the marketing filter and then the automotive, it would display the categories that equal marketing, and automotive. I have no idea how to accomplish this, so I have come to the experts to help me out. This is the script I am working on: <h3>Filter results by:</h3> <a href='pagi_test.php?category=marketing'>marketing</a> <a href='pagi_test.php?category=automotive'>automotive</a> <a href='pagi_test.php?category=sports'>sports</a> <br /> <h3>Results:</h3> <?php //connecting to the database $error = "Could not connect to the database"; mysql_connect('localhost','root','root') or die($error); mysql_select_db('ajax_demo') or die($error); //max displayed per page $per_page = 3; //get start variable $start = $_GET['start']; $category = mysql_real_escape_string($_GET['category']); //count records $record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='$category'")); //count max pages $max_pages = $record_count / $per_page; //may come out as decimal if (!$start) $start = 0; //display data $get = mysql_query("SELECT * FROM explore WHERE category='$category' LIMIT $start, $per_page"); ?> <table width="800px"> <?php while ($row = mysql_fetch_assoc($get)) { // get data $id = $row['id']; $site_name = $row['site_name']; $site_description = $row['site_description']; ?> <tr> <td><?php echo $id; ?></td> <td><?php echo $site_name; ?></td> <td><?php echo $site_description; ?></td> </tr> <?php } //setup prev and next variables $prev = $start - $per_page; $next = $start + $per_page; //show prev button if (!($start<=0)) echo "<a href='pagi_test.php?category=$category&start=$prev'>Prev</a> "; //show page numbers //set variable for first page $i=1; for ($x=0;$x<$record_count;$x=$x+$per_page) { if ($start!=$x) echo " <a href='pagi_test.php?category=$category&start=$x'>$i</a> "; else echo " <a href='pagi_test.php?category=$category&start=$x'><b>$i</b></a> "; $i++; } //show next button if (!($start>=$record_count-$per_page)) echo " <a href='pagi_test.php?category=$category&start=$next'>Next</a>"; ?> Any help on this would be great. Thank you. -- EDIT -- If anyone has a better method of doing a pagination system with multiple filters than the one above, please let me know.

    Read the article

  • submit the value of check box in database

    - by ritu
    <?php $dates = $day1 ; echo "<p>"; for($i=0;$i<count($dates);$i++) { if($i>0 && ($i%3==0)) { echo "</p><p>"; } echo "<input type='checkbox' name='dates[".$dates."]' /><label>".$dates[$i]."</label>"; } echo "</p>"; ?> whan i use this output will be desire Ex 1 2 3 4 5 6 but i want send his data in the databse but only a array will be send into the database $_post[dates so can any on telll me how can i send the data in the database

    Read the article

  • Windows batch file timing bug

    - by elbillaf
    I've used %time% for timing previously - at least I think I have. I have this weird IF NOT "%1" == "" ( echo Testing: %1 echo Start: %time% sqlcmd -S MYSERVER -i test_import_%1.sql -o "test_%1%.log" sleep 3 echo End: %time% ) I run this, and it prints: Testing: pm Start: 13:29:45.30 End: 13:29:45.30 In this case, my sql code is failing (different reason), but I figure the sleep 3 should make the time increment by 3 at least. Any ideas? tx, tff

    Read the article

  • BASH SHELL IF ELSE run command in background

    - by bikerben
    I have used the & command before to make a script run another script in the background like so: #!/bin/bash echo "Hello World" script1.sh & script2.sh & echo "Please wait..." But lets say I have another script with an IF ELSE statment and I would like to set an ELIF statement mid flow as a background task witht the & and then carry on with processing the rest of my script knowing that while rest of the ELIF will carry running in the back ground: #!/bin/bash if cond1; then stuff sleep 10 & stuff stuff elif cond2; then something else else echo "foo" fi stuff echo "Hello World" I really hope this makes sense any help would be greatly appreciated.

    Read the article

  • Dynamically add data stored in php to nested json

    - by HoGo
    I am trying to dynamicaly generate data in json for jQuery gantt chart. I know PHP but am totally green with JavaScript. I have read dozen of solutions on how dynamicaly add data to json, and tried few dozens of combinations and nothing. Here is the json format: var data = [{ name: "Sprint 0", desc: "Analysis", values: [{ from: "/Date(1320192000000)/", to: "/Date(1322401600000)/", label: "Requirement Gathering", customClass: "ganttRed" }] },{ name: " ", desc: "Scoping", values: [{ from: "/Date(1322611200000)/", to: "/Date(1323302400000)/", label: "Scoping", customClass: "ganttRed" }] }, <!-- Somoe more data--> }]; now I have all data in php db result. Here it goes: $rows=$db->fetchAllRows($result); $rowsNum=count($rows); And this is how I wanted to create json out of it: var data=''; <?php foreach ($rows as $row){ ?> data['name']="<?php echo $row['name'];?>"; data['desc']="<?php echo $row['desc'];?>"; data['values'] = {"from" : "/Date(<?php echo $row['from'];?>)/", "to" : "/Date(<?php echo $row['to'];?>)/", "label" : "<?php echo $row['label'];?>", "customClass" : "ganttOrange"}; } However this does not work. I have tried without loop and replacing php variables with plain text just to check, but it did not work either. Displays chart without added items. If I add new item by adding it to the list of values, it works. So there is no problem with the Gantt itself or paths. Based on all above I assume the problem is with adding plain data to json. Can anyone please help me to fix it?

    Read the article

  • Using $this when not in object context

    - by Ken
    I'm creating a function to show blog's. So I made a show blog function but it keeps giving "Using $this when not in object context" error Class Blog{ public function getLatestBlogsBig($cat = null){ $sqlString = "SELECT blog_id FROM jab_blog"; if($cat != null) $sqlString .= " WHERE blog_cat = " . $cat; $sqlString .= " ORDER BY blog_id DESC LIMIT 5"; $blog = mysql_query($sqlString); while($id = mysql_result($blog,"blog_id")){ $this->showBlog($id); //Error is on this line } } function showBlog($id,$small = false){ $sqlString = "SELECT blog_id FROM jab_blog WHERE blog_id=" . $id . ";"; $blog = mysql_query($sqlString); if($small = true){ echo "<ul>"; while($blogItem = mysql_fetch_array($blog)){ echo '<a href="' . $_SESSION['JAB_LINK'] . "blog/" . $blogItem['blog_id'] . "/" . SimpleUrl::toAscii($blogItem['blog_title']) .'">' . $blogItem['blog_title'] . '</a></li>'; } echo "</ul>"; }else{ while($blogItem = mysql_fetch_array($blog)){ ?> <div class="post"> <h2 class="title"><a href="<?php echo $_SESSION['JAB_LINK'] . "blog/" . $blogItem['blog_id'] . "/" . SimpleUrl::toAscii($blogItem['blog_title']);?>"><?php echo $blogItem['blog_title'];?></a></h2> <p class="meta"><span class="date">The date implement</span><span class="posted">Posted by <a href="#">Someone</a></span></p> <div style="clear: both;">&nbsp;</div> <div class="entry"> <?php echo $blogItem['blog_content'];?> </div> </div> <?php } } } }

    Read the article

  • Laravel4: Call to a member function on a non-object

    - by s0hno
    The following code will throw an error Call to a member function `links()` on a non-object routes.php: Route::get('videos', function(){ $data = DB::table('video_data_r')->paginate(5); return View::make('video',$data); }); Corresponding video view: <?php foreach($data as $item): ?> <div class="video_entry"> <a href="<?php echo $item -> url; ?>" target="_blank"><img src="<?php echo $item -> thumb; ?>" /></a> <a href="<?php echo $item -> url; ?>" target="_blank"><?php echo $item -> title; ?>"</a> </div> <?php endforeach; ?> <?php echo $data->links();?> Could you give me a good hint on what looks like a trivial error?

    Read the article

  • Parsing CSV File to MySQL DB in PHP

    - by Austin
    I have a some 350-lined CSV File with all sorts of vendors that fall into Clothes, Tools, Entertainment, etc.. categories. Using the following code I have been able to print out my CSV File. <?php $fp = fopen('promo_catalog_expanded.csv', 'r'); echo '<tr><td>'; echo implode('</td><td>', fgetcsv($fp, 4096, ',')); echo '</td></tr>'; while(!feof($fp)) { list($cat, $var, $name, $var2, $web, $var3, $phone,$var4, $kw,$var5, $desc) = fgetcsv($fp, 4096); echo '<tr><td>'; echo $cat. '</td><td>' . $name . '</td><td><a href="http://www.' . $web .'" target="_blank">' .$web.'</a></td><td>'.$phone.'</td><td>'.$kw.'</td><td>'.$desc.'</td>' ; echo '</td></tr>'; } fclose($file_handle); show_source(__FILE__); ?> First thing you will probably notice is the extraneous vars within the list(). this is because of how the excel spreadsheet/csv file: Category,,Company Name,,Website,,Phone,,Keywords,,Description ,,,,,,,,,, Clothes,,4imprint,,4imprint.com,,877-466-7746,,"polos, jackets, coats, workwear, sweatshirts, hoodies, long sleeve, pullovers, t-shirts, tees, tshirts,",,An embroidery and apparel company based in Wisconsin. ,,Apollo Embroidery,,apolloemb.com,,1-800-982-2146,,"hats, caps, headwear, bags, totes, backpacks, blankets, embroidery",,An embroidery sales company based in California. One thing to note is that the last line starts with two commas as it is also listed within "Clothes" category. My concern is that I am going about the CSV output wrong. Should I be using a foreach loop instead of this list way? Should I first get rid of any unnecessary blank columns? Please advise any flaws you may find, improvements I can use so I can be ready to import this data to a MySQL DB.

    Read the article

  • Using fopen and str_replace to auto fill a select option

    - by Anders Kitson
    Hi Everyone, I have this file 'gardens.php', which pulls data from a table called 'generalinfo' and I use fopen to send that data to a file called 'index.html'. Here is the issue, only one option is filled. I have a demo running here Garden Demo <-- this is a new updated page and location, there are more errors than I have locally If anyone could help me fix them Username:stack1 Password:stack1 Is there a better way to achieve what I want to? Thanks! Always. GARDENS.PHP <?php include("connect.php"); $results = mysql_query("SELECT * FROM generalinfo"); while($row = mysql_fetch_array($results)){ $country = $row['country']; $province = $row['province']; $city = $row['city']; $address = $row['address']; //echo $country; //echo $province; //echo $city; //echo $address; } $fd = fopen("index.html","r") or die ("Can not fopen the file"); while ($buf =fgets($fd, 1024)){ $template .= $buf; } $template = str_replace("<%country%>",$country,$template); echo $template; ?> INDEX.PHP SNIPPET <form name="filter" method="get" action="filter.php"> <select class="country" name="country"> <option><%country%></option> </select> </form>

    Read the article

  • Ajax Request not working. onSuccess and onFailure not triggering

    - by Kye
    Hi all, trying to make a page which will recursively call a function until a limit has been reached and then to stop. It uses an ajax query to call an external script (which just echo's "done" for now) howver with neither onSuccess or onFailure triggering i'm finding it hard to find the problem. Here is the javascript for it. In the header for the webpage there is a script to an ajax.js document which contains the request data. I know the ajax.js works as I've used it on another website var Rooms = "1"; var Items = "0"; var ccode = "9999/1"; var x = 0; function echo(string,start){ var ajaxDisplay = document.getElementById('ajaxDiv'); if(start) {ajaxDisplay.innerHTML = string;} else {ajaxDisplay.innerHTML = ajaxDisplay.innerHTML + string;} } function locations() { echo("Uploading location "+x+" of " + Rooms,true); Ajax.Request("Perform/location.php", { method:'get', parameters: {ccode: ccode, x: x}, onSuccess: function(reply) {alert("worked"); if(x<Rooms) { x++; locations(); } else { x=0; echo("Done",true); } }, onFailure: function() {alert("not worked"); echo("not done"); } }); alert("boo"); } Any help or advice will be most appreciated.

    Read the article

  • Passing variables from PHP to Javascript back to PHP using Ajax.

    - by ObjectiveJ
    I hope this makes sesne, please bare with me. So I have a PHP page that contains variables, I have some radial boxes, and on click of them, it calculates a price for the item you have clicked on. I do this by activating a js function that I have passed some variables to. Like so. PHP: <?php $result = mssql_query("SELECT * FROM Segments ORDER BY 'Squares'"); if (!$result) { echo 'query failed'; exit; } while ($row = mssql_fetch_array($result)) { ?> <span><?php echo $row["Squares"]; ?></span><input name="squares" type="radio" onclick="ajaxCases('<?php echo $row["Squares"]; ?>', '<?php echo $row["StartCaseID"]; ?>', '<?php echo $row["StartMatrixPrice"]; ?>')" value="<?php echo $row["Squares"]; ?>"<?php if ($row["Squares"] == "1") { ?> checked="checked" <?php }else{ ?> checked="" <?php } ?>/> <?php } ?> As you can see onclick it goes to a function called ajaxcases, this function looks like this. function ajaxCases(squares,start,price){ $('#step1').html('<p style="margin:100px 0px 0px 100px"><img src="images/ajax-loader-bigindic.gif" width="32" height="32" alt="" /></p>'); $('#step1').load("ajax-styles.php?squares="+squares); prevId1 = ""; document.varsForm.caseid.value=start; $('#step1price').html('<span style="margin:0px 0px 0px 30px"><img src="images/ajax-loader-price.gif" width="24" height="24" alt="" /></span>'); $('#step1price').load("ajax-step1-price.php?Squares="+Squares); return true; } This then goes to a php page called ajax-step1-price.php and I try to recall the variable Squares. However it doesn't work, I thought it was a GET however that returns undefined. In Summary: I would like to know how to pass a variable from PHP to JS then back to PHP, or if someone could just tell me where I am going wrong that would be greatly appreciated.

    Read the article

  • How to do an if statement on a function in PHP?

    - by Bruce
    I just realized that you can't just use an if statement on a function, for example this doesn't work: function sayHello() { echo "Hello World"; } if(sayHello()) echo "Function Worked"; else echo "Function Failed"; I also saw that a function can't be put as the value of a variable. So how can I do an if statement to check if a function has executed properly and display it to the browser?

    Read the article

  • How do I write this SQL statement to get the ad and posting? (PHP/MySQL)

    - by ggfan
    I am a little confused on the logic of how to write this SQL statement. When a user clicks on a tag, say HTML, it would display all the posts with HTML as its tag. (a post can have multiple tags) I have three tables: Posting--posting_id, title, detail, etc tags--tagID, tagname postingtag--posting_id, tagID I want to display all the title of the post and the date added. global $dbc; $tagID=$_GET['tagID']; //the GET is set by URL //part I need help with. I need another WHERE statment to get to the posting table $query = "SELECT p.title,p.date_added, t.tagname FROM posting as p, postingtag as pt, tags as t WHERE t.tagID=$tagID"; $data = mysqli_query($dbc, $query); echo '<table>'; echo '<tr><td><b>Title</b></td><td><b>Date Posted</b></td></tr>'; while ($row = mysqli_fetch_array($data)) { echo '<tr><td>'.$row['title'].'</td>'; echo '<td>'.$row['date_added'].'</td></tr>'; } echo '</table>'; }

    Read the article

  • How To Call Javascript In Ajax Response? IE: Close a form div upon success...

    - by B.Gordon
    I have a form that when you submit it, it sends the data for validation to another php script via ajax. Validation errors are echo'd back in a div in my form. A success message also is returned if validation passes. The problem is that the form is still displayed after submit and successful validation. I want to hid the div after success. So, I wrote this simple CSS method which works fine when called from the page the form is displayed on. The problem is that I cannot seem to call the hide script via returned code. I can return html like echo "<p>Thanks, your form passed validation and is being sent</p>"; So I assumed I could simply echo another line after that echo "window.onload=displayDiv()"; inside script tags (which I cannot get to display here)... and that it would hide the form div. It does not work. I am assuming that the problem is that the javascript is being returned incorrectly and not being interpreted by the browser... How can I invoke my 'hide' script on the page via returned data from my validation script? I can echo back text but the script call is ineffective. Thanks! This is the script on the page with the form... I can call it to show/hide with something like onclick="displayDiv()" while on the form but I don't want the user to invoke this... it has be called as the result of a successful validation when I write the results back to the div... function displayDiv() { var divstyle = new String(); divstyle = document.getElementById("myForm").style.display; if(divstyle.toLowerCase()=="block" || divstyle == "") { document.getElementById("myForm").style.display = "none"; } else { document.getElementById("myForm").style.display = "block"; } } PS: I am using the mootools.js library for the form validation if this matters for the syntax..

    Read the article

  • sudo changes PATH - why?

    - by Michiel de Mare
    This is the PATH variable without sudo: $ echo 'echo $PATH' | sh /opt/local/ruby/bin:/usr/bin:/bin This is the PATH variable with sudo: $echo 'echo $PATH' | sudo sh /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/X11R6/bin As far as I can tell, sudo is supposed to leave PATH untouched. What's going on? How do I change this? (This is on Ubuntu 8.04). UPDATE: as far as I can see, none of the scripts started as root change PATH in any way. From man sudo: To prevent command spoofing, sudo checks ``.'' and ``'' (both denoting current directory) last when searching for a command in the user's PATH (if one or both are in the PATH). Note, however, that the actual PATH environment variable is not modified and is passed unchanged to the program that sudo executes.

    Read the article

  • Cleaner way of using modulus for columns

    - by WmasterJ
    I currently have a list (<ul>) of people that I have divided up into two columns. But after finishing the code for it I keept wondering if there is a more effective or clean way to do the same thing. echo "<table class='area_list'><tr>"; // Loop users within areas, divided up in 2 columns $count = count($areaArray); for($i=0 ; $i<$count ; $i++) { $uid = $areaArray[$i]; // get the modulus value + ceil for uneven numbers $rowCalc = ($i+1) % ceil($count/2); if ($rowCalc == 1) echo "<td><ul>"; // OUTPUT the actual list item echo "<li>{$users[$uid]->profile_lastname}</li>"; if ($rowCalc == 0 && $i!=0) echo "</ul></td>"; } echo "</tr></table>"; Any ideas of how to make this cleaner or do it in another way?

    Read the article

  • Use of date function in PHP to output a user-friendly date

    - by Jamie
    I have a MySQL database column named DateAdded. I'd like to echo this as a readable date/time. Here is a simplified version of the code I currently have: $result = mysql_query(" SELECT ListItem, DateAdded FROM lists WHERE UserID = '" . $currentid . "' "); while($row = mysql_fetch_array($result)) { // Make the date look nicer $dateadded = date('d-m-Y',$row['DateAdded']); echo $row['ListItem'] . ","; echo $dateadded; echo "<br />"; } Is the use of the date function the best way to output a user-friendly date? Thanks for taking a look,

    Read the article

  • ZendX Jquery Decorator

    - by iJD
    How use partial decorator in Jquery Element I use this code for Form Element: $title = new Zend_Form_Element_Text('title'); $title->setRequired(true) ->setAttrib('class', 'inputbox') ->setLabel('Title'); $title->viewScript = 'RegElement.phtml'; $title->setDecorators( array( array('ViewScript', array('class' => 'RegElement')) ) ); But when i use Jquery Element i dont know how implement it: $datePicker = new ZendX_JQuery_Form_Element_DatePicker( "datePicker1", array("label" => "Date:") ); $datePicker->viewScript = 'RegElement.phtml'; $datePicker->setDecorators( array( array('ViewScript', array('class' => 'RegElement')) ) ); //views/scripts/RegElement.phtml <li class="row <?php echo $this->class ?>"> <div class="cont-error"> <?php echo $this->formErrors($this->element->getMessages()); ?> </div> <div class="rowfields"> <?php echo $this->formLabel($this->element->getName(), $this->element->getLabel()) ?> <?php echo $this->{$this->element->helper}( $this->element->getName(), $this->element->getValue(), $this->element->getAttribs() ) ?> </div> <div class="hint"><?php echo $this->element->getDescription() ?></div> </li> And display this error: Warning: Exception caught by form: Cannot render jQuery form element without at least one decorator implementing the 'ZendX_JQuery_Form_Decorator_UiWidgetElementMarker' interface. I need display datePicker with same format. but idk how implement this interface. thx for your help.

    Read the article

  • how to add multiple filters to pagination?

    - by ClarkSKent
    Hey, I am trying to allow multiple filters to be selected for my pagination script. So in my pagination, when a user clicks the 'marketing' button(link) it queries the database just for the category that = marketing. The same goes for the other 2 filter buttons as seen in the script below. (automotive, sports). The problem is, I want to be able to select multiple filters like only marketing and auomotive or automotive and sports, for example if I click the marketing filter and then the automotive, it would display the categories that equal marketing, and automotive. I have no idea how to accomplish this, so I have come to the experts to help me out. This is the script I am working on (I am very new to PHP and programming in general): <h3>Filter results by:</h3> <a href='pagi_test.php?category=marketing'>marketing</a> <a href='pagi_test.php?category=automotive'>automotive</a> <a href='pagi_test.php?category=sports'>sports</a> <br /> <h3>Results:</h3> <?php //connecting to the database $error = "Could not connect to the database"; mysql_connect('localhost','root','root') or die($error); mysql_select_db('ajax_demo') or die($error); //max displayed per page $per_page = 3; //get start variable $start = $_GET['start']; $category = mysql_real_escape_string($_GET['category']); //count records $record_count = mysql_num_rows(mysql_query("SELECT * FROM explore WHERE category='$category'")); //count max pages $max_pages = $record_count / $per_page; //may come out as decimal if (!$start) $start = 0; //display data $get = mysql_query("SELECT * FROM explore WHERE category='$category' LIMIT $start, $per_page"); ?> <table width="800px"> <?php while ($row = mysql_fetch_assoc($get)) { // get data $id = $row['id']; $site_name = $row['site_name']; $site_description = $row['site_description']; ?> <tr> <td><?php echo $id; ?></td> <td><?php echo $site_name; ?></td> <td><?php echo $site_description; ?></td> </tr> <?php } //setup prev and next variables $prev = $start - $per_page; $next = $start + $per_page; //show prev button if (!($start<=0)) echo "<a href='pagi_test.php?category=$category&start=$prev'>Prev</a> "; //show page numbers //set variable for first page $i=1; for ($x=0;$x<$record_count;$x=$x+$per_page) { if ($start!=$x) echo " <a href='pagi_test.php?category=$category&start=$x'>$i</a> "; else echo " <a href='pagi_test.php?category=$category&start=$x'><b>$i</b></a> "; $i++; } //show next button if (!($start>=$record_count-$per_page)) echo " <a href='pagi_test.php?category=$category&start=$next'>Next</a>"; ?>

    Read the article

  • "Undefined index" when quoting array key

    - by Satish
    I have a form in index.php <?php echo '<form action="update_act.php" method="POST">'; echo '<input type="submit" name="'.$row['act_name'].'" value="edit"> echo </form> ?> Here $row['act_name'] is a value fetched from database. My update_act.php file is <?php echo "Old Activity Name : ".$_POST['$row[\'act_name\']']; ?> But I am getting an error Undefined index: $row['act_name'] in C:\wamp\www\ps\activity\update_act.php. I want to have different names for different submits but I am not able to get its value in another page. Is there any way for it?

    Read the article

  • Using ImageMagick to create an image from a PDF...efficiently

    - by bigsweater
    I'm using ImageMagick to create a tiny JPG thumbnail image of an already-uploaded PDF. The code works fine. It's a WordPress widget, though this isn't necessarily WordPress specific. I'm unfamiliar with ImageMagick, so I was hoping somebody could tell me if this looks terrible or isn't following some best practices of some sort, or if I'm risking crashing the server. My questions, specifically, are: Is that image cached, or does the server have to re-generate the image every time somebody views the page? If it isn't cached, what's the best way to make sure the server doesn't have to regenerate the thumbnail? I tried to create a separate folder (/thumbs) for ImageMagick to put all the images in, instead of cluttering up the WP upload folders with images of PDFs. It kept throwing a permission error, despite 777 permissions on the folder in my testing environment. Why? Do the source/destination directories have to be the same? Am I doing anything incorrectly/inefficiently here that needs to be improved? The whole widget is on Pastebin: http://pastebin.com/WnSTEDm7 Relevant code: <?php if ( $url ) { $pdf = $url; $info = pathinfo($pdf); $filename = basename($pdf,'.'.$info['extension']); $uploads = wp_upload_dir(); $file_path = str_replace( $uploads['baseurl'], $uploads['basedir'], $url ); $dest_path = str_replace( '.pdf', '.jpg', $file_path ); $dest_url = str_replace( '.pdf', '.jpg', $pdf ); exec("convert \"{$file_path}[0]\" -colorspace RGB -geometry 60 $dest_path"); ?> <div class="entry"> <div class="widgetImg"> <p><a href="<?php echo $url; ?>" title="<?php echo $filename; ?>"><?php echo "<img src='".$dest_url."' alt='".$filename."' class='blueBorder' />"; ?></a></p> </div> <div class="widgetText"> <?php echo wpautop( $desc ); ?> <p><a class="downloadLink" href="<?php echo $url; ?>" title="<?php echo $filename; ?>">Download</a></p> </div> </div> <?php } ?> As you can see, the widget grabs whatever PDF is attached to the current page being viewed, creates an image of the first page of the PDF, stores it, then links to it in HTML. Thanks for any and all help!

    Read the article

  • Bizarre error in php

    - by Satish
    I have a form in index.php <?php echo '<form action="update_act.php" method="POST">'; echo '<input type="submit" name="'.$row['act_name'].'" value="edit"> echo </form> ?> Here $row['act_name'] is a value fetched from database. My update_act.php file is <?php echo "Old Activity Name : ".$_POST['$row[\'act_name\']']; ?> But I am getting an error Undefined index: $row['act_name'] in C:\wamp\www\ps\activity\update_act.php. I want to have different names for different submits but I am not able to get its value in another page. Is there any way for it?

    Read the article

  • PHP active page code - I can't figure out parse error

    - by dmschenk
    I'm trying to build an active page menu with PHP and MySQL and am having a difficult time fixing the error. In the while statement I have an if statement that is giving me fits. Basically I think I'm saying that "thispage" is equal to the "title" based on pageID and as the menu is looped through if "thispage" is equal to "title" then echo id="active". Thanks <?php mysql_select_db($database_db_connection, $db_connection); $query_rsDaTa = "SELECT * FROM pages WHERE pagesID = 4"; $rsDaTa = mysql_query($query_rsDaTa, $db_connection) or die(mysql_error()); $row_rsDaTa = mysql_fetch_assoc($rsDaTa); $totalRows_rsDaTa = mysql_num_rows($rsDaTa); $query_rsMenu = "SELECT * FROM menu WHERE online = 1 ORDER BY menuPos ASC"; $rsMenu = mysql_query($query_rsMenu, $db_connection) or die(mysql_error()); $thisPage = ($row_rsDaTa['title']); ?> <link href="../css/MainStyle.css" rel="stylesheet" type="text/css" /> <h2><?php echo $thisPage; ?></h2> <div id="footcontainer"> <ul id="footlist"> <?php while($row_rsMenu = mysql_fetch_assoc($rsMenu)) { echo (" <li" . <?php if ($thisPage==$row_rsDaTa['title']) echo id="active"; ?> . "<a href=\"../" . $row_rsMenu['menuURL'] . "\">" . $row_rsMenu['menuName'] . "</a></li>\n"); } echo "</ul>\n"; ?> </div> <?php mysql_free_result($rsMenu); mysql_free_result($rsDaTa); ?>

    Read the article

  • Symfony sfPHPExcelPlugin in Linux

    - by Tere
    Hi! I'm using the PHPExcel plugin for Symfony 1.4 (sfPHPExcelPlugin) on Ubuntu (using PHP 5.10), with this code for saving the file that I am writing: // Save Excel 2007 file echo date('H:i:s') . " Write to Excel2007 format\n"; $objWriter = PHPExcel_IOFactory::createWriter($objPHPExcel, 'Excel2007'); $objWriter->save(str_replace('.php', '.xlsx', __FILE__)); // Echo done echo date('H:i:s') . " Done writing file.\r\n"; I am sure that the execution is reaching this part of the code, because the echo messages are shown but I am not downloading the Excel file! Could it be because I am trying it in Linux, not in Windows? Thank you!

    Read the article

< Previous Page | 48 49 50 51 52 53 54 55 56 57 58 59  | Next Page >