PHP - Echo Out Whole Content Of Html File + Echo Out Random Lines Of Html File
Hi
I try to echo out random lines of a html file and want after submit password to whole content of the same html file. I have two Problems.
1st Problem When I echo out the random lines of the html file I don't get just the text but the code of the html file as well. I don't want that. I just want the text. How to do that?
for($x = 1;$x<=40;$x++) { $lines = file("$filename.html"); echo $lines[rand(0, count($lines)-1)]."<br>"; }I tried instead of "file("$filename.html");" "readfile("$filename.html");" But then I get the random lines plus the whole content. Is there anything else I can use instead of file so that I get the random lines of text without the html code?P.S file_get_contents doesn't work either have tried that one. 2nd Problem: As you could see in my first problem I have a file called $filename.html. After I submit the value of a password I want the whole content. But it is like the program did forget what $filename.html is. How can I make the program remember what $filename.html is? Or with other words how to get the whole content of the html file? My code: if($_POST['submitPasswordIT']){ if ($_POST['passIT']== $password ){ $my_file = file_get_contents("$filename.html"); echo $my_file; } else{ echo "You entered wrong password"; } }If the password isn't correct I get: You entered wrong password. If the password is correct I get nothing. I probably need to create a path to the file "$filename.html", but I don't know exactly how to do that. Similar TutorialsI have a .txt file containing hundreds of lines of Names (e.g. Brad, Jennifer, etc.) What coding can I use to only echo the first 20? Hi
I echo out images from a folder. The problem is that I have a html file in the same folder but I don't want to echo that out. How can I write my code to get rid of the html in the output?
Here my code:
<?php $filetype = '*.html'; $dirname = substr('$filetype'); $i=0; if (isset($_POST['submit2'])) { //get the dir from POST $selected_dir = $_POST['myDirs']; //now get the files from within the selected dir and echo them to the screen foreach(glob($selected_dir . DIRECTORY_SEPARATOR . '*') as $dirname) { echo substr($dirname, 0, -4); echo '<img src="'.$dirname.'" />'; echo "<label><div class=\"radiobt\"><input type='radio' name='radio1' value='$i'/></div></label>"; } } ?>P.S. when I echo out : substr($dirname, 0, -4); I want to get ride of the html filename there too. How can I do so something like this? Hi
I want to echo out something when I post the value of a radio button and this value matches the same name of a html file. How can I formulate my code to achieve that?
What I have so far:
<?php if (isset($_POST['submitradio'])) { $selected_dir = $_POST['radio1']; echo substr($selected_dir, 0, -4); echo '<img src="'.$selected_dir.'" />'; } ?>So far I get for example the name of the image(order.gif) and the image order. gif. What I now want is to formulate my code so that I can say if I have an image called "order.gif" and a file called "order.html" that I can then echo out some thing. I don't want to actually name the image or the html file. I just want to say if I have two different file types that start with the same name then I can echo out what ever. How can I do that? <td><label for='images'> <b>File to upload:</b> </label></td> <td><input type='file' name = 'drama_image' '<?php echo $row['drama_image']; ?>'/></ </tr> <?php $target_path = "images/"; $target_path = $target_path . basename( $_FILES['images']['name']); if(move_uploaded_file($_FILES['images']['tmp_name'], $target_path)) { echo "The file ". basename( $_FILES['images']['name']). " has been uploaded"; } else{ echo $row['drama_image']; } ?> ['drama_image'] is the name of the file I wanna echo it out in the box of file upload so when I save , the default picture will still be there instead of being overwritten as the box does not have any value in it. Hi, How do I open all HTML file contained in a folder, then search for the title of the page, and save the title in the database? Also, is it possible to do this using functions? Cheers. How would I open a text file from another server and echo it. http://mysite.com/news/feed.txt for($i=0; $i!=count($account_data); $i++){ echo "\n::: ".strtoupper($account_data[$i][0])." :::\n"; if($account_data[$i][1] != "" && stristr($account_data[$i][0], "@") && stristr($account_data[$i][0], ".")){ list($place[0],$place[1],$place[2],$place[3]) = authorize($account_data[$i], $game_link); if($place[2] > 1 && strlen($place[3]) == 40){ $filename = 'auth-keys.txt'; $fp = file($filename,'a+'); $data = $place[2]." ".$place[3]."\r\n"; file_put_contents($filename, $data, FILE_APPEND); echo $data; } else echo "Account Skipped: wrong password/email combo, or a login error...\n"; $echo = file_put_contents($filename, FILE_APPEND); } else echo "Account Skipped: invalid email or blank password...\n"; } echo "\n::: PROGRAM FINISHED, YOU CAN CLOSE THIS AT ANY TIME :::"; sleep(10000); You see where I tried to put $echo file_put_contents etc... On success the acct writes a security id and a security auth key to a file, ever acct puts the tokens on a new line every acct... now when an acct echos account skipped, how can I make it print the echo on the file new line instead of nothing being writen and it just skipping the account Ok, here's my code: How do i make it so that it outputs a maximum of only 5 results from the query where the img file exists. When I add LIMIT to the sql query it doesnt work so I guess its something else, but I have no clue. Can anyone help? $query = mysql_query("SELECT * FROM table WHERE date >= '$now' ORDER BY max(date) desc"); while ($row = mysql_fetch_assoc($query)) { $cid = $row['cat_id']; $title = $row['name']; $seoname = $row['seourl']; $img = 'images/'.$seoname.'.jpg'; if (!file_exists($img)) { $img = ''; } else { $img = 'images/'.$seoname.'.jpg'; } if (!empty($img)) { echo '<a href="images/'.$seoname.'/"><img src="'.$img .'" style="width:528px;" alt="" /></a>' . "\n"; } } Thanks. I have an edit page but whenever I click edit the pathname of the original file is not in the file box so everytime I hit the save button , it will be considered as no file so my previous image or file will be overwritten by "nothing". Hi, Have a text file that is being read. I want to print the contents however I want each line of the text file to appear on a new line in the browser. So far $str = file_get_contents('records.txt'); $lines = explode("\n", $str); echo $lines[0]; Help appreciated Hi, Im trying to pull an email from an url, and then echo the email as value into a email form-- under the "from" heading. The echo is not working i get blanks. here is the code: <?php $thisurl='$_SERVER['REQUEST_URI']'; $thepart = explode('/', $thisurl); $themail='$thepart[2]'; ?> <form method="post" action="sendit.php"> To: <br><input type="text" name="to" size="44" value="<?php for($i=1; $i<=$lines; $i=$i+1) { $names[$i] = str_replace("<br>", "", $names[$i]); $names[$i] = str_replace("\n", "", $names[$i]); echo $names[$i].';'; } ?>"><br> From: <br><input type="text" name="email" size="44" value="<?php echo '$themail'; ?>" readonly><br> Subject: <br><input type="text" name="subject" size="44" /><br><br> I have written this code but im not used to echoing html in php so I have errors. echo " <option value= '{$county['county']}'" if ($User['county'])==$county['county']) {echo "selected";} echo">{$county['county']}</option>"; } the server error is; syntax error, unexpected T_IF, expecting ',' or ';' Can somebody please point out how I have written it wrong? Hi im trying to echo out some html within php based on an if statement. Basically if you the is part of a club within a list item list the clubs, if they are not echo none within the list. Here is the code and below will be the error from the server. <?php // Query for finding out if the signed in user is part of any clubs and then display them here. $qMyClub = "SELECT clubID, name FROM clubs WHERE memberID = ".$User['memberID'].""; $rMyClub = mysql_query($qMyClub); $NumClubs = mysql_num_rows($rMyClub); if ($NumClubs>=1) { while($Clubs = mysql_fetch_assoc($rMyClub)){ echo"<li>My Clubs <ul> <li><a href=\"members/myclub.php?club='$Club['clubID']'\">$Club['name']</a></li> </ul> </li>"; } else ($NumClubs==0) { echo" <li>My Clubs <ul> <li>None</li> </ul> </li>"; } } ?> and the error; [error] [client ] PHP Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /home/sites/index.php on line 78, referer: its referencing to the following line; <li><a href=\"members/myclub.php?club='$Club['clubID']'\">$Club['name']</a></li> I dont understand php enough to fix this can anyone help? Hi everyone, First time post for me I am quite new to PHP so excuse the beginner question but I can't find and answer for it. I have a backend application that allows an admin to update the frontend homepage with a WYSIWYG html editor (Xinha). I store the html into mysql and retrieve the data for display on the front end homepage. The problem is the html is not rendering and displaying with the tags as text. When saving the data to the db I use htmlentities($data) and when retrieving I used html_entity_decode($data) before passing it to the template to be rendered. I am using a simple <?php echo $data ?> to display the data. On the same page I have <?php $testingHTML = "<p>This is some text with a <b>bold</b> word in it</p>" ?> followed down the page by an <?php echo $testingHTML ?> and that renders perfectly. So I must be doing something wrong? Thank you in advance and I hope that makes sense. Steve I'm trying to learn php, and I've picked up a php/mysql member system type script, and I'm trying to put the login/sign-up forms into my own pages. The problem I'm having though is getting PHP and HTML to work in the same echo""; A week ago i thought i saw something about another way to display a form in the php manual but cant find it anymore. so im wondering is there another way to echo a form with out echoing every html line. Hi everyone, i have a user posting system and i want to echo data so that any HTML tags of any kind will be regarded as plaintext, so that it will have no effect on the webpage, thanks Jack Hey, I posted a thread before about a news cms I've been developing. And right now I'm trying to flesh out the css of the articles I post on the home page. Now the problem I having is everytime I place a <span class=""></span> or <p class=""></p> or even <div class=""></div>, my web page does NOT load, and instead gives me an error. Code: [Select] Parse error: syntax error, unexpected T_STRING in /home/content/d/j/s/djsmiley/html/index.php on line 405 Now, I know it's possible to add css to a PHP echo statement, but I'm clearly doing something wrong. Here's the code for the php echo statement with the css included: Code: [Select] <? include("dbconnect.php"); //include the file to connect to the database $getnews = mysql_query("SELECT * FROM mynews ORDER BY id DESC"); //query the database for all of the news while($r=mysql_fetch_array($getnews)){ //while there are rows in the table extract($r); //remove the $r so its just $variable echo("<br><span class="NewsID">$type</span> <span class=h2>$title</span><br><br> <em>posted by <strong>$user</strong> | added on $time</em><br><br> $message<br><br> <label class="fltlft2"><img src="../../Websites/DJSmiley.Net/images/Icons/Arrows/Right.png" width="20" height="20"/></label><a href=$url>Read more - $url</a> <div class="newsLikeShareRate"> <table width="100%" border="0"> <tr> <td width="3%" height="21"><script src="http://connect.facebook.net/en_US/all.js#xfbml=1"></script> <fb:like href="$url" show_faces="true" width="450" font="arial"></fb:like> </td> <td width="65%"><a name="fb_share" id="fb_share4" type="icon_link" share_url="$url">Share</a> <script src="http://static.ak.fbcdn.net/connect.php/js/FB.Share" type="text/javascript"></script></td> <td width="32%">Rate this article: </td> </tr> </table> </div>"); } ?> If you're wondering why there's a facebook script in there, I made it to where everytime a new post is added, a facebook like and share button are added, and are assigned the url I specify when the article is posted. Well, I think that's it for the code. Can anyone tell me what I'm doing wrong? Hi all. I have this code. <?php $ext = ' <form action="/members/clubs/view-activity.php" method="post" name="venuefrm"> <button type="submit" id="close" class="link"><span>View Activity</span></button> <input name="activity" type="hidden" value="<?php echo $activites['activityID']; ?>" /> </form>'; ?> The code falls down at the value="<?php echo $activites['activityID']; ?>" part. Can someone please show me how I break into php to write this part properly as I dont understand an it is obviously wrong. Thanks Danny |