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PHP - Stristr Problem
Is there a better way to specify the "if (strsistr...)" line?
Suppose I want to check 4 or 5 file ext's instead of just 2? <?php if ($handle = opendir('pics')) { echo 'Directory handle: ' . $handle . '<br />'; echo 'Files: <br />'; while (false !== ($file = readdir($handle))) { if ( stristr($file,("jpg")) || stristr($file,("png")) ) { echo $file . '<br />'; } } closedir($handle); } ?> thanks, Keith Similar TutorialsI have an array with the following elements define("LIST_OF_SCHOOLS", "st patricks,Piarsaigh,Phiarsaigh,St Vincents,Glanmire,Farranree,North Presentation,north mon,north monestrey,Scoil iosagain,St aidans,knocknaheeny,mon"); $schoolList = explode(',', LIST_OF_SCHOOLS); function checkForSchool($page, $schoolList) { preg_match('%(Studied at|Went to|Goes to) \\\u003ca href=\\\\"http:\\\/\\\/www.facebook.com\\\/pages\\\/[a-zA-Z-]*\\\/\d*\\\\" data-hovercard=\\\\"\\\/ajax\\\/hovercard\\\/page.php\?[a-zA-Z=0-9]*\\\\">([a-zA-Z\s]*)\\\u003c\\\/a>%', $page, $match); if($match && count($match)>0) { normal($match[2]); for($cnt=0; $cnt<count($schoolList); $cnt++) { $school = trim($schoolList[$cnt]); green(stristr($match[2],$school)); if($school!="" && stristr($match[2], $school)) { blue($match[2].": Match for school found"); unset($school); return true; } } } else { red("No school set on profile"); } unset($school); return false; How come my function is returning true for this match mond Community High School Inveralmond Community High School: Match for school found I only want it too return true if all the match corresponds with one of the element in the array $schoolList Hello, stristr('http://www.google.com/', 'google') doesn't work but stristr('http://localhost/test.html', 'sandro') works why is this? and how can I solve this problem? Hello I have multiple stristr condition (300+) involved in my php script (in 1 script only). I would like to know if that is okay or should i move to other solution like Mysql table search data and than direct it to accordingly. Kindly suggest some thing... the 300 condition are stored in include files (every file there is condition i.e there are 300 files in a directory which is included.) Kindly suggest some thing efficient and fast. Hi! I'm attempting to do this and feel that I'm fairly close. Would you take a look and throw me any bones? PHP Code: Code: [Select] <?php if (isset($_POST['ppassword']) && (stristr($string,'') === TRUE)) { switch($_POST['ppassword']) { case "Google": header("Location: http://www.google.com"); exit(); case "Yahoo": header("Location: http://www.yahoo.com"); exit(); case "Bing": header("Location: http://www.bing.com"); exit(); } } ?> The Form: Code: [Select] <form name="portal" id="portal" method="post" action=""> <input name="ppassword" type="text" id="ppassword" maxlength="25" onfocus="this.select()" onblur="this.value=!this.value?'company name':this.value;" value="company name" onclick="this.value=''" /> <input name="login" type="submit" id="login" value="Continue" /> </form> I appreciate any of your help. I feel close and this part Code: [Select] (stristr($string,'') feels like the missing piece of the puzzle. Guys thanks for helping me solve the problem i had but now i have another problem and i am lost. i have included the code below for you to have overview.
This is the code:
<table style="width:100%; margin-left:auto; margin-right:auto"> I have having a problem getting a mysql query to work. If I just use mysql_query it looks like this and works fine: INSERT INTO schedule (schedule_pk, schedule_month, schedule_day, schedule_year, schedule_hour, schedule_minute, schedule_type, employeenumber) VALUES ( NULL, 10, 10, 1999, 12, 6, 'DayIn', 3); If I put it through mysql_real_escape_string it turns it to this and does not work when I put it into mysql_query: INSERT INTO schedule (schedule_pk, schedule_month, schedule_day, schedule_year, schedule_hour, schedule_minute, schedule_type, employeenumber) VALUES ( NULL, 10, 10, 1999, 12, 6, \'DayIn\', 3); aka: $query = "INSERT INTO...." mysql_query($query); that works, but: $query = "INSERT INTO...." $sqlQuery = mysql_real_escape_string($query, $con); mysql_query($sqlQuery); results in the error. mysql_query($query); The error is: 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 '\'DayIn\', 3)' at line 1 Hows it going guys. I am currently having a problem. I had a program that worked. I migrated the website and now the program is broken. Any time i try to run it, i get this error: Quote Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/thegoo20/public_html/class/wordgame/wordgame.php on line 31 Here is the code at that point: Code: [Select] function changeword(){ $result = mysql_query("Select words from CurrentWords"); $totalwords = array(); while ($row = mysql_fetch_array($result)){ $totalwords[] = $row["words"]; } $_SESSION['word'] = $totalwords[rand(0, count($totalwords)-1)]; $_SESSION['scrambled'] = str_shuffle($_SESSION['word']); } where line 31 is the while statement. Any help would be appreciated. Thanks Hi Guys, I am having an issue with an if statement, I cant get it to go true, even though it clearly is! Please excuse the messy code, im pulling my hair out here! $itemprice[0] = 8.99; $itemprice[1] = 19.95; $itemprice[2] = 8.99; print_r($itemprice); echo "<BR>"; $x=0; foreach($itemprice AS $val) { echo gettype($val) .$val. "<br>"; if($val == 8.99) { $x++; $freecount++; } } echo $x; outputs correctly: Array ( => 8.99 [1] => 19.95 [2] => 8.99 ) double8.99 double19.95 double8.99 x = 2 replacing: $newtotal = add_to_price($id, $non_disc); $itemprice[] = $newtotal - $non_disc; print_r($itemprice); echo "<BR>"; $x=0; foreach($itemprice AS $val) { echo gettype($val) .$val. "<br>"; if($val == 8.99) { $x++; $freecount++; } } echo $x; The add to price function returns a running total, the take away gives me the current item price in a for loop to build the array. Output: Array ( => 8.99 [1] => 19.95 [2] => 8.99 ) double8.99 double19.95 double8.99 x = 1 The problem? Well when I build the array up automatically it still prints the same, has the same datatype, yet the if statement does not catch the second 8.99 value? Please accept my apologies for this post, it is my first ever and i've been doing this for 10 years now, i have never been this stumped on something so simple, I can only think it is a datatype error but i have tried apostrophes in the if, make no difference. All help massively gratefully received. Please email me if you want to see it in action... I do not receive email for my published php file which is: <?php ///// easend.php ///// $youremail = "acdelco40108@yahoo.com"; /*put the email address here, between quotes, the email address you want the message sent to*/ $to = $youremail; $email = $_POST['EMail']; $name2 = $_POST['Name']; $street2 = $_POST['Street']; $city2 = $_POST['City']; $state2 = $_POST['State']; $zip2 = $_POST['Zip']; $Phone = $_POST['Home_Phone']; $Cell = $_POST['Cell_Phone']; $education = $_POST['email1']; $comments = $_POST['email2'] ; $headers = "From:" . $email; $fields = array(); $fields{"Name"} = "Name"; $fields{"Street"} = "Street"; $fields{"City"} = "City"; $fields{"State"} = "State"; $fields{"Zip"} = "Zip"; $fields{"Home_Phone"} = "Home Phone"; $fields{"Cell_Phone"} = "Cell Phone"; $fields{"EMail"} = "Email"; $fields{"email1"} = "Education"; $fields{"email2"} = "Comments"; $subject = "We have received the following information from your employment application"; $body = "We have received the following information from your employment application:\n\n"; foreach($fields as $a => $b) { $body .= sprintf("%20s: %s\n",$b,$_POST[$a]); } mail ($to, $subject, $body, $headers); //send mail to owner #end create email vars $headers = "From:" . $to; mail ($email, $subject, $body, $headers); //send mail to user #end create email vars echo "<head><META HTTP-EQUIV=\"Refresh\" CONTENT=\"2; URL=ThankYou.html\"></head>"; ?> Okey so i made a table that you put your name,author,and message when you submit it , it echoes a table with the name,author and message written (and it also echoes a delete buttom,so delete this post oif necessary) im new to php and i have been with this problem for a couple of weeks so i guess its time to ask for some help the problem is that i dunnot know how to make my delete buttom work! i tried if statement but it dosent work i also tried ternary operation and dint work :S i read that there is something like $post[ID](and this is supposed to get the ID of the post submmited, and delete it) im not sure, im so confused! help! XD this is the code <?php $tittle=$_POST['tittle']; $author=$_POST['author']; $message=$_POST['message']; if ($_POST['submitnews']){ $currentdate= date("y-m-d"); $currenttime=date("H:i:s",strtotime("-6 hours")); $post=mysql_query("INSERT INTO news VALUES('','$tittle','$author','$message','$currentdate','$currenttime')"); echo"Posted!"; } $select=mysql_query("SELECT * FROM news ORDER BY id DESC"); while ($row= mysql_fetch_assoc($select)) { $id=$row['id']; $tittle=$row['tittle']; $author=$row['author']; $message=$row['message']; $date=$row['date']; $time=$row['time']; if ($_SESSION['admin']) { echo " <table width='488px' id='news_table'> <tr> <td> </td> <td> <center><font size='5'>$tittle</font></center><br> </td> </tr> <tr> <td> </td> <td> $message </td> </tr> <tr> <td> </td> <td> <font size='1'>Posted By:<font color='green'>$author</font> on <font color='gray'>$date</font> at <font color='gray'>$time</font></font> <input name='delete' type='submit' value='delete'> <td> </td> </tr><br><br> </table>"; The problem is whit the var $threadid . It dosent work here(it dosen't display the data from database): $dbh = "SELECT *FROM comments WHERE threadid = '".$threadid."' "; But if I assign a value before the that code the html page show correctly. Code: [Select] include("config.php"); $name = $_POST['name']; $comment = $_POST['comment']; $threadid = $_POST['threadid']; if($name & $comment) { $dbh="INSERT INTO comments (name,comment,threadid) VALUES ('$name','$comment','$threadid') "; mysql_query($dbh); } $dbh = "SELECT *FROM comments WHERE threadid = '".$threadid."' "; $req = mysql_query($dbh); while($row=mysql_fetch_array($req)) { echo "<li>"; echo "<br>"."<b>".$row['name']."</b></br>"."<br>".$row['comment']."</br>"; echo "</li>"; } Code: [Select] $('.submit').click(function(){ location.reload(); var name = $("#name").val(); var comment = $("#comment").val(); var threadid = $("#v").val(); var dat = 'name='+name+'&comment='+comment+'&threadid='+threadid; $.ajax({ type:"post", url:"comment.php", data:dat, success:function(){ console.log("dat"); } }); return false; }); I Have An error on line 42 not sure what it is can any one tell? $foundnum = mysql_num_rows($run); You may see a lot of things wrong with the code below, but please bring it to my attention! I'm trying to advance my knowledge on OOP, but I'm trying my best to get the basics down. The below code, after selecting the two characters to fight, the page returns nothing, just blank. I'm pretty sure the error lyes within class_li.php, but I'm not exactly sure. Script is attached. I am trying to understand PHP OOP and I have some code that is not working. here is the error code. Code: [Select] Parse error: parse error, expecting `T_FUNCTION' in C:\wamp\www\testing\armor_lib.php on line 11 And here is the armor_lib.php file: <?php class armor { // Head public $head; public $torso; public $pants; public $gloves; public $boots; // new stuff here class head extends armor { function __construct($head){ $this->set_head($head); } } // Torso class torso extends armor { function __construct($torso){ $this->set_torso($torso); } } // Pants class pants extends armor { function __construct($pants){ $this->set_pants($pants); } } // Gloves class gloves extends armor { function __construct($gloves){ $this->set_gloves($gloves); } } // Boots class boots extends armor { function __construct($boots){ $this->set_boots($boots); } } } ?> And here is the php in the armor.php file I have: <?php $head = new armor("Leather Helm"); $torso = new armor("Leather Shirt"); $pants = new armor("Leather Pants"); $gloves = new armor("Chain Mail Gloves"); $boots = new armor("Leather Boots"); echo "You are wearing: " . $head->get_head() . "on your Head"; echo "<br />"; echo "You are wearing: " . $torso->get_torso() . "on your Torso"; echo "<br />"; echo "You are wearing: " . $pants->get_pants() . "on your Legs"; echo "<br />"; echo "You are wearing: " . $gloves->get_gloves() . "on your Hands"; echo "<br />"; echo "You are wearing: " . $boots->get_boots() . "on your Feet"; ?> Any Help in understanding this will be much appreciated. Thanks. I am trying to save this an an xml document but am getting this error when I try to open the xml file "feed.xml" - "XML Parsing Error: no element found" $xml = '<rss version="2.0"> <channel> <title> RSS Feed</title> <link></link> <description>the best industry-lead opinions</description> <language>en-us</language> </channel> </rss>"; $xml2 = new DOMDocument('1.0'); $xml2->Load($xml); $xml2->save("feed.xml"); Hi all. When I typed symbol ' in my textarea , after I submit it and view for what I typed , it will automatically add a slash in front of the ' . Such as the message is "I'm fine" , then it will turn out as "I\'m fine". Can I know what is the problem ? and how can I solve it? Thanks for every reply . Code: [Select] $sql = "SELECT softwareID,softwareName,softwareType,softwareDesc,softwarePath,ITOnly FROM software WHERE softwareName LIKE '%($searchSoftware)%' ORBER BY softwareName"; $result = mysqli_query($cxn,$sql) or die(mysqli_error()); There is something wrong with my LIKE statement because it's not pulling it since I'm either formatting it wrong or something. Can anyone catch it? Hi. I am from poland. I am 17 old age and like webmastering. I write my social network. I have new server 10 GB VPS and my script no runing In my server. in my server do can not login. No runing function Code: [Select] mb_strtolower();as delete function mb_strtolower() ; this login runing. In my server haven`t installing liberary GD and no runing upload avatars. my code upload is: elseif ($_GET['act'] == "upload") { echo <<< KONIEC <hr> <center> <p> <h2>Dodaj zdjęcie</h2> <ul class="gallery clearfix"> <li class="extra"> KONIEC; error_reporting(E_ALL); // we first include the upload class, as we will need it here to deal with the uploaded file include('include/class.upload.php'); // retrieve eventual CLI parameters $cli = (isset($argc) && $argc > 1); if ($cli) { if (isset($argv[1])) $_GET['file'] = $argv[1]; if (isset($argv[2])) $_GET['dir'] = $argv[2]; if (isset($argv[3])) $_GET['pics'] = $argv[3]; } // set variables $dir_dest = (isset($_GET['dir']) ? $_GET['dir'] : 'test'); $dir_pics = (isset($_GET['pics']) ? $_GET['pics'] : $dir_dest); if (!$cli) { } // we have three forms on the test page, so we redirect accordingly if ((isset($_POST['action']) ? $_POST['action'] : (isset($_GET['action']) ? $_GET['action'] : '')) == 'simple') { // ---------- SIMPLE UPLOAD ---------- // we create an instance of the class, giving as argument the PHP object // corresponding to the file field from the form // All the uploads are accessible from the PHP object $_FILES $handle = new Upload($_FILES['my_field']); // then we check if the file has been uploaded properly // in its *temporary* location in the server (often, it is /tmp) if ($handle->uploaded) { // yes, the file is on the server // now, we start the upload 'process'. That is, to copy the uploaded file // from its temporary location to the wanted location // It could be something like $handle->Process('/home/www/my_uploads/'); $handle->Process($dir_dest); function TestProcess(&$handle, $title = 'test', $details='') { global $dir_pics, $dir_dest; $unlink = 'test/'. $handle->file_dst_name; $myid = $_SESSION['id']; $opis = htmlspecialchars(stripslashes(strip_tags(trim($_POST["opis"]))), ENT_QUOTES); $handle->Process($dir_dest); if ($handle->processed) { unlink("$unlink"); $link = ''.$dir_pics.'/' . $handle->file_dst_name .''; $addphoto = mysql_query("INSERT INTO photo VALUES('', '$myid', '$link', '$opis')"); echo <<< KONIEC <br> <center> <img src="$link" > <br> <br /> </center> KONIEC; } else { echo '<fieldset class="classuploadphp">'; echo ' <legend>' . $title . '</legend>'; echo ' Error: ' . $handle->error . ''; if ($details) echo ' <pre class="code php">' . htmlentities($details) . '</pre>'; echo '</fieldset>'; } } if (!file_exists($dir_dest)) mkdir($dir_dest); // ----------- $handle->image_convert = 'jpg'; $handle->image_resize = true; $handle->image_ratio_y = true; $handle->image_x = 500; $handle->image_precrop = 15; $handle->image_watermark = "watermark_large.png"; $handle->image_watermark_x = 20; $handle->image_watermark_y = -20; TestProcess($handle, '15px pre-cropping (before resizing 800 wide), large watermark automatically reduced, position 20 -20', "\$foo->image_convert = 'jpg';\n\$foo->image_resize = true;\n\$foo->image_ratio_y = true;\n\$foo->image_x = 800;\n\$foo->image_precrop = 15;\n\$foo->image_watermark = 'watermark_large.png';\n\$foo->image_watermark_x = 20;\n\$foo->image_watermark_y = -20;"); } else { // one error occured echo '<fieldset>'; echo ' <legend>file not uploaded to the wanted location</legend>'; echo ' Error: ' . $handle->error . ''; echo '</fieldset>'; } // we copy the file a second time // we delete the temporary files $handle-> Clean(); } else { // if we are here, the local file failed for some reasons echo '<fieldset>'; echo ' <legend>local file error</legend>'; echo ' Error: ' . $handle->error . ''; echo '</fieldset>'; } echo <<< KONIEC <li> </ul> </p> </center> KONIEC; is it the only problem that isn`t installing GD as i have something off in php.ini ? Plise Help me |
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