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PHP - Hello, Call_user_func_array Problem, Help!
Hello i have two questions, regarding my problem.
1. I have a script which uses call_user_func_array , i heard it is not the best solution, here i post my script, please tell me what can i use instead of call_user_func_array, of course if the rumours are true that it is not the best solution. 2. My second question is that i get a warning message, if i turn error_reporting on: 'Warning: Wrong parameter count for array_intersect() in index.php on line xxx' However the site displays properly, but i want to get rid of this error message, why is it appearing what is the problem. My script: Code: [Select] $result = mysql_query("select id from table where word = '". $value ."'); while ($row = mysql_fetch_array($result, MYSQL_NUM)) { $query_results[$value][] = $row[0]; } $ids = call_user_func_array('array_intersect', $query_results); Similar TutorialsIs there a call_user_func_array() alternative other than eval()? With the shear genius of some of the individuals on this forum, I'd expect that if there is an alternative, someone here would know it. Any tips or insight greatly appreciated. Thanks Background (blah blah blah) I'm being kinda picky on efficiency and performance since I will be looking at possible hundreds of executions a request... so I'd like to avoid using the above mentioned functions since they take ~3x and ~10x longer (according to a this comment). Well in the class UserValidator I have a public method validate(), which checks the validation type and then calls specific private methods to execute the validation script. It gets a bit tricky here since the method validate() has to decide what private method to call. I am thinking about using call_user_func_array(), the script currently looks like this below: Code: [Select] public function validate(){ // The core method validate, it sends requests to different private methods based on the type if(empty($this->type)) throw new Exception('The validation type is empty, something must be seriously wrong...'); if(is_array($this->type) and !is_array($this->value)) throw new Exception('Cannot have scalar value if the type is an array.'); if(!is_array($this->type) and is_array($this->value)) throw new Exception('Cannot have scalar type if the value is an array.'); // Now we are validating our user data or input! $validarray = array("register", "login", "password", "session", "email", "reset", "profile", "contacts", "friends"); foreach($this->type as $val){ $method = "{$val}validate"; if(in_array($val, $validarray)) call_user_func_array(array($this, $method), array()); } } The method to execute depends on the validation type passed into the validator class, so it is somewhat like a dynamic function call. The problem is, well, I am not sure if I am using call_user_func_array() properly. I read it from php manual that it needs to accept a class instance, but there is no such instance inside a class method so I use $this in the argument. Is this the correct way of using call_user_func_array()? If not, how am I supposed to do this? thanks. The page that I'm currently working on has a form which allows, through a series of check boxes and text entry fields, a dynamic mysql query to be created. The php code constructs the query based on which fields the user updated in order to find a specific user in the database. Everything was working fine until I started to only display part of the result set. In order to display a message similar to "Displaying 12-25 of 65 results." I ran the query twice. Once to get the total number of results, and once to only get the specified set. The code for the section producing the error is included below. The elseif statement works as desired, as does the initial query of the if statement. if (isset($sql) && $sql !=""){ $sql = "SELECT DISTINCT users.StudentID FROM users lEFT JOIN events ON users.StudentID=events.StudentID WHERE" . $sql; $stmt = $mysqli->prepare($sql); call_user_func_array(array(&$stmt, 'bind_param'), $array_of_params); $stmt->execute(); $stmt->store_result(); $total= $stmt->num_rows; $stmt->close(); $sql = "SELECT DISTINCT users.StudentID,users.FirstName,users.LastName,users.Email,users.Phone,users.Texting,users.Admin,users.HDHS,users.Sex,users.Confirmation FROM users lEFT JOIN events ON users.StudentID=events.StudentID WHERE" . $sql . " ORDER BY LastName,FirstName LIMIT " . $_GET['start'] . ",12"; $stmt = $mysqli->prepare($sql); call_user_func_array(array(&$stmt, 'bind_param'), $array_of_params); $stmt->bind_result($row['StudentID'],$row['FirstName'],$row['LastName'],$row['Email'],$row['Phone'],$row['Texting'],$row['Admin'],$row['HDHS'],$row['Gender'],$row['Confirmation']); } elseif(isset($_GET['submit']) || isset($_GET['earlier']) || isset($_GET['later'])){ $stmt = $mysqli->prepare("SELECT StudentID,FirstName,LastName,Email,Phone,Texting,Admin,HDHS,Sex,Confirmation FROM users ORDER BY LastName,FirstName"); $stmt->bind_result($row['StudentID'],$row['FirstName'],$row['LastName'],$row['Email'],$row['Phone'],$row['Texting'],$row['Admin'],$row['HDHS'],$row['Gender'],$row['Confirmation']); $stmt->execute(); $stmt->store_result(); $total= $stmt->num_rows; $stmt = $mysqli->prepare("SELECT StudentID,FirstName,LastName,Email,Phone,Texting,Admin,HDHS,Sex,Confirmation FROM users ORDER BY LastName,FirstName LIMIT " . $_GET['start'] . ",12"); $stmt->bind_result($row['StudentID'],$row['FirstName'],$row['LastName'],$row['Email'],$row['Phone'],$row['Texting'],$row['Admin'],$row['HDHS'],$row['Gender'],$row['Confirmation']); $stmt->execute(); } If anyone would be willing to help me out here it would be greatly appreciated. If you need more information let me know. Thank you! 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"> Hi guys, I'm working on a side-project whilst my work at Uni is at a low but I'm extremely rusty on PHP; I've not touched it since around '07. Before I get on to my project, which will lift information from external websites, I'm just trying to acquaint myself with cURL again. Before I get flamed, yes this code is largely lifted from a website, but I pretty much understand it. My aim is to be able to get the title of a website, namely Facebook. Whilst this code is fine on seemingly any other website, it doesn't seem to be able to work for Facebook. I've checked the source code, and I really can't see the problem. Can anyone shed some light for me? Code: [Select] <?php $url = 'http://www.facebook.com/'; $webinfo = get_data($url); $title = get_match('/<title>(.*)<\/title>',$webinfo); $content= '<h2>Title of webpage: </h2><p>'.$name.'</p>'; echo $content; function get_match($regex,$content) { preg_match($regex,$content,$matches); return $matches[1]; } //gets the data from a URL function get_data($url) { $ch = curl_init(); $timeout = 5; curl_setopt($ch,CURLOPT_URL,$url); curl_setopt($ch,CURLOPT_RETURNTRANSFER,1); curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,$timeout); $data = curl_exec($ch); curl_close($ch); return $data; } ?> Thanks!!! Not exactly sure how to do this. What I need is when a person clicks on the link ie: page.php?nid=1 then it loads what needs to be loaded but if a user clicks and there is no nid ie: page.php?nid= Then it loads something else. if (nid==$nid){ echo "This is the data to show"; }else{ Show this data instead } Both if and else both show different mysql data. Hopefully I explained myself well. Code: [Select] $sql = "SELECT softwareID,softwareName,softwareType,softwareDesc,softwarePath,ITOnly FROM software WHERE softwareName LIKE '%($searchSoftware)%' ORBER BY softwareName"; $result = mysqli_query($cxn,$sql) or die(mysqli_error()); There is something wrong with my LIKE statement because it's not pulling it since I'm either formatting it wrong or something. Can anyone catch it? I'm trying to run a process using shell_exec, however the program i'm trying to run must communicate with a process that is not a system process, and everytime i run it, it fails, does anyone know how to make a apache/php server run active windows not in the system process? Hi. I am from poland. I am 17 old age and like webmastering. I write my social network. I have new server 10 GB VPS and my script no runing In my server. in my server do can not login. No runing function Code: [Select] mb_strtolower();as delete function mb_strtolower() ; this login runing. In my server haven`t installing liberary GD and no runing upload avatars. my code upload is: elseif ($_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 Hi there, I have a map of America made in flash, where you click on a state and the page should display the SQL database information for that state in the HTML table - but instead all it shows is the first entry of the database regardless of which state you click and it doesn't display the 2 radio buttons. My code is as follows Flash Actionscript 3 (just showing one state) Code: [Select] function waClick(event:MouseEvent):void { var waURL:URLRequest = new URLRequest("restaurants.php?state=Washington"); navigateToURL(waURL, "_self"); } wa_btn.addEventListener(MouseEvent.CLICK, waClick); PHP code Code: [Select] <?php include("mvfconnect.php"); $theChoice = $_GET['state']; $query = "SELECT * FROM restaurants WHERE" .$theChoice; $result = @ mysql_query($query); if (!$result) { $message="Unfortunately we are having problems with this page, we promise to have it fixed as soon as possible"; die($message); } $num = mysql_num_fields($result); $i=0; while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){ $show1=substr($row['state'],0,50)."..."; $show2=substr($row['city'],0,50)."..."; $show3=substr($row['rname'],0,50)."..."; $show4=substr($row['address'],0,50)."..."; $show5=substr($row['pnum'],0,50)."..."; $show6=substr($row['web'],0,50)."..."; $show7=substr($row['dishes'],0,50)."..."; $show8=substr($row['dish_details'],0,50)."..."; $show9=substr($row['challenges'],0,50)."..."; $show10=substr($row['challenge_details'],0,50)."..."; $show11=substr($row['youtube'],0,50)."..."; $show12=substr($row['images'],0,50)."..."; echo "<tr>". "<td>".$row['city']."</td>". "<td>".$row['rname']."</td>". "<td>".$row['address']."</td>". "<td>".$row['pnum']."</td>". "<td>".$row['web']."</td>". "<td>".$row['dishes']."</td>". "<td>".$row['challenges']."</td>". "<td id='state".$i."' style='display:none'>".$row['state']."</td>". "<td id='city".$i."' style='display:none'>".$row['city']."</td>". "<td id='rname".$i."' style='display:none'>".$row['rname']."</td>". "<td id='address".$i."' style='display:none'>".$row['address']."</td>". "<td id='pnum".$i."' style='display:none'>".$row['pnum']."</td>". "<td id='web".$i."' style='display:none'>".$row['web']."</td>". "<td id='dishes".$i."' style='display:none'>".$row['dishes']."</td>". "<td id='dish_details".$i."' style='display:none'>".$row['dish_details']."</td>". "<td id='challenges".$i."' style='display:none'>".$row['challenges']."</td>". "<td id='challenge_details".$i."' style='display:none'>".$row['challenge_details']."</td>". "<td id='youtube".$i."' style='display:none'>".$row['youtube']."</td>". "<td id='images".$i."' style='display:none'>".$row['images']."</td>". "<td><input type='radio' name='vid' id='vid".$i."' onclick='openVideo(".$i.")' /></td>". "<td><input type='radio' name='pic' id='pic".$i."' onclick='openImage(".$i.")' /></td>". "<td class='last'style='display:none'>".$show1." ".$show2." ".$show3." ".$show4." ".$show5." ".$show6." ".$show7." ".$show8." ".$show9." ".$show10."</td>". "</tr>"; $i++; } ?> I have used this code before and it has worked fine, can someone please help me out! i'm getting this erro "Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 3 in /***/***/*****.php on line 278 not an active session, starts new" this is the code: if ($phase = mysql_result(mysql_query("SELECT `phase` FROM `as_support` WHERE `mission` = '$var2'"),0,0)) { if ($phase=='5') { // mysql_query("DELETE FROM `as_support` WHERE `mission` = '$var2'"); // header("Location: https://www.**********/*******/*******.php"); } else { echo date('h:i:s') . ": PHASE is $phase<br />"; echo "\$var1=$var1,\$var2=$var2<br />"; } } else { echo "not an active session, starts new<br />"; echo "\$var1=$var1,\$var2=$var2<br />"; mysql_query("INSERT INTO `as_support` (`mission`, `phase`) VALUES ('$var2', '0')") OR DIE(mysql_error()); the first line is line no. 278 can someone please tell me what to do? tnx shirley Elo, I'm creating a very simple image list that is function generated which pulls the source from a table in my database, this is my code Code: [Select] $output = "<ul>"; $img_src = get_all_image_src($sel_event['id']); while($src = mysql_fetch_array($img_src)) { for($i=0;$i <= 9; $i++) { $output .= "<li>"; $output .= "<a"; $output .= " class ='album'"; $output .= " rel='group'"; $output .= " href='/images/gallery/'" . $src['img_src'] . $i; $output .= ".jpg"; $output .= " title='sample 1'"; $output .= ">"; //Image Thumb $output .= "<img"; $output .= " src='/images/gallery/'"; $output .= $src['img_thumb_src']. $i . ".jpg"; $output .= "alt='image01' "; $output .= "title='sample title' />"; $output .= "tae"; $output .= "</a>"; $output .= "</li>"; } } $output .= "</ul>"; return $output; and the img src doesnt seems to add up... my folder structure is Sandbox>image>gallery>2011>Mall>Event>Thumbs where Sandbox is the main directory. i tried echoing the $src['img_src'] and it displays correctly the data to be added (which is 2011/Mall/Event/ plus $i and a .jpg) .. sorry noob here... any thoughts? am i missing a simple detail? 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... 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 . 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"); 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($query); The error is: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '\'DayIn\', 3)' at line 1 Hi, i use mysql_real_escape_string() function when insert in db, but when echo i use htmlentities($var, ENT_QUOTES, "UTF-8") Now problem is '\ in that text. In db it write text with single quote with \, example That\'s wrong, and when echo it show just the same like in db.. How can i fix this ? 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>"; ?> |
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