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PHP - Php Function In Mysql Db Field And Execute It? Is This Possible?
Let's say I have a mysql table named 'function' and a field named 'body' with 1 row.
The 'body' field contains 'Echo out the number 1000 with a comma. ".number_format(1000).". Simple Example.' As you can see there is a function; number_format(); in the database, but is it possible to execute it onto a live webpage and be displayed correctly? Similar TutorialsHi, I'm experimenting with CURL. I'm trying to post data with CURL. Everything went fine until I found a form where the site submits data using AJAX technology. The data is still in <input> tag, so I can "type" it there using CURL. However, I can't submit the data as it is executed via <a> tag with Javascript (AJAX). Could you guys help me with this? Is it possible for PHP to execute AJAX function(with parameters sending)? Is there any similar example somewhere? Regards, Hi Guys, I have a webpage which has subsequent pages stored in a database e.g. index.php?id=1 The problem being, is that id=1 has it's data pulled from a database. This was fine & dandy until I started to insert PHP...I am trying to get the below to executre <?php $rater_id=1; $rater_item_name='Rate This'; include("rater.php");?> However nothing shows & nothing happens, I know eval can be used but have not been succesfull in implementing this, can someone please help! Hi: I'm going crazy trying to do the following: I'm making a job registration process where the user registers on one php page to the website, must acknowlege and email receipt using an activate php page, then is directed to upload their C.V. (resume) based on the email address they enter in the active page output. I then run an upload page to store the resume in teh MySQL db based on the users email address in the same record. If I isolate the process of the user registering to the db, it works perfectly. If I isolate the file upload process into the db, it works perfect. I simply cannot upload teh file to the existing record based on teh email form field matching the user_email field in the db. With the processes together, teh user is activated, but teh file is not uploaded. Maybe I've simply been at this too long today, but am compeled to get through it by end day. If anyone can help sugest a better way or help me fix this, I will soo greatly appreciate it. My code is as follows for the 2 pages. ---------activate.php------- <?php session_start(); include ('reg_dbc.php'); if (!isset($_GET['usr']) && !isset($_GET['code']) ) { $msg = "ERROR: The code does not match.."; exit(); } $rsCode = mysql_query("SELECT activation_code from subscribers where user_email='$_GET[usr]'") or die(mysql_error()); list($acode) = mysql_fetch_array($rsCode); if ($_GET['code'] == $acode) { mysql_query("update subscribers set user_activated=1 where user_email='$_GET[usr]'") or die(mysql_error()); echo "<h3><center>Thank You! This is step 2 of 3. </h3>Your email is confirmed. Please upload your C.V. now to complete step 3.</center>"; } else { echo "ERROR: Incorrect activation code... not valid"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta content="text/html; charset=utf-8" http-equiv="Content-Type" /> <title>Job application activation</title> </head> <body> <center> <br/><br/><br/> <p align="center"> <form name="form1" method="post" action="upload.php" style="padding:5px;"> <p>Re-enter you Email : <input name="email" type="text" id="email"/></p></form> <form enctype="multipart/form-data" action="upload.php" method="POST"> <input type="hidden" name="MAX_FILE_SIZE" value="4000000"> Upload your C.V.: <input name="userfile" type="file" id="userfile"> <input name="upload" type="submit" id="upload" value="Upload your C.V."/></form> </p> </center> </body> </html> --------upload.php---------- <?php session_start(); if (!isset($_GET['usr']) && !isset($_GET['code']) ) { $msg = "ERROR: The code does not match.."; exit(); } if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0) { $fileName = $_FILES['userfile']['name']; $tmpName = $_FILES['userfile']['tmp_name']; $fileSize = $_FILES['userfile']['size']; $fileType = $_FILES['userfile']['type']; $email = $_POST['email']['user_email']; $fp = fopen($tmpName, 'r'); $content = fread($fp, filesize($tmpName)); $content = addslashes($content); fclose($fp); if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); } include 'reg_dbc.php'; $query = "UPDATE subscribers WHERE $email = user_email (name, size, type, content ) ". "VALUES ('$fileName', '$fileSize', '$fileType', '$content')"; mysql_query($query) or die('Error, query failed'); mysql_close($dbname); } ?> <center> <br/> <br/> <br/> <br/> Thank you for uploading your <?php echo "$fileName"; ?> file, completing your registration, and providing us your C.V. for this position. <br/> <br/> <br/> We will contact you if your canditature qualifies. </center> Hi! I was wondering if there is a way to execute php code which is stored in mysql database using php. At the minute I am using a echo to try and run php code stored in a mysql database but this just displays the code and does not run the php code. Thanks for any help! Hey all. I am trying to make a mod for our fantasy football site but keep getting an error. Fatal error: Call to a member function Execute() on a non-object in C:\Inetpub\wwwroot\phpffl\program_files\autorun\general\top_team.php on line 8. Here is the code. <?php global $PHP_SELF, $DB; $leagues_ID='1'; $week='2'; $week_ID=$week -1; $game_ID=$week_ID; $sql="select ID, team_name from teams where leagues_ID='$leagues_ID';"; [b] $teams_rs=$DB->Execute($sql);[/b] while (!$teams_rs->EOF) { $top_team_ID=$teams_rs->fields('ID'); $top_team_name=$teams_rs->fields('team_name'); $top_team_points=get_total_points_game($top_team_ID, $week_ID, $week_ID, $leagues_ID); //echo $top_team_points." "; $top_team_points_array[$top_team_name]=$top_team_points; $teams_rs->MoveNext(); } $top_scoring_team_ID=""; $top_scoring_team_name=""; $top_scoring_points=""; foreach(array_keys($top_team_points_array) as $key) { if ($top_team_points_array[$key] > $top_scoring_points) { $top_scoring_points=$top_team_points_array[$key]; //$top_scoring_team_ID=$top_team_ID; $top_scoring_team_name=$key; } } $sql="insert into total_scores (team_ID, teams_name, week_id, total_score, leagues_ID) values('$top_team_ID', '$top_scoring_team_name', '$week_ID', '$top_scoring_points', '$leagues_ID');"; $rs=$DB->Execute($sql); echo "<br>"; echo "<br>"; echo "Weekly Top Scoring Team for Week $week_ID is: $top_scoring_team_name with a score of $top_scoring_points points!"; echo "<br>"; echo "<br>"; //print_r ($top_team_points_array); echo "<br>"; echo "<br>"; $weekly_top_score=max($top_team_points_array); ?> I have a function in the global file looks like this. /************** BEGIN ADDITIONAL FUNCTIONS **************************/ function get_top_scoring_teams() { global $PHP_SELF, $DB, $PHPFFL_IMAGE_PATH; $sql="select * from total_scores order by week_ID DESC limit 1;"; $teams_rs=$DB->Execute($sql); while (!$teams_rs->EOF) { $team_ID=$teams_rs->fields('team_ID'); $team_name=$teams_rs->fields('teams_name'); $team_points=$teams_rs->fields('total_score'); $week_ID=$teams_rs->fields('week_ID'); echo "The Top Scoring Team for <strong>Week $week_ID</strong> was <br><img src='{$PHPFFL_IMAGE_PATH}team_logos_standings/$team_ID.gif'><br> <strong>$team_name</strong> with a score of <strong>$team_points</strong> points. <hr>"; $teams_rs->MoveNext(); } } /************** BEGIN ADDITIONAL FUNCTIONS**************************/ Any one see what the issue might be? I am running this on a windows 2003 server with iis 6 php 5.3 and mysql 5.1.5 Any help would be appreciated. Thanks Hitster4 I have this script running: <?php $query = mysql_query("SELECT * FROM jobs WHERE event = 'Yes' ORDER BY title"); while ($row = mysql_fetch_array($query)) { ?> How do I check if a field is empty and NOT display it...? For instance it has a 'applied' field, if that is empty I dont want it to display, however I still need the event = 'Yes' part. Hi i have the following code Code: [Select] $orderFetch->order="3"; mysql_query("UPDATE `categories` SET order=order-1 WHERE `order` > $orderFetch->order"); i wish to minus -1 from all the "order" fields in categories that are over a certain value i.e 3. But however this code is not working can anyone tell me where i'm going wrong? thank you I have made a simple form where users who have been subscribed and unsubscribe by inserting their email address. In my database using PHPMyAdmin, my database to store the emails is 'Links', the table is 'email' and the fields are the 'id' and 'emailaddress'. What I have tried is making a text input field, where the user ill insert his or her email address, to unsubscribe on the website. As a result the user's field for his or her email address will be delete in the database which is saving the emails for all users who have subscribed. My HTML codes a Code: [Select] <p>Subscribe for newsletters:</p> <img src="images/k-newsletter-icon.png" width="96" height="96" alt="subscri"/> <form action="index.php" method="post"> <input type="text" size="25" placeholder="Your email address..." name="enter"/> <input class="submit" type="submit" value="Subscribe" name="subscribe"/> </form> My PHP codes a Code: [Select] <?php if ($_SERVER['REQUEST_METHOD'] == 'POST') { $email = $_POST['enter']; @mysql_connect ('localhost', 'root', '') or die ('Error'); @mysql_select_db ('links') or die ('Error'); if (!filter_var($email, FILTER_VALIDATE_EMAIL)) { echo "Not an email"; return false; } else { mysql_query("DELETE FROM email WHERE emailaddress ='$email'"); echo "deleted"; } } ?> When I test it,it is not working, as I see the email which was saved, is still in the database! Help! I have a mysql field that I want to store a php variable in and then retrieve it.
Example:
Color table has the field named colorBackground with the value of #FF0000
CSS table has the field named cssBackground with the value of .background { background-color: #444444; }
What I want to do is make Color/colorBackground a variable {{$backgroundcolor = rsColor['colorBackground']
Then I want to change the CSS/cssBackground value to .background { background-color:$backgroundcolor; }
Creating the $backgroundcolor works fine. I'm just not sure how to put the $backgroundcolor in my CSS/cssBackground field.
In my mysql database i have a field that records peoples details one part of the details is their county/region they live in. It records the county in a normal varchar field however i need a simple php script that searching the database and finds the most common county so i can then return to the screen where the most popular region for my members. Thank you for reading. I'm trying to create a search form where members of a site can search other members whose user information is stored in a mysql database. One of the search parameters is gender which searches for records from a column in the database called gender where members can either have the value "man" or "woman". The user can choose between 3 options from a pull down menu named gender (man, woman, both). Then I'm thinking about using a select query to search the database upon submission of the form, which goes something like this: Code: [Select] $sql= SELECT* FROM members WHERE gender ='{$_POST['gender']}' ; Now I can easily assign the value "man" for the option man in the pull down menu and "woman for the option woman, to match their corresponding names in the database. The problem lies with the "both" option which has to be either man or woman. I'm thinking about assigning it the value "man OR woman" so that when the form is submitted, the query would read: SELECT*FROM members WHERE gender ='{$_POST[man OR woman']}; I just don't know if this would be a right usage of the OR keyword and if such a query would work. Before trying it out, I'd like to know if this makes any sense and if not, what's an alternative way to work around this? Can anyone tell me the best way to handle empty search field on a mysql Query? The query below if $name is left blank but the other fields are filled it shows no results, currently i use if.... else.... but that means i have a lot of code replicated, is there a better way? $sql = "SELECT private.username, se.age, se.ro, se.suc, se.bu, se.fu, com.in, com.ol, se.di, se.bo, se.uin FROM Reg_Profile_public AS se INNER JOIN Reg_Profile_Private AS private USING (uin) INNER JOIN Reg_Profile_public_Com AS com USING (uin) WHERE se.age BETWEEN '$low' AND '$high' AND se.ro=''$name"; Hi Guys, Can someone please help me before I kill myself! I have a table holding details of images to be displayed in a gallery. The table has columns as 'id', 'image_caption', 'file_path', and 'userid'. The 'id' column is the unique key, and the 'userid' column is obviously id of the user who uploaded the image. What I would like to do is select and display one image from each unique userid and then display each image along with the details in my gallery. It doesn't really matter which users image is selected aslong as there is only one for each unique user. I have been trying a number of ways to do this (DISTINCT / subqueries) but I just cant get anything to work. Could someone please advise me on how I could get this to work. Cheers in advance. Hi. The idea: I want to pull a client record and simply place all the row columns into editable form field text boxes. Once the user reviews and edits a submit button will update the record. What I have: Code: [Select] $Query01 = "SELECT * FROM `CustomerSignups` WHERE `Id` = '$_GET[Id]'"; $Result01 = mysql_query($Query01) or die("Error 01: " . mysql_error()); while ($get_info = mysql_fetch_row($Result01)) { foreach ($get_info as $field) echo "<p>$field</p>"; It displays as expected, a line for each field. How can I get this to pull the field names also and then I can use those as text box input fields...? Hi, when I submit a form I want to update a field in the "daterange" table. In the form the id field (RID) from "daterange" is selected prior to submit. Basically, I want to change the STATUS field to B (from A). The structure of daterange table is as follows: RID (key field) DEND MONTH DATE SITE PRICE STATUS What is the easiest way to accomplish this? Here is the code from the form (if it will help)... Code: [Select] <?php //************************************** // Page load dropdown results // //************************************** function getTierOne() { $result = mysql_query("SELECT DISTINCT MONTH FROM daterange") or die(mysql_error()); while($tier = mysql_fetch_array( $result )) { echo '<option value="'.$tier['MONTH'].'">'.$tier['MONTH'].'</option>'; } } //************************************** // First selection results // //************************************** if($_GET['func'] == "drop_1" && isset($_GET['func'])) { drop_1($_GET['drop_var']); } function drop_1($drop_var) { include_once('db.php'); $result = mysql_query("SELECT * FROM daterange WHERE DEND > DATE(NOW()) AND STATUS='A' AND MONTH='$drop_var' ORDER BY DATE, SITE") or die(mysql_error()); echo '<select name="RID"> <option value=" " disabled="disabled" selected="selected">Choose a Reservation</option>'; while($drop_2 = mysql_fetch_array( $result )) { echo '<option value="'.$drop_2['RID'].'">'.$drop_2 ['DATE']. ', '.$drop_2 ['SITE']. ', '.$drop_2 ['PRICE'].'</option>'; } echo '</select> '; echo "<br />"; echo "</select><p align=left><label><font size=\"2\" face=\"Arial\">First Name: <input type=\"text\" name=\"FNAME\" size=\"50\" maxlength=\"50\" tabindex=\"1\"<br />"; echo "<p align=left><label>Last Name: <input type=\"text\" name=\"LNAME\" size=\"50\" maxlength=\"50\" tabindex=\"2\"<br />"; echo "<p align=left><label>Address Line 1: <input type=\"text\" name=\"ADDR1\" size=\"50\" maxlength=\"50\" tabindex=\"3\"<br />"; echo "<p align=left><label>Address Line 2: <input type=\"text\" name=\"ADDR2\" size=\"50\" maxlength=\"50\" tabindex=\"4\"<br />"; echo "<p align=left><label>City: <input type=\"text\" name=\"CITY\" size=\"50\" maxlength=\"50\" tabindex=\"5\"<br />"; echo "<p align=left><label>State (abbrev.): <input type=\"text\" name=\"STATE\" size=\"2\" maxlength=\"2\" tabindex=\"6\"<br />"; echo "<p align=left><label>Zip Code: <input type=\"text\" name=\"ZIP\" size=\"5\" maxlength=\"5\" tabindex=\"7\"<br />"; echo "<p align=left><label>Contact Phone Number: (<input type=\"text\" name=\"PHONE1\" size=\"3\" maxlength=\"3\" tabindex=\"8\""; echo "<label>)<input type=\"text\" name=\"PHONE2\" size=\"3\" maxlength=\"3\" tabindex=\"9\""; echo "<label>-<input type=\"text\" name=\"PHONE3\" size=\"4\" maxlength=\"4\" tabindex=\"10\"<br />"; echo "<p align=left><label>Email: <input type=\"text\" name=\"EMAIL\" size=\"50\" maxlength=\"50\" tabindex=\"11\"<br />"; echo '<input type="submit" name="submit" value="Book Now!" /><br />'; echo '<input type="reset" name="submit" value="Reset" /><br />'; } ?> I have a MySQL table called "products": Quote +------+-------+-------+-------+ | id | sizes | sizem | sizel | +------+-------+-------+-------+ | 1 | 2 | 0 | 1 | | 2 | 3 | 1 | 0 | +------+-------+-------+-------+ What I am wanting to know is if I can make PHP print a value from it, but only if it is not a zero.. something like: $sizeL = (table(products)id1.sizel); if (table(products)id1.sizel == "0") {echo "";} else {echo "$sizeL";} Yeah I know the above code is far from beeing valid, but is the best I could come up with. heh Following from my previous post, i have a questionnaire and within each question it has the options to pick from raddio button (strongly agree, agree, disagree, N/A). For each value of each question i want to pass it into the relevant Field in my database but rather than overwriting the current value i want to be added on to it. (For example if the current value is 2, once the new value (1) has added it will make the new value of 3.). I have a script that uploads and images, creates 2 versions of it, re-sizes the images and inserts all the info into a DB. The odd thing is that if I don't have an image to upload, it will still run the script but it will completely skip over the description. If I upload an image, it works fine. Not too sure what the issue is. Is it because my variable of $desc causing a conflict? Here's my code: Code: [Select] <?php ini_set("display_errors",1); include 'dbconn.php'; $change=""; $abc=""; define ("MAX_SIZE"," 10000"); function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } $errors=0; if($_SERVER["REQUEST_METHOD"] == "POST") { $image =$_FILES["file"]["name"]; $uploadedfile = $_FILES['file']['tmp_name']; if ($image) { $filename = stripslashes($_FILES['file']['name']); $extension = getExtension($filename); $extension = strtolower($extension); if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { $change='<div class="msgdiv">Unknown Image extension </div> '; $errors=1; } else { $size=filesize($_FILES['file']['tmp_name']); if ($size > MAX_SIZE*1024) { $change='<div class="msgdiv">You have exceeded the size limit!</div> '; $errors=1; } if($extension=="jpg" || $extension=="jpeg" ) { $uploadedfile = $_FILES['file']['tmp_name']; $src = imagecreatefromjpeg($uploadedfile); } else if($extension=="png") { $uploadedfile = $_FILES['file']['tmp_name']; $src = imagecreatefrompng($uploadedfile); } else { $src = imagecreatefromgif($uploadedfile); } echo $scr; list($width,$height)=getimagesize($uploadedfile); $newwidth=150; $newheight=($height/$width)*$newwidth; $tmp=imagecreatetruecolor($newwidth,$newheight); $newwidth1=50; $newheight1=($height/$width)*$newwidth1; $tmp1=imagecreatetruecolor($newwidth1,$newheight1); imagecopyresampled($tmp,$src,0,0,0,0,$newwidth,$newheight,$width,$height); imagecopyresampled($tmp1,$src,0,0,0,0,$newwidth1,$newheight1,$width,$height); $famname = $_POST['famname']; $desc = $_POST['description']; $petname = $_POST['petname']; $letter = $_POST['letter']; $filename = "images/". $_FILES['file']['name']; $filename1 = "images/small". $_FILES['file']['name']; imagejpeg($tmp,$filename,100); imagejpeg($tmp1,$filename1,100); imagedestroy($src); imagedestroy($tmp); imagedestroy($tmp1); }} } //If no errors registred, print the success message if(isset($_POST['Submit']) && !$errors) { $insert="insert ignore into tbl_tribue (petname,familyname,letter,description,imgpath,thumbpath) values ('$petname','$famname','$letter','$desc','$filename','$filename1')"; mysql_query($insert); //mysql_query("update gallery set imgpath='$filename'"); $change=' <div class="msgdiv">Pet Added Successfully!</div>'; print $insert; } ?> I had turned on the error and the print to see the output. Anyone have any ideas? Thanks! I see some of the fields in mysql show NULL and some are just empty, wonder what is the difference, is empty field still NULL? Basically I have a date of birth field (date) now how would I go about splitting that date field into 3 variables $year $month and $day? Any small push forward is much appreciated, I have searched but it seems that I might not be putting it into words correctly. Actually while writing this I suppose I might of thought of the solution Code: [Select] <?php $year = date('Y', strtotime($row['dob'])); $month = date('m', strtotime($row['dob'])); $day = date('d', strtotime($row['dob'])); ?> Would that be how you would go about it? |
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