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PHP - Query To Return Value After A Certain Character In A Field
ok, this is clearly 1st grade code for some, but i'm not there yet - I'm querying posts in WordPress, and the post_content will always have an image in the beginning of the post followed by the content. i don't want to get the image, just the content that's after the image, which is wrapped in anchor tags, of course.
Code: [Select] <a href="http://path/to/image.jpg"><img src="http://path/to/image.jpg" /></a> <p>Post content yadda, yadda, hoowie</p> obviously a character count won't work, so i need to get anything that follows the first "</a>", say...? is this the best way, or is there an easier way? thanks for anyone's help. GN Similar TutorialsI have a PHP Form where users have to enter and upload a file but when they click upload everything is working except that it is only sending first character of text field to Database! Can any one help?? Thanks File upload.php form Code: [Select] <form action="add_file.php" method="post" enctype="multipart/form-data"> <p><br> JobID: <input name="JobID" type="text" value="<?php echo $row_Recordset1['JobID']; ?>" readonly="readonly" /> </p> <p>Company ID: <input name="CompanyID" type="text" value="<?php echo $row_Recordset1['CompanyID']; ?>" readonly="readonly" /> </p> <p>UserID: <input name="UsersID" type="text" /> </p> <p>Select File to upload: <input type="file" name="uploaded_file" /> </p> <p>Make Sure All Details Are Correct before Submitting!</p> <p> <input type="submit" value="Submit Application" /> </p> <form> add_file.php // Check if a file has been uploaded if(isset($_FILES['uploaded_file'])) { // Make sure the file was sent without errors if($_FILES['uploaded_file']['error'] == 0) { // Connect to the database $dbLink = new mysqli('localhost', 'user', 'password', 'phpsite'); if(mysqli_connect_errno()) { die("MySQL connection failed: ". mysqli_connect_error()); } // Gather all required data $JobID = $dbLink->real_escape_string($_POST['JobID']['JobID']); $CompanyID = $dbLink->real_escape_string($_POST['CompanyID']['CompanyID']); $UsersID = $dbLink->real_escape_string($_POST['UsersID']['UsersID']); $name = $dbLink->real_escape_string($_FILES['uploaded_file']['name']); $mime = $dbLink->real_escape_string($_FILES['uploaded_file']['type']); $data = $dbLink->real_escape_string(file_get_contents($_FILES ['uploaded_file']['tmp_name'])); $size = intval($_FILES['uploaded_file']['size']); // Create the SQL query $query = " INSERT INTO `application` ( `JobID`, `CompanyID`, `UsersID`, `name`, `mime`, `size`, `data`, `N_DateAndTime` ) VALUES ( '{$JobID}', '{$CompanyID}', '{$UsersID}', '{$name}', '{$mime}', {$size}, '{$data}', NOW() )"; // Execute the query $result = $dbLink->query($query); // Check if it was successfull if($result) { echo 'Success! Your job application was successfully sent!'; } else { echo 'Error! Failed to insert the file, please try again' . "<pre>{$dbLink->error}</pre>"; } } else { echo 'An error accured while the file was being uploaded, please try again. ' . 'Error code: '. intval($_FILES['uploaded_file']['error']); } // Close the mysql connection $dbLink->close(); } else { echo 'Error! A file was not sent! Please try again!'; } // Echo a link back to the main page echo '<p>Click <a href="year1index.php">here</a> to go back</p>'; ?> MOD EDIT: [code] . . . [/code] BBCode tags added. I have the following mysql query: Code: [Select] $lead_query=$this->db->query(" SELECT `first_name`, `last_name` , `state` FROM leads WHERE `lead_id`='$lead_id' "); $this->view->lead_query=$lead_query->fetchALL(); Now when I retrieve the above details I need to return the first_name last_name together(separated as space) and the name of the key as client_name in the array. I need it that way because when i return a json_encode($lead_query), I want to return the first_name.last_name as client_name. How should it look if I want to make a query where a field value should be lower than another field? Code: [Select] ....WHERE total_points<max_points"); Hi I have a query where it returns a few fields based on the location. I created a class for the function and an index page. It does not return any values. Can someone please advise/ THE CLASS Code: [Select] <?php /**************************************** * * WIP Progress Class * * ****************************************/ class CHWIPProgress { var $conn; // Constructor, connect to the database public function __construct() { require_once "/var/www/reporting/settings.php"; define("DAY", 86400); if(!$this->conn = mysql_connect(DB_HOST, DB_USERNAME, DB_PASSWORD)) die(mysql_error()); if(!mysql_select_db(DB_DATABASE_NAME, $this->conn)) die(mysql_error()); } public function ListWIPOnLocation($location) { $sql = "SELECT `ProgressPoint.PPDescription` AS Description ,`Bundle.WorksOrder` AS WorksOrder, `Bundle.BundleNumber` AS Number, `Bundle.BundleReference` AS Reference,`TWOrder.DueDate` AS Duedate FROM `TWOrder`,`Bundle`,`ProgressPoint` WHERE `Bundle.CurrentProgressPoint`=`ProgressPoint.PPNumber` AND `TWOrder.Colour=Bundle.Colour` AND `TWOrder.Size=Bundle.Size` AND `TWOrder.WorksOrderNumber`=`Bundle.WorksOrder` AND `ProgressPoint.PPDescription` LIKE '" . $location . "%' ORDER BY TWOrder.DueDate DESC"; mysql_select_db(DB_DATABASE_NAME, $this->conn); $result = mysql_query($sql, $this->conn); echo $sql; while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { $return[] = $row; } return $return; } } ?> The index page Code: [Select] <?php // First of all initialise the user and check for permissions require_once "/var/www/users/user.php"; $user = new CHUser(7); // Initialise the template require_once "/var/www/template/template.php"; $template = new CHTemplate(); // And create a cid object require_once "/var/www/WIPProgress/DisplayWIPOnLocation.php"; $WIPProgress= new CHWIPProgress(); $content = "Check WIP Status on Location <br>"; $content = "<form action='index.php' method='get' name ='location'> <select id='location' > <option>Skin Room</option> <option>Clicking</option> <option>Kettering</option> <option>Closing</option> <option>Rushden</option> <option>Assembly</option> <option>Lasting</option> <option>Making</option> <option>Finishing</option> <option>Shoe Room</option> </select> <input type='submit' /> </form>"; $wip = $WIPProgress->ListWIPOnLocation($_GET['location']); // Now show the details $content .= "<h2>Detail</h2> <table> <tr> <th>PPDescription</th> <th>Works Order</th> <th>Bundle Number</th> <th>Bundle Reference</th> <th>Due Date</th> </tr>"; foreach($wip as $x) { $content .= "<tr> <td>" . $x['Description'] . "</td> <td>" . $x['WorksOrder'] . "</td> <td>" . $x['Number'] . "</td> <td>" . $x['Reference'] . "</td> <td>" . $x['DueDate'] . "</td> </tr>"; } $template->SetTag("content", $content); echo $template->Display(); ?> thank you Hi guys, im trying to connect to a database and get the value for the user in the row called 'user_credit', if it equals 1 or more then i want to show the ''You have £ ....'' bit in the script. Problem is nothing shows at all, even without the if statement. I have changed the value for me in the database so in user_credit the value is 100, which is more than 1 so it should appear. I have probably done something wrong. Any ideas? Code: [Select] <? include '../admin/database/membership_dbc.php'; $r = mysql_query("SELECT * FROM users WHERE user_name='".safe($_SESSION['user_name'])."'") or die ("Cannot find table"); while( $cred = mysql_fetch_array($r) ) { if ($cred >= '1' ) { ?> <p>You have £<? echo $cred['user_credit']; ?> available on you account, would you like to use it on this order?<br> <label for="credit"></label> <select name="credit" id="credit"> <option value="Y" selected>Yes, use credit</option> <option value="N">No, save credit</option> </select> </p> <? } } ?> Given the following query,
SELECT t1.a, t1.b, t2.c FROM t1 INNER JOIN t2 ON t2.id=t1.t2_id WHERE t1.pk=123;I get the following three records:
array(
array('a'=>1,'b'=>2,'c'=>4),
array('a'=>1,'b'=>2,'c'=>5),
array('a'=>1,'b'=>2,'c'=>8)
)
What would be the best way to get just one record such as the following?
array('a'=>1,'b'=>2, 'c'=>array(4,5,8))
My thoughts were to use MySQL's GROUP_CONCAT, and then use implode() to turn it into an array, but didn't know if there was a better way.
Thanks
Hi all
we have an ajax query which queries a MySQL database for cities around the world for our hotel search. We are running on a Linux web hosted server.
The problem we are having is when you first visit out site and type in Brisbane or Sydney it takes around 15 seconds maybe to return. After that what ever city you query is lightening fast. I have had a few friends try this and when they first visit our site they experience the same then second time around it is fast.
Could this be some sort of caching issue or setting somewhere that I am missing?
I have attached a file so you can see what I mean
Thanks in advance
Attached Files
ajaxqry.jpg 169.31KB
0 downloads hi there alll newbie here lol just needed help with the code for the next page of results ? IE i have say 30 results returned from a sql query but only want to show ten at a time i no how to limit the result with the LIMIT 10 but how do i get it to put link show the other ten and so on. the code i have is here at the bottom of page http://www.phpfreaks.com/forums/index.php/topic,307971.0.html thanks in advance for any help kaine Hi i have this edit form that allows user to mofy data but the problems on the text box is that it deletes the rest of the data after the space from the first word i tried to increase the size of the varChars on mysql but did no work why it happens how can i stop from happening?? this the form input <input type="text" name="name" id="name" class='text_box' value="<?php echo $_GET['name'];?>"/> Given the below 3 database tables I am trying to construct a SQL query that will give me the following result: customer_favourites.cust_id customer_favourites.prod_id OR product.id product.code product.product_name product.hidden product_ section.section_id (MUST BE ONLY THE ROW WITH THE LOWEST SECTION ID, I.E. ROW ID #44108) product_ section.catpage (MUST BE ONLY THE ROW WITH THE LOWEST SECTION ID, I.E. ROW ID #44108) product_ section.relative_order (MUST BE ONLY THE ROW WITH THE LOWEST SECTION ID, I.E. ROW ID #44108) I currently have.... SELECT customer_favourites.cust_id, customer_favourites.prod_id, product.code, product.product_name, product.hidden, product_section.section_id, product_section.relative_order, product_section.catpage FROM customer _favourites INNER JOIN product ON customer_favourites.prod_id = product.id INNER JOIN product_section ON product_section.product_code = product.code WHERE `cust_id` = '17' AND `hidden` = '0' GROUP BY `code` ORDER BY `section_id` ASC, `relative_order` ASC, `catpage` ASC LIMIT 0,30 This gives me what I want but only sometimes, at other times it randomly selects any row from the product_section table. I was hoping that by having the row I want as the last row (most recent added) in the product_section table then it would select that row by default but it is not consistent. Somehow, I need to be able to specify which row to return in the product_section table, it needs to be the row with the lowest section_id value or it should by the last row (most recent). Pulling my hair out so any help is gratefully received. customer_favourites id cust_id prod_id 70 4 469 product id code product_name hidden 469 ABC123 My Product 0 product_section id section_id catpage product_code relative_order recommended 44105 19 232 ABC123 260 1 44106 3 125 ABC123 87 1 44107 2 98 ABC123 128 1 44108 1 156 ABC123 58 0 hello in the attached code, how do i add another field to the query? ie, i have all the code to create the result and would like to add further values such as $dept? what is the correct way to code query? i must stress that i am using php 4.4.7 so json_encode is out. the code is working but just need to find a way to add the $dept to the qeury? many thanks $dept = array(); $box = array(); while ($row = mysql_fetch_array($result)) { $dept[] = $row['department']; $box[] = $row['custref']; } /*$items = rtrim($_POST['items'],","); $sql = "UPDATE `boxes` SET status = 'Out' WHERE Id IN ($items)"; $result = runSQL($sql);*/ $total = count(explode(",",$items)); $result = runSQL($sql); $total = mysql_affected_rows(); /// Line 18/19 commented for demo purposes. The MySQL query is not executed in this case. When line 18 and 19 are uncommented, the MySQL query will be executed. header("Expires: Mon, 26 Jul 1997 05:00:00 GMT" ); header("Last-Modified: " . gmdate( "D, d M Y H:i:s" ) . "GMT" ); header("Cache-Control: no-cache, must-revalidate" ); header("Pragma: no-cache" ); header("Content-type: text/x-json"); $json = ""; $json .= "{\n"; $json .= "box: [\"". implode('","', $box) ."\"]\n"; $json .= "}\n"; echo $json; $sql = "INSERT INTO `act` (`item`) VALUES ('". implode("'),('", $box) . "')"; $result = runSQL($sql); Hi there. I'm totally new (about a week!) with php and mysql and am encountering a problem that perhaps someone can help me with?
I've looked through to see if a similar problem has appeared or been solved, but without success, so apologies if I am repeating something.
In php I am trying to update 7 fields from a form from which a user has edited/modified any of the fields in a chosen record (except id).
Here is the code:
$id=$_GET['id']; Hi, I'm trying to set up a page which first queries for mySQL record results matching a country that the user selects. This works fine, but in the event of there being no records for that country, I want it to look at another field, "Region" and pick the records matching that Region instead. For example, a user searches for "Australia" but there are no matching records. So, I want it to pick all the records for the region of Australasia, and display records for Australia, New Zealand, Papua New Guinea and so on. I had created the following: $query = "SELECT * FROM specialists WHERE Country LIKE '$country' ORDER BY SpecialistName"; // specify the table and field names for the SQL query //$query .= " limit $s,$limit"; $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); if ( $numrows == 0 ) { echo '<p>We don\'t have any results for specialists in countries near to yours at the moment. Please try <a href="specialists.php" style="text-decoration:underline;">searching a different country</a></p>'; } // get results $result = mysql_query($query) or die("Couldn't execute query"); // display the results returned while ($row= mysql_fetch_array($result)) { $region = $row["Region"]; $count++ ; } // next determine if s has been passed to script, if not use 0 if (empty($s)) { $s=0; } echo '<table width="600" class="cardisplay"><tr>'; $count = 1 + $s ; echo $region; // Build SQL Query $query2 = "SELECT * FROM specialists WHERE Region LIKE '$region' ORDER BY SpecialistName"; // specify the table and field names for the SQL query //$query .= " limit $s,$limit"; $numresults=mysql_query($query2); $numrows=mysql_num_rows($numresults); // get results $result = mysql_query($query2) or die("Couldn't execute query"); // display the results returned while ($row= mysql_fetch_array($result)) { $title1 = $row["ID"]; $specialistname = $row["SpecialistName"]; $address1 = $row["Address1"]; $address2 = $row["Address2"]; $address3 = $row["Address3"]; $address4 = $row["Address4"]; $address5 = $row["Address5"]; $postcode = $row["Postcode"]; $country = $row["Country"]; $region = $row["Region"]; $website = $row["Website"]; $email = $row["Email"]; $telephone = $row["Telephone"]; $businesstype = $row["BusinessType"]; //followed by echoing out the various data etc. etc. But the problem is that $region is always blank / empty in the cases where $query is empty, so I can't pull the value out and therefore $query2 is also empty... Any ideas? Hi, I am new to the boards and php and mysql. I have created a database and can add entries via a form. I can query the database with another form and get the results to display in a table. All good so far as that is what I was hoping to achieve but one of the fields I want to display as a hyperlink but I am having problems with the syntax as I keep getting errors when I try to wrap the variables in <a href > tag. Code: [Select] <tr> <td><?php echo $fquery['prefix']; echo $fquery['website']; ?><?php echo $fquery['website']; ?></td> </tr> now that displays e.g. http://www.example.comwww.example.com. so I feel I am nearly there I just need advice as to how to construct a hyperlink from the fields queried. I created a drop-down menu using a MySQL statement in php for a form. My drop down menu works fine, but I want to assign a default value to it (normal text) the value will never change, thus I do not need to extract the default value from the table, I already know the value. Here is the basic snippet from the script: <?php include("../includes/xxx.php"); $cxn = mysqli_connect($host,$user,$password,$dbname); $query = "SELECT DISTINCT `plant_id` FROM `plant` ORDER BY `plant_id`"; $result = mysqli_query($cxn,$query); while($row = mysqli_fetch_assoc($result)) { extract($row); echo "<option value='$plant_id'>$plant_id</option>\n"; } ?> I have a form that allows my client to update some products. Now the products are simple just basic info and 1 picture. I have set this up so they can edit the products and change the information, having done this many times in the past, but now hit a puzzling block that I am baffled. The client when editing is presented with the form with the information pulled from the database and the form fields loaded with that data ready to edit. The image can either be left alone or they can choose to upload a new image. They are shown the image they currently have stored in the database. The problem I have is EVEN if they decide not to upload an image and change other information, when the submit the form it must be sending a blank value for the image somewhere as it is updating the database and removing the image reference as if it has been removed. I have an if/else statement based on the form to perform 2 different queries for the update in mysql. Here is the code for the form update, as you can see the image should not update?? Please help?? if ($_SERVER['REQUEST_METHOD'] =='POST') { //This stops SQL Injection in POST vars foreach ($_POST as $key => $value) { $_POST[$key] = mysql_real_escape_string($value); } // **************************** THIS IS FOR NO NEW IMAGE ******************************** if ($_SERVER['REQUEST_METHOD'] =='POST' && empty($_FILES['product_image']['name'])) { # setup SQL statement for no new image $SQL = " UPDATE products SET product_title = '{$_POST['product_title']}', product_description = '{$_POST['product_description']}', standard_price = '{$_POST['standard_price']}', deluxe_price = '{$_POST['deluxe_price']}' WHERE product_id = '{$_REQUEST['product_id']}' "; } // **************************** THIS IS FOR A NEW IMAGE ******************************** else { // Check the image type is a jpeg or gif for the image. if (($_FILES['product_image']['type'] != "image/gif") && ($_FILES['product_image']['type'] != "image/jpeg") && ($_FILES['product_image']['type'] != "image/pjpeg")) { echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\">You have chosen not to upload a <b>Product Image</b>.<BR></SPAN>" ; } elseif ($_FILES['product_image']['size'] > 100000) { echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\">The file size is bigger than 300kb.<BR></SPAN>" ; } else { move_uploaded_file($_FILES['product_image']['tmp_name'], "/httpdocs/product_images/".$_FILES['product_image']['name']) ; echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\"><B>Your front image has successfully uploaded.</B><BR></SPAN>" ; } } # setup SQL statement for update $SQL = " UPDATE products SET product_title = '{$_POST['product_title']}', product_description = '{$_POST['product_description']}', standard_price = '{$_POST['standard_price']}', deluxe_price = '{$_POST['deluxe_price']}', product_image = '{$_FILES['product_image']['name']}' WHERE product_id = '{$_REQUEST['product_id']}' "; } #execute SQL statement $result = mysql_db_query( *****,"$SQL",$connection ); # check for error if (!$result) { echo("ERROR: " . mysql_error() . "\n$SQL\n"); } I have a MySQL table with a list of albums and there is a field called "views" with the number of views each album has received. I'm looking to generate an array of all the albums in the table and sort the array by the number of views (descending). I have a list of functions defined in a ContentController.php file. I created a new function called build_albumlist, which I've pasted below. The function "get_ip_log" already exists and works, and I used it as a template to create the "build_albumlist" function: public function build_albumlist(){ return $this->select_raw("SELECT * FROM albums WHERE deleted = '0' ORDER BY views DESC",array(),'all'); } public function get_ip_log(){ return $this->select_raw("SELECT * FROM sessions ORDER BY ID DESC",array(),'all'); } When I use the function, I get this warning: Warning: mysql_real_escape_string() expects parameter 1 to be string, array given inC:\xampp\htdocs\Controllers\DBController.php on line 10 [font=cabin, 'trebuchet ms', helvetica, arial, sans-serif]The "select_raw" function that I used in "build_albumlist" is defined in the DBController.php file, and is defined as below:[/font] private function clean_array($params){ $final=array(); foreach($params as $param){ $final[]=mysql_real_escape_string($param); } return $final; } public function select_raw($query,$params,$type=''){ $query=str_replace("?","'%s'",$query); $final_query= call_user_func_array('sprintf', array_merge((array)$query, $this->clean_array($params))); if($type==''){ $result=mysql_query($final_query) or die(mysql_error()); return mysql_fetch_assoc($result); } elseif($type=='all'){ $result=mysql_query($final_query) or die(mysql_error()); $final=array(); while($row=mysql_fetch_assoc($result)){ $final[]=$row; } return $final; } Does anyone know why the "build_albumlist" function is generating this warning, while the "get_ip_log" is not? Any help would be great, as I am obviously pretty new to this. Hi all, I have the following MySQL insert query: Code: [Select] $insert= mysql_query ("INSERT INTO tablename (column1,`".$EXPfields."`) VALUES ('$something','".$EXPvalues."')"); where $EXPfields is an array of table-field-names and $EXPvalues is an array of table-field-values. Now I want to write an equivalent query, but using UPDATE instead of INSERT INTO, but I don't want to write out all the field names/values separately, but again want to use $EXPfields and $EXPvalues. So something like this: Code: [Select] $update = mysql_query ("UPDATE tablename SET (column1,`".$EXPfields."`) = ('$something','".$EXPvalues."') WHERE .... "); Is this possible? If so, what is the proper syntax? Thanks! This is my first real jump into PHP, I created a small script a few years ago but have not touched it since (or any other programming for that matter), so I'm not sure how to start this. I need a script that I can run once a day through cron and take the date from one table/filed and insert it into a different table/field, converting the human readable date to a Unix date. Table Name: Ads Field: endtime_value (human readable date) to Table Name: Node Field: auto_expire (Converted to Unix time) Both use a field named "nid" as the key field, so the fields should match each nid field from one table to the next. Following a tutorial I have been able to insert into a field certain data, but I don't know how to do it so the nid's match and how to convert the human readable date to Unix time. Thanks in advance!. 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> |
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