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PHP - String Problems When Trying To Insert To Sql!
Hi all,
I'm having a hard time with strings... I am pulling data from an XML feed, and then trying to insert into my database. All the code is wrote, the problem is that some of the descriptions contain both ' and " so it is messing with my insert statement. I can get over the ' by doing this: \"$newstring\" however it then fails on " I have tried every function I can think of: str_replace, mysql_real_escape_string, htmlspecialchars, addslashes, stripslashes etc. etc. nothing seems to do the trick! I have attached the code in full, would someone be able to take a look and help me out? Any help would be greatly appreciated! many thanks Greens85 Similar TutorialsHi everyone this time I want to do something everyone here knows it is possible: when posting a message at a forum you can press buttons and text appears at your pointer. Now that is not exactly what I want to do. at some forum sites (not phpfreaks) when pressing a button while posting you get a popup with buttons which make things appear in the not popup-window. that is exactly what I need to do. This is not about posting messages, btw, but a list of contacts of which you can select 2 in a popup. any ideas? any help is appreciated I am trying to add a value, input into a form, to a MySQL database. However, something must be wrong with the casting, because if there is a space in the form value, then I get an error, as in: //$_POST['string'] == '1blah 2blah'; sql = "INSERT INTO table (some_string) VALUES ($_POST[string])"; $sql_result = mysql_query($sql) or die ('The error is as follows: ' . mysql_error() . '<br /><br />Value could not be added.'); Then I get the following error: The error is as follows: 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 '2blah' at line 1 I've entered paragraphs into a database before, so this error is now to me. The column 'some_value' is a type: varchar(50). Hi,
I currently have some html stored in a variable as below:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
Hello world
</body>
</html>
This requires a "<div class='container'>" to be inserted directly after the opening body tag. My original thinking was to use str_replace on the <body> to be <body><div class="container"> however note that occasionally my body will require an ID or Class. I therefore need a way to make it appear after the body tag, even if it has an ID or Class on it :-\ . I assume this may be reg-ex, but am not sure if there is a simple way around this? Would anyone be able to clarify and point me in the right direction.
Much appreciated.
MoFish
Hello all, been trying for over a day to get something to work and after searching and searching I could not find what I was looking for. I found people looking to do something similar, but the answers seemed more complicated than I would think they need to be. First off, I'm still horrible at arrays and loops so try not to laugh too hard if my code is waaaayyyy off...haha I am essentially trying to tie products and quantities to a reservation. The user has the ability to add items to the form. I'm using jquery .clone to create each new item product and quantity form fields. When the user submits the form jquery inputs the values of the cloned fields into 2 hidden text fields called "productsField" and "quantitiesField" The values of these fields look something like value="1,2,2,1,3" depending on how many products were added. What I want to do is explode both and then enter them into the database. Seems simple enough, but apparently I am not getting it. I thought something like this would work, but it is not. if (isset($_POST["reserve_button"])) { $theReservationID = $_SESSION['createdReservationID']; $items = explode (",",$_POST['productsField']); $quantities = explode (",",$_POST['quantitiesField']); // loop through array $number = count($items); for ($i=0; $i<=$number; $i++) { // store a single item number and quantity in local variables $itno = $items[$i]; $quant = $quantities[$i]; if ($items[$i] <> '') { mysql_query('INSERT INTO reservation_items (reservationID,productID,productQuantity) VALUES($theReservationID,$itno,$quant)'); } } } Any help would be greatly appreciated. Thanks in advance, Twitch Hi guys, I have a very simple add.php to add data to a mySQL db. I have a menu/list drop down as one of my fields on my form and this shows an array of results from another table (ranks of the RAF) within my db. When I click the save button I have it process a INSERT INTO command but all i get inputted into my staff table is the first word... eg if I chose "Pilot Officer" from the list menu and then click save all that would appear in my db is "Pilot". Any clues? I will paste the php below... Code: [Select] <?php include('config.php'); ?> <form action='' method='POST' enctype='multipart/form-data'> <p><b>Rank:</b><br /> <select name="rank" id="rank"> <option selected>Please Select</option> <?php $query = "SELECT * FROM ranks ORDER BY rank ASC"; $result = mysql_query($query); while($row = mysql_fetch_array($result)) { echo "<option value=". $row["rank"] .">". $row["rank"] ."</option>"; } ?> </select> <p><b>Forename:</b><br /> <input name="forename" type="text" id="forename" value="" size="40"> <p><b>Surname:</b><br /><input name='surname' type='text' id="surname" value='' size="40" /> <p><b>Category:</b><br /> <select name="category" id="category"> <option selected>Please Select</option> <?php $query = "SELECT * FROM categories"; $result = mysql_query($query); while($row = mysql_fetch_array($result)) { echo "<option value=". $row["category"] .">". $row["category"] ."</option>"; } ?> </select> <p><b>Email:</b><br /><input name='email' type='text' id="email" value='' size="50" /> <p><b>Mobile:</b><br /> <input name='mobile' type='text' id="mobile" value='' size="40" /> </p> <input type='submit' value='Save' /> <input type='hidden' value='1' name='submitted' /> </form> <?php if (isset($_POST['submitted'])) { $rank = mysql_real_escape_string($_POST['rank']); $forename = mysql_real_escape_string($_POST['forename']); $surname = mysql_real_escape_string($_POST['surname']); $category = mysql_real_escape_string($_POST['category']); $email = mysql_real_escape_string($_POST['email']); $mobile = mysql_real_escape_string($_POST['mobile']); $sql = "INSERT INTO `staff` (`rank` , `forename` , `surname` , `category` , `email` , `mobile` ) VALUES ( '$rank' , '$forename' , '$surname' , '$category' , '$email' , '$mobile')"; mysql_query($sql) or die(mysql_error()); echo (mysql_affected_rows()) ? "Staff Added":"Nothing Added"; } ?> Hi to everybody I need ur help because I’m trying to make a script in php to write the same data but with different date in MySQL depending on the splitting ($data06).... like a schedule...
For example if $data06 = annual and $data03 = “2019/01/01” programm must create in MySQL : 2019/01/01 2020/01/01 2021/01/01 2022/01/01 2023/01/01
if $data06=“half year” will create 10 date , increasing 6 moths ...
But I have a problem at finish of code switch ($data06) { case 'annual': $numrate = 5; $aumdata = "+12 months"; break; case 'half year': $numrate = 10; $aumdata = "+6 month"; break;
default: echo "error"; $sql = "INSERT into $nometb ( name, scadenza ) values ( '$data01', '$data03' )"; if ($conna->query($sql) === TRUE) {} else {die('ERROR'. $conna->error); }
//IMPORT NEXT AND NEW DATE $newDate = date_create($data03); for ($mul = 2; $mul <= $numrate; ++$mul) {
$datanuova = date_create($data03); $datanuova->modify($aumdata); $datanuova->format('yy/m/d'); $newDate = $datanuova->format('yy/m/d');
$sql = "INSERT into $nometb ( name, scadenza ) values ( '$data01', '$newDate' )"; if ($conna->query($sql) === TRUE) {} else {die('ERROR'. $conna->error); }
//HERE THERE IS ERROR $data03 = $newDate;
if ($conna->query($sql) === TRUE) {} else {die('ERRORE NELL\'IMPORTAZIONE'. $conna->error); } }
} }
}
How I can resolve ?
many thanks Francesco Hi everyone, I'm new to this group and new to php. I have created a multi-part form that allows the user the option to add multiple input fields to a form to upload images. Here is the form structu Code: [Select] <form action="Scripts/processreports2.php" method="post" enctype="multipart/form-data" name="report_form" target="uploader" class="reportfrm"> <fieldset> <legend>Upload your images</legend> <ol id="add_images"> <li> <input type="file" class="input" name="files[]" /> </li> <li> <input type="file" class="input" name="files[]" /> </li> <li> <input type="file" class="input" name="files[]" /> </li> </ol> <input type="button" name="addFile" id="addFile" value="Add Another Image" onclick="window.addFile(this);"/> </fieldset> <p>* indicates a required field.</p> <input name="submit" type="submit" id="submit" value="Send Info!" /> </form> Through php a maximum of three input fields are fed into an array that checks to make sure that the uploaded files are images and not some other type of file. The uploading porition of this script works. Now I am trying to get the values of the input fields as a string and insert them into the database as one record. Let's say the user has three files they want to upload. I have managed to get the files as a string ie; file1.jpg, file2.jpg, file3.jpg but what is happening is that I am getting three separate records with file1.jpg, file2.jpg, file3.jpg in them. If the user has only two files to upload then I get two separate records with file1.jpg and file2.jpg in them. (Hope that makes sense). I want one record. I have been struggling with this since Monnday and while every day I get closer, this is as close as I can get. Below is the php code. #connect to the database mysql_connect("localhost", "root", ""); mysql_select_db("masscic"); //Upload Handler to check image types function is_image($file) { $file_types = array('jpeg', 'gif', 'bmp'); //acceptable file types if ($img = getimagesize($file)){ //echo '<pre>'; //print_r($_FILES); used for testing //print_r($img); used for testing if(in_array(str_replace('image/', '', $img['mime']), $file_types)) return $img; } return false; } //form submission handling if(isset($_POST['submit'])) { //file variables $fname = $_FILES['files']['name']; $ftype = $_FILES['files']['type']; $fsize = $_FILES['files']['size']; $tname = $_FILES['files']['tmp_name']; $ferror = $_FILES['files']['error']; $newDir = '../uploads/'; //relative to where this script file resides for($i = 0; $i < count($fname); $i++) { //echo 'File name ' . $fname[$i] . ' has size ' . $fsize[$i]; used for testing if ($ferror[$i] =='UPLOAD ERR OK' || $ferror[$i] ==0) { if(is_image($tname[$i])) { //append the tmp_name($tname) to the file name ($fname) and upload to the server move_uploaded_file($tname[$i], ($newDir.time().$fname[$i])); echo '<li><span class="success">'.$fname[$i].' -- image has been accepted<br></span></li>'; }else echo '<li><span class="error">'.$fname[$i].' -- is not an accepted file type<br></span></li>'; } if (is_array($fname)) $files = implode(', ',$fname); //else $files = $fname; $sqlInsert = mysql_query("INSERT INTO files (file_names) VALUES('$files')") or die (mysql_error()); } } Hi People. Thanks for all the help with my ongoing project. Please can you tell me what my problem is with this error? Error During Insert : 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 ') VALUES ('test', 'tesast', 'tesat','tear', 'sest', 'none whatsoever', '', 'ter'' at line 1 Error occured running the following code : INSERT INTO HelperFormData (surname, firstname, phone1, phone2, location, qualifications, expertise, assistance, languages, e_mail, consent, published, comments, other_info,) VALUES ('test', 'tesast', 'tesat','tear', 'sest', 'none whatsoever', '', 'ter', 'my language', 'vince@vince.com', 'No', 'No', 'ateaglea', 'ttaelgae') Here is the code to my insert form Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Stranded Flyer</title> </head> <body> <?php $host = 'localhost'; $usr = "username"; $password = 'password'; $db_name = 'stranded'; //connect to database mysql_connect ("$host","$usr","$password") or die ('Error During Connect:<br>'.mysql_error()); mysql_select_db ("$db_name") or die ('Error Selecting DB:<br>'.mysql_error()); $surname = mysql_real_escape_string($_POST['surname']); $firstname = mysql_real_escape_string($_POST['firstname']); $phone1 = mysql_real_escape_string($_POST['phone1']); $phone2 = mysql_real_escape_string($_POST['phone2']); $location = mysql_real_escape_string($_POST['location']); $qualifications = mysql_real_escape_string($_POST['qualifications']); $expertise = mysql_real_escape_string($_POST['expertise']); $assistance = mysql_real_escape_string($_POST['assistance']); $languages = mysql_real_escape_string($_POST['languages']); $e_mail = mysql_real_escape_string($_POST['e_mail']); $consent = mysql_real_escape_string($_POST['consent']); $published = mysql_real_escape_string($_POST['published']); $comments = mysql_real_escape_string($_POST['comments']); $other_info = mysql_real_escape_string($_POST['other_info']); $errorstring = ""; // default value of errorstring function makeSafe($value) { if (get_magic_quotes_gpc()) { $value = stripslashes($value); } $value = mysql_real_escape_string($value); return $value; } if(isset($_POST['submit'])) { // Validate all the code inputs // Captcha Validation require_once('recaptchalib.php'); $privatekey = "MyPrivateKeyRemovedForTheForumPost"; $resp = recaptcha_check_answer ($privatekey, $_SERVER["REMOTE_ADDR"], $_POST["recaptcha_challenge_field"], $_POST["recaptcha_response_field"]); if (!$resp->is_valid) { // What happens when the CAPTCHA was entered incorrectly $errorstring = $errorstring. "Invalid CAPTCHA, please try again"; } else { if ($surname =="") $errorstring = $errorstring. "Surname<br>"; if ($phone1 =="") $errorstring = $errorstring. "phone1<br>"; if ($phone2 =="") $errorstring = $errorstring. "phone2<br>"; if ($location =="") $errorstring = $errorstring. "location<br>"; if ($qualifications =="") $errorstring = $errorstring. "qualifications<br>"; if ($assistance =="") $errorstring = $errorstring. "assistance<br>"; if ($languages =="") $errorstring = $errorstring. "languages<br>"; if ($e_mail =="") $errorstring = $errorstring. "e_mail<br>"; if ($consent =="") $errorstring = $errorstring. "consent<br>"; if ($published =="") $errorstring = $errorstring. "published<br>"; if ($comments =="") $errorstring = $errorstring. "comments<br>"; if ($other_info =="") $errorstring = $errorstring. "other info<br>"; // does the errorstring = "nothing"? } // Figure out which error message to show i.e. field validation or CAPTCHA if ($errorstring !="") if (strstr($errorstring,"CAPTCHA")) echo $errorstring; else echo "You have not put anything in the following fields: <br><br> $errorstring"; //echo "If you have nothing to put in the box please type the word \"None\" or \"N\/A\""; //die ("Please try again, ensuring that you fill out all the fields!"); else { //echo "Your data has been saved"; $insert_query = "INSERT INTO HelperFormData (surname, firstname, phone1, phone2, location, qualifications, expertise, assistance, languages, e_mail, consent, published, comments, other_info,) VALUES ('$surname', '$firstname', '$phone1','$phone2', '$location', '$qualifications', '$expertise', '$assistance', '$languages', '$e_mail', '$consent', '$published', '$comments', '$other_info')"; $insert_action = mysql_query($insert_query) or die ('Error During Insert :<br>'.mysql_error().'<br><br>Error occured running the following code :<br>'.$insert_query); $id = mysql_insert_id(); echo "Thank you, Your details have been submitted."; //include "resultcard.php"; // Output what the form looks like // End of how the form looks } } if(!isset($_POST['submit']) || (isset($_POST['submit']) && !empty($errorstring))) { ?> <form name = "form1" method ="post" action=""> <table width="700" border="0" cellspacing="5" cellpadding="5" bgcolor = "#dddddd"> <caption> Submit Your Details </caption> <tr> <td width = "50"> </td> <td width = "240"> </td> <td width = "250"> </td> <td width = "160"><b>Example Input</b></td> </tr> <tr> <td> </td> <td>Surname</td> <td><input type='text' name='surname' size = '40' maxlength='30' value = '<?php echo $surname; ?>'></td> <td>Smith</td> </tr> <tr> <td> </td> <td>First Name</td> <td><input type='text' name='firstname' size = '40' maxlength='30' value = '<?php echo $firstname; ?>'></td> <td>Bill</td> </tr> <tr> <td> </td> <td>Phone 1</td> <td><input type='text' name='phone1' size = '40' maxlength='30' value = '<?php echo $phone1; ?>'></td> <td>07777 777777</td> </tr> <tr> <td> </td> <td>Phone 2</td> <td><input type='text' name='phone2' size = '40' maxlength= '30'value = '<?php echo $phone2; ?>'></td> <td>0118 123 4567</td> </tr> <tr> <td> </td> <td>Location</td> <td><input type='text' name='location' size = '40' maxlength='40'value = '<?php echo $location; ?>'></td> <td>52.289071,-1.952004</td> </tr> <tr> <td> </td> <td>Qualifications</td> <td><input type='text' name='qualifications' size = '40' maxlength='30' value = '<?php echo $qualifications; ?>'></td> <td>Electrician</td> </tr> <tr> <td> </td> <td>Expertise</td> <td><input type='text' name='qualifications' size = '40' maxlength='30' value = '<?php echo $expertise; ?>'></td> <td>Flexwings</td> </tr> <tr> <td> </td> <td>Assistance Offered</td> <td><input type='text' name='assistance' size = '40' maxlength='50' value = '<?php echo $assistance; ?>' /></td> <td>Trailer and Tow Car</td> </tr> <tr> <td> </td> <td>Language Skills</td> <td><input type='text' name='languages' size = '40' maxlength='50' value = '<?php echo $languages; ?>' /></td> <td><p>English / Spanish</p></td> </tr> <tr> <td> </td> <td>e_mail</td> <td><input type='text' name='e_mail' size = '40' maxlength='50' value = '<?php echo $e_mail; ?>' /></td> <td><p>example@mail.net</p></td> </tr> <tr> <td> </td> <td>Consent</td> <td> <select name = "consent"> <option value = "Yes" <?php if ($_POST['consent'] == 'Yes') { echo 'selected="selected"'; } ?>>Yes</option> <option value = "No" <?php if ($_POST['consent'] == 'No') { echo 'selected="selected"'; } ?>>No</option> </select> </td> <td>Yes</td> </tr> <tr> <td> </td> <td>Published</td> <td> <select name = "published"> <option value = "Yes" <?php if ($_POST['published'] == 'Yes') { echo 'selected="selected"'; } ?>>Yes</option> <option value = "No" <?php if ($_POST['published'] == 'No') { echo 'selected="selected"'; } ?>>No</option> </select> </td> <td>Yes</td> </tr> <tr> <td> </td> <td>Comments</td> <td><textarea name= "comments" input type = 'text' rows = "5" cols = "29" /><?php echo $comments; ?></textarea></td> <td>As much info as possible 500 characters max</td> </tr> <tr> <td> </td> <td>Other Info</td> <td><textarea name= "other_info" input="input" type = 'text' rows = "5" cols = "29" /><?php echo $other_info; ?></textarea></td> <td>As much info as possible 500 characters max</td> </tr> <tr> <td> </td> <td><input type='submit' name='submit' value='Submit Your Info' /></td> <td colspan="2"> <? require_once('recaptchalib.php'); $publickey = "MyPublicKeyRemovedFromTheForumPost"; // you got this from the signup page echo recaptcha_get_html($publickey); ?></td> </tr> </table> </form> </body> </html> <?php } ?> I am having a problem inserting information into my database. It will not create a record, no matter what I do. I have used the exact same template for two other pages, and it works fine, but for this one, it simply will not work. If I echo each variable, it will print out, but even if I reduce the insert to one field, it still will not work. Please help... Here is my code.
<?php
<html>
<?php
$ID = $_POST['ID'];
if ($conn->query($sql) == TRUE) {
</body> It seems like every new language I learn, I get this exact same issue. I want to make a clickable link that will go to "viewpost?postid=" and then add the value of a variable. The variable is an array. Here's what I'm trying. I'm not really sure why it doesn't work: <?php $poststr = "<a href=\'viewpost.php?"+$postid[$num]+"\'>View This Post</a>"; echo $poststr; ?> $postid[$num] in this example should be an integer. But what it outputs is really weird. It's not "View This Post" as a link, which is what I want. It just outputs the value of the variable $postid[$num] with no link. Can someone tell me what my problem is? Hi all, I use a MySQL insert query of the following form: Code: [Select] <?php $insert1= mysql_query ("INSERT INTO tablename (H1, H2, H3) VALUES '$V1','$V2','$V3')"); ?> However, I encounter problems when $V1, $V2 or $V3 contain certain symbols, such as quotes ("'). What is the best way to avoid this? Thanks! I have a table with postcodes in it. At some point some of the codes will be entered with spaces (LA12 5TH) and some without (LA17HU) what i need is a function to check if there is a space, if not - count back 3 characters from the end of the string and insert a space. I don't know which string function to use - looked at pad but that seems to be for either end of string and not the middle - any ideas? Can anyone tell me why this is not INSERTing? My array data is coming out just fine.. I've tried everything I can think of and cannot get anything to insert.. Ahhhh! <?php $query = "SELECT RegionID, City FROM geo_cities WHERE RegionID='135'"; $results = mysqli_query($cxn, $query); $row_cnt = mysqli_num_rows($results); echo $row_cnt . " Total Records in Query.<br /><br />"; if (mysqli_num_rows($results)) { while ($row = mysqli_fetch_array($results)) { $insert_city_query = "INSERT INTO all_illinois SET state_id=$row[RegionID], city_name=$row[City] WHERE id = null" or mysqli_error(); $insert = mysqli_query($cxn, $insert_city_query); if (!$insert) { echo "INSERT is NOT working!"; exit(); } echo $row['City'] . "<br />"; echo "<pre>"; echo print_r($row); echo "</pre>"; } //while ($rows = mysqli_fetch_array($results)) } //if (mysqli_num_rows($results)) else { echo "No results to get!"; } ?> Here is my all_illinois INSERT table structu CREATE TABLE IF NOT EXISTS `all_illinois` ( `state_id` varchar(255) NOT NULL, `city_name` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Here is my source table geo_cities structu CREATE TABLE IF NOT EXISTS `1` ( `CityId` varchar(255) NOT NULL, `CountryID` varchar(255) NOT NULL, `RegionID` varchar(255) NOT NULL, `City` varchar(255) NOT NULL, `Latitude` varchar(255) NOT NULL, `Longitude` varchar(255) NOT NULL, `TimeZone` varchar(255) NOT NULL, `DmaId` varchar(255) NOT NULL, `Code` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; I'm missing something here. I have a form, and when the submit is pressed, the relevant post data inserts into table one, then I want the last insert id to insert along with other form data into a second table. The first table's still inserting fine, but I can't get that second one to do anything. It leapfrogs over the query and doesn't give an error. EDIT: I forgot to add an error: I get: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'usage, why VALUES ('14', '', '123', '','1234', '', '')' at line 1 query:INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('test 14', 'asdfa', 'asdf', 'adf','asdf', '', '', '', '123', '', '') Code: [Select] if (empty($errors)) { require_once ('dbconnectionfile.php'); $query = "INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('$description12', '$sn', '$description4', '$cne','$description5', '$description6', '$description7', '$description8', '$description9', '$description10', '$description11')"; $result = @mysql_query ($query); if ($result) { $who_donated=mysql_insert_id(); $query2 = "INSERT INTO tbl_donation (donor_id, donor_expyear, donor_cvv, donor_cardtype, donor_authorization, amount, usage, why) VALUES ('$who_donated', '$donate2', '$donate3', '$donate4','$donate5', '$donate6', '$donate7')"; $result2 = @mysql_query ($query2); if ($result2) {echo "Info was added to both tables! yay!";} echo "table one filled. Table two was not."; echo $who_donated; //header ("Location: http://www.twigzy.com/add_plant.php?var1=$plant_id"); exit(); } else { echo 'system error. No donation added'; Hi, I am trying to make some adjustments to uploadify.php which comes with the latest version of uploadify (3.0 beta), so that it works with a session variable that stores the login username and adds it to the path for uploads. Here is uploadify.php as it currently looks: Code: [Select] <?php session_name("MyLogin"); session_start(); $targetFolder = '/songs/' . $_SESSION['name']; // Relative to the root if (!empty($_FILES)) { $tempFile = $_FILES['Filedata']['tmp_name']; $targetPath = $_SERVER['DOCUMENT_ROOT'] . $targetFolder; $targetFile = rtrim($targetPath,'/') .'/'. $_FILES['Filedata']['name']; // Validate the file type $fileTypes = array('m4a','mp3','flac','ogg'); // File extensions $fileParts = pathinfo($_FILES['Filedata']['name']); if (in_array($fileParts['extension'],$fileTypes)) { move_uploaded_file($tempFile,$targetFile); echo '1'; } else { echo 'Invalid file type.'; } } echo $targetFolder; ?> I added Code: [Select] echo $targetFolder; at the bottom so that I could make sure that the string returned was correct, and it is, i.e. '/songs/nick'. For some reason though, uploads are not going to the correct folder, i.e. the username folder, but instead are going to the parent folder 'songs'. The folder for username exists, with correct permissions, and when I manually enter Code: [Select] $targetFolder = '/songs/nick';all works fine. Which strikes me as rather strange. I have limited experience of using php, but wonder how if the correct string is returned by the session variable, the upload works differently than with the manually entered string. Any help would be much appreciated. It's the last issue with a website that was due to go live 2 days ago! Thanks, Nick This topic has been moved to PHP Regex. http://www.phpfreaks.com/forums/index.php?topic=326004.0 Hello all, I'm trying to change the end of a javascript call based on the end of the url string. The common part of all the url strings is sobi2Id=, I'm trying to do this with strstr but am having no luck. I'm new to php so my syntax knowledge is terrible! at the moment i've got Code: [Select] <?php $url = $_SERVER['REQUEST_URI']; $tag = strstr ($url, 'sobi2Id='); echo $tag; ?> but this returns an unexpected T_STRING, expecting ',' or ';' Can anyone debug this? I may well be being really silly! Hey there, Thanks for taking the time to read my thread. My issue is that I can't think of a way to edit a XML file using PHP's XML functionality and then assign the edited contents to a string instead of saving the file. Because my issue is that I have to edit the XML file based upon a string brought from a remote location then give it back to that remote location using a string again, to be exact I am doing it via Linux command line utilizing SSH2. This is what I managed to complete on my own. function CheckIVMPConfig($ServerID) { global $Panel; if(is_numeric($ServerID) && $this->IsValidServer($ServerID)) { // We select the game server that the FTP account was created for. $Servers = mysql_query("SELECT * FROM control_servers WHERE server_id = '".mysql_real_escape_string($FTPAccount['ftp_server'])."'"); $Server = mysql_fetch_array($Servers); // Here we select the Box ID that the game server is on. $Boxs = mysql_query("SELECT * FROM control_machines WHERE machine_id = '".$Server['server_machine']."'"); $Box = mysql_fetch_array($Boxs); // Now we select the required package for the box. $Packages = mysql_query("SELECT * FROM control_packages WHERE package_id = '".$Server['server_package']."'"); $Package = mysql_fetch_array($Packages); // Retrive the file. $Config = $CProtocol->exec("cat /home/{$Server['server_id']}/{$Package['package_config']}"); $Parse = SimpleXMLElement($Config); foreach($Parse as $Entry) // loop through our books { if($Entry->port != $Server['server_port']) { // edit the value } else if($Entry->maxplayers > $Server['server_slots']) { // edit the value } } } } I have the following function, which takes a string with commas in it and attempts remove those commas. The way I have it here is that I use explode to take out the commas, which makes an array, and then iterate through that array to put a string back together without the commas. function tags_to_sort_by( $sortMeta ) { $sortByArray = explode(",", $sortMeta); $sortByCount = count($sortByArray); for ($i = 0; $i < $sortByCount; $i++) { $sortByString += $sortByArray[$i]; } return $sortByString; } What does not work is the following line: $sortByString += $sortByArray[$i]; Somehow that outputs a 0, which I don't understand. It should output something like: arrayItem1 arrayItem2 array3 etc. My question is if there either is an easier way to remove the commas from the original string or what I am doing wrong in this line: $sortByString += $sortByArray[$i]; // I am trying to concatenate each part of the array back into the string. Thanks a lot for help with this! |
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