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  • User Getting Logged Out After Making First Comment

    - by John
    Hello, I am using a login system that works well. I am also using a comment system. The comment function does not show up unless the user is logged in (as shown in commentformonoff.php below). When a user makes a comment, the info is passed from the function "show_commentbox" to the file comments2a.php. Then, the info is passed to the file comments2.php. When the site is first pulled up on a browser, after logging in and making a comment, the user is logged out. After logging in a second time during the same browser session, the user is no longer logged out after making a comment. How can I keep the user logged in after making the first comment? Thanks in advance, John Commentformonoff.php: <?php if (!isLoggedIn()) { if (isset($_POST['cmdlogin'])) { if (checkLogin($_POST['username'], $_POST['password'])) { show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl); } else { echo "<div class='logintocomment'>Login to comment</div>"; } } else { echo "<div class='logintocomment'>Login to comment</div>"; } } else { show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl); } ?> Function "show_commentbox": function show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl) { echo '<form action="http://www...com/.../comments/comments2a.php" method="post"> <input type="hidden" value="'.$_SESSION['loginid'].'" name="uid"> <input type="hidden" value="'.$_SESSION['username'].'" name="u"> <input type="hidden" value="'.$submissionid.'" name="submissionid"> <input type="hidden" value="'.stripslashes($submission).'" name="submission"> <input type="hidden" value="'.$url.'" name="url"> <input type="hidden" value="'.$submittor.'" name="submittor"> <input type="hidden" value="'.$submissiondate.'" name="submissiondate"> <input type="hidden" value="'.$countcomments.'" name="countcomments"> <input type="hidden" value="'.$dispurl.'" name="dispurl"> <label class="addacomment" for="title">Add a comment:</label> <textarea class="checkMax" name="comment" type="comment" id="comment" maxlength="1000"></textarea> <div class="commentsubbutton"><input name="submit" type="submit" value="Submit"></div> </form> '; } Included in comments2a.php: $uid = mysql_real_escape_string($_POST['uid']); $u = mysql_real_escape_string($_POST['u']); $query = sprintf("INSERT INTO comment VALUES (NULL, %d, %d, '%s', NULL)", $uid, $subid, $comment); mysql_query($query) or die(mysql_error()); $lastcommentid = mysql_insert_id(); header("Location: comments2.php?submission=".$submission."&submissionid=".$submissionid."&url=".$url."&submissiondate=".$submissiondate."&comment=".$comment."&subid=".$subid."&uid=".$uid."&u=".$u."&submittor=".$submittor."&countcomments=".$countcomments."&dispurl=".$dispurl."#comment-$lastcommentid"); exit(); Included in comments2.php: if($_SERVER['REQUEST_METHOD'] == "POST"){header('Location: http://www...com/.../comments/comments2.php?submission='.$submission.'&submissionid='.$submissionid.'&url='.$url.'&submissiondate='.$submissiondate.'&submittor='.$submittor.'&countcomments='.$countcomments.'&dispurl='.$dispurl.'');} $uid = mysql_real_escape_string($_GET['uid']); $u = mysql_real_escape_string($_GET['u']);

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  • Check for a unique value within a count, but get all results

    - by pedalpete
    I'm trying to create a single query which, similar to stack overflow, will give me the number of votes, but also make sure that the currently viewing user can't upvote again if they've already upvoted. my query currently looks like SELECT cid, text, COUNT(votes.parentid) FROM comments LEFT JOIN votes ON comments.cid=votes.parentid AND votes.type=3 WHERE comments.type=0 AND comments.parentid='$commentParentid' GROUP BY comments.cid But I'm completely stumpted on how to add the check to see if the userid is in the votes table. The other option is to add a seperate query where SELECT COUNT(*) FROM votes WHERE userid='$userid' AND parentid='$commentParentid' AND type=3 I'm just realizing I'm so lost with this that I don't even really know what tags to provide.

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  • Compare structures of two databases?

    - by streetparade
    Hello, I wanted to ask whether it is possible to compare the complete database structure of two huge databases. We have two databases, the one is a development database, the other a production database. I've sometimes forgotten to make changes in to the production database, before we released some parts of our code, which results that the production database doesn't have the same structure, so if we release something we got some errors. Is there a way to compare the two, or synchronize?

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  • AIR: sync gui with data-base?

    - by John Isaacks
    I am going to be building an AIR application that shows a list (about 1-25 rows of data) from a data-base. The data-base is on the web. I want the list to be as accurate as possible, meaning as soon as the data-base data changes, the list displayed in the app should update asap. I do not know of anyway that the air application could be notified when there is a change, I am thinking I am going to have to poll the data-base at certain intervals to keep an up to date list. So my question is, first is there any way to NOT have to keep checking the data-base? or if I do keep have to keep checking the data-base what is a reasonable interval to do that at? Thanks.

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  • Create fulltext index on a VIEW

    - by kylex
    Is it possible to create a full text index on a VIEW? If so, given two columns column1 and column2 on a VIEW, what is the SQL to get this done? The reason I'd like to do this is I have two very large tables, where I need to do a FULLTEXT search of a single column on each table and combine the results. The results need to be ordered as a single unit. Suggestions? EDIT: This was my attempt at creating a UNION and ordering by each statements scoring. (SELECT a_name AS name, MATCH(a_name) AGAINST('$keyword') as ascore FROM a WHERE MATCH a_name AGAINST('$keyword')) UNION (SELECT s_name AS name,MATCH(s_name) AGAINST('$keyword') as sscore FROM s WHERE MATCH s_name AGAINST('$keyword')) ORDER BY (ascore + sscore) ASC sscore was not recognized.

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  • PHP + MYSQLI: Variable parameter/result binding with prepared statements.

    - by Brian Warshaw
    In a project that I'm about to wrap up, I've written and implemented an object-relational mapping solution for PHP. Before the doubters and dreamers cry out "how on earth?", relax -- I haven't found a way to make late static binding work -- I'm just working around it in the best way that I possibly can. Anyway, I'm not currently using prepared statements for querying, because I couldn't come up with a way to pass a variable number of arguments to the bind_params() or bind_result() methods. Why do I need to support a variable number of arguments, you ask? Because the superclass of my models (think of my solution as a hacked-up PHP ActiveRecord wannabe) is where the querying is defined, and so the find() method, for example, doesn't know how many parameters it would need to bind. Now, I've already thought of building an argument list and passing a string to eval(), but I don't like that solution very much -- I'd rather just implement my own security checks and pass on statements. Does anyone have any suggestions (or success stories) about how to get this done? If you can help me solve this first problem, perhaps we can tackle binding the result set (something I suspect will be more difficult, or at least more resource-intensive if it involves an initial query to determine table structure).

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  • adding other parameter to function

    - by Ronnie Chester Lynwood
    hello. i got a function that listing downloads in a table with foreach. it's also lists searched term, search type. public function fetchDownloads($displaySite=true) { $downloads = array(); $sqlWhere = ""; if(isset($this->q)) { if(strlen($this->q) <= $this->recents_length && !empty($this->q)) { $insertRecent = $this->processDataHook("insertRecent",$this->q); if($insertRecent) { if(!@mysql_query("INSERT INTO wcddl_recents (query) VALUES ('".$this->qSQL."')")) { @mysql_query("UPDATE wcddl_recents SET searches = searches+1 WHERE query = '".$this->qSQL."'"); } } } if($this->search_type == "narrow") { $sqlWhere = " WHERE title LIKE '%".mysql_real_escape_string(str_replace(" ","%",$this->q))."%'"; } elseif($this->search_type == "wide") { $qExp = explode(" ",$this->q); $sqlWhere = array(); foreach($qExp as $fq) $sqlWhere[] = "title LIKE '%".mysql_real_escape_string($fq)."%'"; $sqlWhere = implode(" OR ",$sqlWhere); $sqlWhere = " WHERE (".$sqlWhere.")"; } } if(isset($this->type)) { if(!empty($sqlWhere)) { $sqlWhere .= " AND type = '".$this->typeSQL."'"; } else { $sqlWhere = " WHERE type = '".$this->typeSQL."'"; } } $sqlWhere = $this->processDataHook("fetchDownloadsSQLWhere",$sqlWhere); $this->maxPages = mysql_query("SELECT COUNT(*) FROM wcddl_downloads".$sqlWhere.""); $this->maxPages = mysql_result($this->maxPages,0); $this->numRows = $this->maxPages; $this->maxPages = ceil($this->maxPages/$this->limit); $sqlMain = "SELECT id,sid,title,type,url,dat,views,rating FROM wcddl_downloads".$sqlWhere." ORDER BY ".(isset($this->sqlOrder) ? mysql_real_escape_string($this->sqlOrder) : "id DESC")." LIMIT ".$this->pg.",".$this->limit.""; $sqlMain = $this->processDataHook("whileFetchDownloadsSQL",$sqlMain); $sqlMain = mysql_query($sqlMain); $this->processHook("whileFetchDownloads"); while($row = mysql_fetch_assoc($sqlMain)) { if($displaySite) { $downloadSite = mysql_query("SELECT name as sname, url as surl, rating as srating FROM wcddl_sites WHERE id = '".$row['sid']."'"); $downloadSite = mysql_fetch_assoc($downloadSite); $row = array_merge($row,$downloadSite); } $downloads_current = $this->mapit($row,array("stripslashes","strip_tags")); $downloads_current = $this->processDataHook("fetchDownloadsRow",$downloads_current); $downloads[] = $downloads_current; } $this->pageList = $this->getPages($this->page,$this->maxPages); $this->processHook("endFetchDownloads"); return $downloads; } I want to add if $_REQUEST['site'] is set, order downloads by sname that catching from wcddl_sites.

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  • uploading image & getting back from database

    - by Anup Prakash
    Putting a set of code which is pushing image to database and fetching back from database: <!-- <?php error_reporting(0); // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","root","")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("test"); $q = <<<CREATE create table image ( pid int primary key not null auto_increment, title text, imgdata longblob, friend text) CREATE; @mysql_query($q); // Insert any new image into database if (isset($_POST['submit'])) { move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($instr) < 149000) { $image_query="insert into image (title, imgdata,friend) values (\"". $_REQUEST['title']. "\", \"". $image. "\",'".$_REQUEST['friend']."')"; mysql_query ($image_query) or die("query error"); } else { $errmsg = "Too large!"; } $resultbytes=''; // Find out about latest image $query = "select * from image where pid=1"; $result = @mysql_query("$query"); $resultrow = @mysql_fetch_assoc($result); $gotten = @mysql_query("select * from image order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; $resultbytes = $row[imgdata]; $friend=$row[friend]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../wellimg/ctco.jpg","rb"); $bytes = fread($instr,filesize("../wellimg/ctco.jpg")); } if ($resultbytes!='') { echo $resultbytes; } } ?> <html> <head> <title>Upload an image to a database</title> </head> <body bgcolor="#FFFF66"> <form enctype="multipart/form-data" name="file_upload" method="post"> <center> <div id="image" align="center"> <h2>Heres the latest picture</h2> <font color=red><?php echo $errmsg; ?></font> <b><?php echo $title ?></center> </div> <hr> <h2>Please upload a new picture and title</h2> <table align="center"> <tr> <td>Select image to upload: </td> <td><input type="file" name="imagefile"></td> </tr> <tr> <td>Enter the title for picture: </td> <td><input type="text" name="title"></td> </tr> <tr> <td>Enter your friend's name:</td> <td><input type="text" name="friend"></td> </tr> <tr> <td><input type="submit" name="submit" value="submit"></td> <td></td> </tr> </table> </form> </body> </html> --> Above set of code has one problem. The problem is whenever i pressing the "submit" button. It is just displaying the image on a page. But it is leaving all the html codes. even any new line message after the // Printing image on browser echo $resultbytes; //************************// So, for this i put this set of code in html tag: This is other sample code: <!-- <?php error_reporting(0); // Connect to database $errmsg = ""; if (! @mysql_connect("localhost","root","")) { $errmsg = "Cannot connect to database"; } @mysql_select_db("test"); $q = <<<CREATE create table image ( pid int primary key not null auto_increment, title text, imgdata longblob, friend text) CREATE; @mysql_query($q); // Insert any new image into database if (isset($_POST['submit'])) { move_uploaded_file($_FILES['imagefile']['tmp_name'],"latest.img"); $instr = fopen("latest.img","rb"); $image = addslashes(fread($instr,filesize("latest.img"))); if (strlen($instr) < 149000) { $image_query="insert into image (title, imgdata,friend) values (\"". $_REQUEST['title']. "\", \"". $image. "\",'".$_REQUEST['friend']."')"; mysql_query ($image_query) or die("query error"); } else { $errmsg = "Too large!"; } $resultbytes=''; // Find out about latest image $query = "select * from image where pid=1"; $result = @mysql_query("$query"); $resultrow = @mysql_fetch_assoc($result); $gotten = @mysql_query("select * from image order by pid desc limit 1"); if ($row = @mysql_fetch_assoc($gotten)) { $title = htmlspecialchars($row[title]); $bytes = $row[imgdata]; $resultbytes = $row[imgdata]; $friend=$row[friend]; } else { $errmsg = "There is no image in the database yet"; $title = "no database image available"; // Put up a picture of our training centre $instr = fopen("../wellimg/ctco.jpg","rb"); $bytes = fread($instr,filesize("../wellimg/ctco.jpg")); } } ?> <html> <head> <title>Upload an image to a database</title> </head> <body bgcolor="#FFFF66"> <form enctype="multipart/form-data" name="file_upload" method="post"> <center> <div id="image" align="center"> <h2>Heres the latest picture</h2> <?php if ($resultbytes!='') { // Printing image on browser echo $resultbytes; } ?> <font color=red><?php echo $errmsg; ?></font> <b><?php echo $title ?></center> </div> <hr> <h2>Please upload a new picture and title</h2> <table align="center"> <tr> <td>Select image to upload: </td> <td><input type="file" name="imagefile"></td> </tr> <tr> <td>Enter the title for picture: </td> <td><input type="text" name="title"></td> </tr> <tr> <td>Enter your friend's name:</td> <td><input type="text" name="friend"></td> </tr> <tr> <td><input type="submit" name="submit" value="submit"></td> <td></td> </tr> </table> </form> </body> </html> --> ** But in this It is showing the image in format of special charaters and digits. 1) So, Please help me to print the image with some HTML code. So that i can print it in my form to display the image. 2) Is there any way to convert the database image into real image, so that i can store it into my hard-disk and call it from tag? Please help me.

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  • What's wrong with this SQL query?

    - by ThinkingInBits
    I have two tables: photographs, and photograph_tags. Photograph_tags contains a column called photograph_id (id in photographs). You can have many tags for one photograph. I have a photograph row related to three tags: boy, stream, and water. However, running the following query returns 0 rows SELECT p.* FROM photographs p, photograph_tags c WHERE c.photograph_id = p.id AND (c.value IN ('dog', 'water', 'stream')) GROUP BY p.id HAVING COUNT( p.id )=3 Is something wrong with this query?

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  • How do I use Django to insert a Geometry Field into the database?

    - by alex
    class LocationLog(models.Model): user = models.ForeignKey(User) utm = models.GeometryField(spatial_index=True) This is my database model. I would like to insert a row. I want to insert a circle at point -55, 333. With a radius of 10. How can I put this circle into the geometry field? Of course, then I would want to check which circles overlap a given circle. (my select statement)

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  • can you make an sql query for this situation?

    - by saurav
    i have a table as below. name and 10 cities in which he lived during his lifetime. name , city1 , city2 , city3 ,city4 , city5 ,city6 , city7 , city8 , city9 city10 suppose for a particular name i want to fetch other names in table matching with maximum number of cities. for example if i want to fetch other people who have lived in three or more cities lived by this person.

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  • 'Attempt to call private method' error when trying to change change case of db entires in migration

    - by Senthil
    class AddTitleToPosts < ActiveRecord::Migration def self.up add_column :posts, :title, :string Post.find(:all).each do |post| post.update(:title => post.name.upcase) end end def self.down end end Like you can nothing particularly complicated, just trying to add new column title by changing case of name column already in db. But I get attempt to call private method error. I'm guessing it has something to do with 'self'? Thanks for your help.

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  • Lost with hibernate - OneToMany resulting in the one being pulled back many times..

    - by Andy
    I have this DB design: CREATE TABLE report ( ID MEDIUMINT PRIMARY KEY NOT NULL AUTO_INCREMENT, user MEDIUMINT NOT NULL, created TIMESTAMP NOT NULL, state INT NOT NULL, FOREIGN KEY (user) REFERENCES user(ID) ON UPDATE CASCADE ON DELETE CASCADE ); CREATE TABLE reportProperties ( ID MEDIUMINT NOT NULL, k VARCHAR(128) NOT NULL, v TEXT NOT NULL, PRIMARY KEY( ID, k ), FOREIGN KEY (ID) REFERENCES report(ID) ON UPDATE CASCADE ON DELETE CASCADE ); and this Hibernate Markup: @Table(name="report") @Entity(name="ReportEntity") public class ReportEntity extends Report{ @Id @GeneratedValue(strategy = GenerationType.AUTO) @Column(name="ID") private Integer ID; @Column(name="user") private Integer user; @Column(name="created") private Timestamp created; @Column(name="state") private Integer state = ReportState.RUNNING.getLevel(); @OneToMany(mappedBy="pk.ID", fetch=FetchType.EAGER) @JoinColumns( @JoinColumn(name="ID", referencedColumnName="ID") ) @MapKey(name="pk.key") private Map<String, ReportPropertyEntity> reportProperties = new HashMap<String, ReportPropertyEntity>(); } and @Table(name="reportProperties") @Entity(name="ReportPropertyEntity") public class ReportPropertyEntity extends ReportProperty{ @Embeddable public static class ReportPropertyEntityPk implements Serializable{ /** * long#serialVersionUID */ private static final long serialVersionUID = 2545373078182672152L; @Column(name="ID") protected int ID; @Column(name="k") protected String key; } @EmbeddedId protected ReportPropertyEntityPk pk = new ReportPropertyEntityPk(); @Column(name="v") protected String value; } And i have inserted on Report and 4 Properties for that report. Now when i execute this: this.findByCriteria( Order.asc("created"), Restrictions.eq("user", user.getObject(UserField.ID)) ) ); I get back the report 4 times, instead of just the once with a Map with the 4 properties in. I'm not great at Hibernate to be honest, prefer straight SQL but I must learn, but i can't see what it is that is wrong.....? Any suggestions?

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  • Unknown column even thoug it exits

    - by george
    I have SELECT servisler.geo_location, servisler.ADRES_MERKEZ, servisler.ADRES_ILCE, servisler.ADRES_IL, servisler.FIRMA_UNVANI, servisler.ADRES_ISTEL, servisler.YETKILI_ADISOYADI, urun_gruplari.GRUP_ADI FROM servisler INNER JOIN urun_gruplari ON kullanici_cihaz.URUN_GRUP_NO= urun_gruplari.RECNO INNER JOIN kullanici ON kullanici.SERVIS_RECNO = servisler.RECNO INNER JOIN kullanici_cihaz ON kullanici.RECNO = kullanici_cihaz.KUL_RECNO AND kullanici_cihaz.URUN_GRUP_NO = urun_gruplari.RECNO where kullanici.kullanici = 'MAR.EDI.003' but it says [Err] 1054 - Unknown column 'kullanici_cihaz.URUN_GRUP_NO' in 'on clause' enen though the column exits. What is its problem? schema Server version: 5.1.33-community-log

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  • Correlate GROUP BY and LEFT JOIN on multiple criteria to show latest record?

    - by Sunbird
    In a simple stock management database, quantity of new stock is added and shipped until quantity reaches zero. Each stock movement is assigned a reference, only the latest reference is used. In the example provided, the latest references are never shown, the stock ID's 1,4 should have references charlie, foxtrot respectively, but instead show alpha, delta. How can a GROUP BY and LEFT JOIN on multiple criteria be correlated to show the latest record? http://sqlfiddle.com/#!2/6bf37/107 CREATE TABLE stock ( id tinyint PRIMARY KEY, quantity int, parent_id tinyint ); CREATE TABLE stock_reference ( id tinyint PRIMARY KEY, stock_id tinyint, stock_reference_type_id tinyint, reference varchar(50) ); CREATE TABLE stock_reference_type ( id tinyint PRIMARY KEY, name varchar(50) ); INSERT INTO stock VALUES (1, 10, 1), (2, -5, 1), (3, -5, 1), (4, 20, 4), (5, -10, 4), (6, -5, 4); INSERT INTO stock_reference VALUES (1, 1, 1, 'Alpha'), (2, 2, 1, 'Beta'), (3, 3, 1, 'Charlie'), (4, 4, 1, 'Delta'), (5, 5, 1, 'Echo'), (6, 6, 1, 'Foxtrot'); INSERT INTO stock_reference_type VALUES (1, 'Customer Reference'); SELECT stock.id, SUM(stock.quantity) as quantity, customer.reference FROM stock LEFT JOIN stock_reference AS customer ON stock.id = customer.stock_id AND stock_reference_type_id = 1 GROUP BY stock.parent_id

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  • sql query question / count

    - by scheibenkleister
    Hi, I have houses that belongs to streets. A user can buy several houses. How do I find out, if the user owns an entire street? street table with columns (id/name) house table with columns (id/street_id [foreign key] owner table with columns (id/house_id/user_id) [join table with foreign keys] So far, I'm using count which returns the result: select count(*), street_id from owner left join house on owner.house_id = house.id group by street_id where user_id = 1 count(*) | street_id 3 | 1 2 | 2 A more general count: select count(*) from house group by street_id returns: count(*) | street_id 3 | 1 3 | 2 How can I find out, that user 1 owns the entire street 1 but not street 2? Thanks.

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  • How to add space between the images being fetched using database through php

    - by ParveenArora
    I am using following code to fetch images using database with php. while($row = mysql_fetch_array($result)) //To excute result query { echo "<a href='http://".$row['website']."' target='_blank'><img src=\"" . $PathImage . $row['logo'] . "\" height = $FooterWidth /></a>XX; } Here I am using $row[logo] is fetching the path of images stored on the server and XX to put the spaced between the images having the same color of text XX as background, and but I want to use the proper method I know this can be done using table but I want to do it without using table. Any Suggestions?

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  • Prevent two users from editing the same data

    - by Industrial
    Hi everyone, I have seen a feature in different web applications including Wordpress (not sure?) that warns a user if he/she opens an article/post/page/whatever from the database, while someone else is editing the same data simultaneously. I would like to implement the same feature in my own application and I have given this a bit of thought. Is the following example a good practice on how to do this? It goes a little something like this: 1) User A enters a the editing page for the mysterious article X. The database tableEvents is queried to make sure that no one else is editing the same page for the moment, which no one is by then. A token is then randomly being generated and is inserted into a database table called Events. 1) User B also want's to make updates to the article X. Now since our User A already is editing the article, the Events table is queried and looks like this: | timestamp | owner | Origin | token | ------------------------------------------------------------ | 1273226321 | User A | article-x | uniqueid## | 2) The timestamp is being checked. If it's valid and less than say 100 seconds old, a message appears and the user cannot make any changes to the requested article X: Warning: User A is currently working with this article. In the meantime, editing cannot be done. Please do something else with your life. 3) If User A decides to go on and save his changes, the token is posted along with all other data to update the database, and toggles a query to delete the row with token uniqueid##. If he decides to do something else instead of committing his changes, the article X will still be available for editing in 100 seconds for User B Let me know what you think about this approach! Wish everyone a great weekend!

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  • Unnecessary Error Message Being Displayed

    - by ThatMacLad
    I've set up a form to update my blog and it was working fine up until about this morning. It keeps on turning up with an Invalid Entry ID error on the edit post page when I click the update button despite the fact that it updates the homepage. All help is seriously appreciated. <html> <head> <title>Ultan's Blog | New Post</title> <link rel="stylesheet" href="css/editpost.css" type="text/css" /> </head> <body> <div class="new-form"> <div class="header"> </div> <div class="form-bg"> <?php mysql_connect ('localhost', 'root', 'root') ; mysql_select_db ('tmlblog'); if (isset($_POST['update'])) { $id = htmlspecialchars(strip_tags($_POST['id'])); $month = htmlspecialchars(strip_tags($_POST['month'])); $date = htmlspecialchars(strip_tags($_POST['date'])); $year = htmlspecialchars(strip_tags($_POST['year'])); $time = htmlspecialchars(strip_tags($_POST['time'])); $entry = $_POST['entry']; $title = htmlspecialchars(strip_tags($_POST['title'])); if (isset($_POST['password'])) $password = htmlspecialchars(strip_tags($_POST['password'])); else $password = ""; $entry = nl2br($entry); if (!get_magic_quotes_gpc()) { $title = addslashes($title); $entry = addslashes($entry); } $timestamp = strtotime ($month . " " . $date . " " . $year . " " . $time); $result = mysql_query("UPDATE php_blog SET timestamp='$timestamp', title='$title', entry='$entry', password='$password' WHERE id='$id' LIMIT 1") or print ("Can't update entry.<br />" . mysql_error()); header("Location: post.php?id=" . $id); } if (isset($_POST['delete'])) { $id = (int)$_POST['id']; $result = mysql_query("DELETE FROM php_blog WHERE id='$id'") or print ("Can't delete entry.<br />" . mysql_error()); if ($result != false) { print "The entry has been successfully deleted from the database."; exit; } } if (!isset($_GET['id']) || empty($_GET['id']) || !is_numeric($_GET['id'])) { die("Invalid entry ID."); } else { $id = (int)$_GET['id']; } $result = mysql_query ("SELECT * FROM php_blog WHERE id='$id'") or print ("Can't select entry.<br />" . $sql . "<br />" . mysql_error()); while ($row = mysql_fetch_array($result)) { $old_timestamp = $row['timestamp']; $old_title = stripslashes($row['title']); $old_entry = stripslashes($row['entry']); $old_password = $row['password']; $old_title = str_replace('"','\'',$old_title); $old_entry = str_replace('<br />', '', $old_entry); $old_month = date("F",$old_timestamp); $old_date = date("d",$old_timestamp); $old_year = date("Y",$old_timestamp); $old_time = date("H:i",$old_timestamp); } ?> <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <p><input type="hidden" name="id" value="<?php echo $id; ?>" /> <strong><label for="month">Date (month, day, year):</label></strong> <select name="month" id="month"> <option value="<?php echo $old_month; ?>"><?php echo $old_month; ?></option> <option value="January">January</option> <option value="February">February</option> <option value="March">March</option> <option value="April">April</option> <option value="May">May</option> <option value="June">June</option> <option value="July">July</option> <option value="August">August</option> <option value="September">September</option> <option value="October">October</option> <option value="November">November</option> <option value="December">December</option> </select> <input type="text" name="date" id="date" size="2" value="<?php echo $old_date; ?>" /> <select name="year" id="year"> <option value="<?php echo $old_year; ?>"><?php echo $old_year; ?></option> <option value="2004">2004</option> <option value="2005">2005</option> <option value="2006">2006</option> <option value="2007">2007</option> <option value="2008">2008</option> <option value="2009">2009</option> <option value="2010">2010</option> </select> <strong><label for="time">Time:</label></strong> <input type="text" name="time" id="time" size="5" value="<?php echo $old_time; ?>" /></p> <p><strong><label for="title">Title:</label></strong> <input type="text" name="title" id="title" value="<?php echo $old_title; ?>" size="40" /> </p> <p><strong><label for="password">Password protect?</label></strong> <input type="checkbox" name="password" id="password" value="1"<?php if($old_password == 1) echo " checked=\"checked\""; ?> /></p> <p><textarea cols="80" rows="20" name="entry" id="entry"><?php echo $old_entry; ?></textarea></p> <p><input type="submit" name="update" id="update" value="Update"></p> </form> <p><strong>Be absolutely sure that this is the post that you wish to remove from the blog!</strong><br /> </p> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <input type="hidden" name="id" id="id" value="<?php echo $id; ?>" /> <input type="submit" name="delete" id="delete" value="Delete" /> </form> </div> </div> </div> <div class="bottom"></div> </body> </html>

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  • Why is the ( ) mandatory in the SQL statement select * from gifts INNER JOIN sentgifts using (giftID

    - by Jian Lin
    Why is the ( ) mandatory in the SQL statement select * from gifts INNER JOIN sentgifts using (giftID); ? The ( ) usually is for specifying grouping of something. But in this case, are we supposed to be able to use 2 or more field names... in the example above, it can be all clear that it is 1 field, is it just that the parser is not made to bypass the ( ) when it is all clear? (such as in the language Ruby).

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  • PHP coding a price comparaison tool

    - by Tristan
    Hello, it's the first time I developp such tool you all know (the possibility to compare articles according to price and/or options) Since I never did that i want to tell me what do you think of the way i see that : On the database we would have : offer / price / option 1 / option 2 / option 3 / IDseller / IDoffer best buy / 15$ / full FTP / web hosting / php.ini / 10 / 1 .../..../.... And the request made by the client : "SELECT * FROM offers WHERE price <= 20 AND option1 = fullFTP"; I don't know if it seems OK to you. Plus i was wondering, how to avoid multiples entries for the same seller. Imagine you have multiple offers with a price <= 20 with the option FullFTP for the same seller, i don't want him to be shown 5 times on the comparator. If you have any advices ;) Thanks

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