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  • Upload files, form within form

    - by Alexd2
    Hello everyone and thanks in advance. I have a problem and I have 2 form into one another, the domestic form is to perform a file upload. As I can do to make when sending in internal form not run the main form. <form name="x" method="post" action="xxx.php"> .... <form action="" method="post" enctype="multipart/form-data" target="xxx"> <input type="file" /> <input type="submit" /> </form> <iframe id="xxx" src="process.php"> </iframe> .... <input type="submit" name="pro" value="Register user"/ > </form> Doing this does not work, as this within another form. Any help or possible solution.

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  • How can I store and access SQLite databases from a central location?

    - by Brian Ramsay
    I have installed SQLite on my UNIX system, and am planning on accessing it from PHP. However, the database is created in the directory I initialize it in, and if a script runs in a different directory a new database is created in the same directory. Is there an option for SQLite (or the PHP wrappers) to create the databases in one location, and have those databases accessible just by name outside of that directory? Ideally, I'd like to be able to do something like $db=new SQLiteDatabase("db.test"); in any directory and have it reference the same database, if that makes sense.

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  • CakePHP accessing other controllers

    - by James
    CakePHP Newbie :) I am having trouble accessing another controller and passing that data to a view in one of my controllers: In controllers/landings_controller.php: var $uses = 'User'; function home() { $userdata = $this->User->read(); $this->set(compact('userdata')); } In views/landings/home.ctp: <?php echo $this->userdata; ?> When accessing /landings/home I get the following error: Notice (8): Undefined property: View::$userdata [APP/views/landings/home.ctp, line 38] I don't know what I am doing wrong. Any help? Thanks!

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  • PDOException “could not find driver”

    - by Mike
    I have just installed Debian Lenny with Apache, MySQL, and PHP and I am receiving a PDOException could not find driver. This is the specific line of code it is referring to: $dbh = new PDO('mysql:host=' . DB_HOST . ';dbname=' . DB_NAME, DB_USER, DB_PASS) DB_HOST, DB_NAME, DB_USER, and DB_PASS are constants that I have defined. It works fine on the production server (and on my previous Ubuntu Server setup). Is this something to do with my PHP installation? Searching the internet has not helped, all I get is experts-exchange and examples, but no solutions.

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  • Is there any way to use imagegrabwindow() with Firefox or Chrome?

    - by patternsofchaos
    I'm trying to generate website thumbnails programatically in PHP. To do this, I'm using imagegrabwindow() with a COM object: $browser = new COM("InternetExplorer.Application"); $handle = $browser->HWND; $browser->Visible = true; $browser->Navigate($pre.$URL); while ($browser->Busy) { com_message_pump(4000); } $img = imagegrabwindow($handle); What I'm wondering is if there is any way to do the same thing with Firefox or Chrome? Can I invoke either of them with PHP COM?

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  • Error in header function

    - by user1178695
    I'm using the php header function for the redirection but it is not working.I'm using the following code. $sql=mysql_query("select * from password where username='$email' and password1 = '$pwd'"); //echo "selct * from password where username='$email' and password = '$pwd'"; $row=mysql_fetch_row($sql); $fieldset=mysql_num_rows($sql); $host=$_SERVER['HTTP_HOST']."/beta/"; if($fieldset>0 && $conEmail !="") { $_SESSION['email']=$email; $_SESSION['Email']=$email; $_SESSION['memberID']=$id; $_SESSION['status']='Admin'; header("location:http://".$host."member.php"); }

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  • ajax how to read out $_post variable

    - by alex
    Hi, I am trying to filter/search a database with ajax $.ajax({ type: "POST", url: "filterSearch.php", queryString: qry, success: function(data){ alert( "Data Saved: " + data ); $('#searchResult').html(data); // Fill the search results box } }); Now in filterSearch.php i have the following test codes if(isset($_POST['queryString'])) { echo "TEST"; } if($_POST['runquery']==1) { $sql = "SELECT * FROM fs_vacatures WHERE here-the-like-query?"; $msg = $sql; echo $msg; die(); } die(); But nor TEST or the $sql is return in the alert??

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  • Problem with onsubmit ID

    - by Liso22
    I need an onsubmit to return false if what's written on the first field is a certain PHP value however the field has a strange id format and I'm not quite sure how to add it in the onsubmit. This is the form, I didn't add the element ID to it: <form class="questionform" name="questionform-0" id="questionform-0" onsubmit="if (document.getElementById('').value == '<?php echo $casi; ?>') return false;" > <textarea class="question-box" style="width:97%;" cols="20" rows="4" id="question-box-' . $questionformid . '" name="title" type="text" maxlength="80" size="28" value=""></textarea> I tried many times but couldn't add it. How should I do it? Thanks

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  • Uploading files wont work on production server

    - by Camran
    I have a php script which uploads images to a temporary folder on the server. This works on my local computer with the local (virtual) server. (wampserver). However, on the production server (linux) I cant get it working. Everything is loading as it should, except for that the file doesn't show up on the server. My Q is, is there anything I should think about when moving to a production server with uploading images or files? I have set permissions on the folder where the images are supposed to go to 777 and also the php-code which uploads them to 777. Thanks

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  • Injecting single-use object into class

    - by moteutsch
    I have the following code: <?php class X { public function do($url) { $httpRequest = new \HttpRequest\Curl($url, $this->getOptions()); $httpRequest->fire(); // etc. } // ... } In order to be able to unit test this class, I'd like to inject a mocked HttpRequest class. One way to do this would be as follows: <?php class X { private $httpRequestClass; public function __construct($httpRequestClass = '\HttpRequest\Curl') { $this->httpRequestClass = $httpRequestClass; } public function do($url) { $httpRequest = new $this->httpRequestClass($url, $this->getOptions()); $httpRequest->fire(); // etc. } // ... } But this doesn't seem right. Any other ideas?

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  • Stopping users posting more than once

    - by user342391
    Before posting my form I am checking the database to see if there are any previous posts from the user. If there are previous posts then the script will kick back a message saying you have already posted. The problem is that what I am trying to achieve isn't working it all goes wrong after my else statement. It is also probable that there is an sql injection vulnerability too. Can you help??4 <?php include '../login/dbc.php'; page_protect(); $customerid = $_SESSION['user_id']; $checkid = "SELECT customerid FROM content WHERE customerid = $customerid"; if ($checkid = $customerid) {echo 'You cannot post any more entries, you have already created one';} else $sql="INSERT INTO content (customerid, weburl, title, description) VALUES ('$_POST[customerid]','$_POST[webaddress]','$_POST[pagetitle]','$_POST[pagedescription]')"; if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } echo "1 record added"; ?>

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  • How to access constant defined in child class from parent class functions?

    - by kavoir.com
    I saw this example from php.net: <?php class MyClass { const MY_CONST = "yonder"; public function __construct() { $c = get_class( $this ); echo $c::MY_CONST; } } class ChildClass extends MyClass { const MY_CONST = "bar"; } $x = new ChildClass(); // prints 'bar' $y = new MyClass(); // prints 'yonder' ?> But $c::MY_CONST is only recognized in version 5.3.0 or later. The class I'm writing may be distributed a lot. Basically, I have defined a constant in ChildClass and one of the functions in MyClass (father class) needs to use the constant. Any idea?

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  • 1054 - Unknown column 'apa_calda' in 'where clause'

    - by sebastian
    Hi, I keep getting this error in mysql. Here is the query: SELECT user_id FROM detalii_contor WHERE tip_contor=apa_calda i want to use this query in a php file but it doesn't give any result. so i tried to write it in the sql command prompt. here is what i tried in the php file: $Q = "SELECT id_contor, den_contor FROM detalii_contor WHERE tip_contor='".$contor."'"; $Q = "SELECT id_contor, den_contor FROM detalii_contor WHERE tip_contor='$contor'"; even without "" or without '' i wanted to get $contor from a form, i also tried with $_POST['util'] and {$_POST['util']} i've also tried to set $contor the value i need, but no result. please help. thanks, Sebastian

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  • how to list out the submited data in same page where form submitted?

    - by OM The Eternity
    I have a form with 3 text values and one image.. I want to save these values such that i can display these records in the list below.. how can i do that... I am using osCommerce For example: <form method="post" id="fm-form" action ="" enctype="multipart/form-data"> <label>Name:</label> <input type="text" id="fm-name" name="fm-name" value="" /> <label>Email:</label> <input type="text" id="fm-email" name="fm-email" value="" /> <label>Birthdate:</label> <input type="text" id="fm-birthdate" name="fm-birthdate" value="" /> <input type="file" id="fm-image" name="fm-image"/> <input type="submit" id="fm-submit" value="Save it"> </form> <table width="100%" border="0" cellspacing="0" cellpadding="0"> <tr > <td align="center" class="productListing-heading">Product(s)</td> <td align="center" class="productListing-heading">Edit</td> <td align="center" class="productListing-heading">Delete</td> </tr> <?php for($i=0;$i<$count_image;$i++){?> <tr> <td align="left" class="productListing-data1"> <?php echo tep_image(DIR_WS_IMAGES . $file_realname, $save_image[$i], '110', '110');?> </td> <td align="center" class="productListing-data1">Edit</td> <td align="center" class="productListing-data1">Delete</td> </tr> <tr><td>&nbsp;</td></tr> <?php }?> </table> In the above format as the form is submitted the image has to be stored in a count_image array variable... and the on its count, the list below the form is displayed.. but i cannot get it worked.. could u pls help in doing this...

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  • How is this problem usually solved without using sever-side scripting?

    - by cyggi
    Simple problem I have so far always solved via PHP: You have a site with header, menu, footer and content field. Header, menu and footer are usually the same for each page. How do you, without PHP or any other server-side language, have the header, menu and footer data exist only in one file? So that for example you don't have ten sites (like home.html, products.html, about.html, ..) all having a copy of the static header and menu in their html files. Now if you want to change the header you have to change ten files. I hope I made my question clear enough, if not please leave a comment :)

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  • Does ini_set('session.save_path', 'custom path'); effect the session garbage cleaner?

    - by newbtophp
    Hi! Does ini_set('session.save_path', 'custom path'); effect the session garbage cleaner? As I'm setting a custom directory for the sessions, because I've read from various php security guides, that setting a custom directory on shared hosting for sessions; can improve session security. But the problem is I've read somewhere that PHP does/handles the session garbage cleaning only when the session_save_path is the default and not modified (ie. using a custom directory)? - is this true, if so is their a solution for this?. (take into consideration I'm using shared hosting). Appreciate all help!

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  • Creating CFArray from MySQL Result Array

    - by Andrew
    Is there an easy way to dump an array returned from mysql_fetch_row into a CFArray? (part of the PHP implementation of CFPropertyList) I'm bummed by the lack of documentation on CFPropertyList for PHP. Iterating through each item in the array seems inefficient. I'm open to using a different mysql_fetch_... command. I'd like to just say: $NewArray = new CFArray( $ResultArray ) But that deosn't seem to work. This is my current code: $plist = new CFPropertyList(); $ResultRow = mysqli_fetch_row( $result ); $plist-add( $TableRow = new CFArray() ); foreach ( $ResultRow as $Item ){ $TableRow-add( new CFString( $Item ) ); }

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  • MySQL search help.

    - by Emrul Hasan
    Hi, I am using php/mysql for a search. My table is 'height' and data type = varchar(10) which contains value like (5ft 2in, 5ft 3in,...and so on). While searching I got 2 values - height1 and height2 which are basically the ranges. How can i search in that table with the ranges? say - i will give ranges 5ft 1in to 5ft 10in and want to get the data between those values. I am using php. Please help me about this. thanks.

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  • mysql_real_escape_string & slashes (again, oh yes)

    - by Fizzadar
    Righto, firstly magic quotes & runtime are disabled correctly in php.ini, and confirmed by phpinfo(). PHP version: 5.3.4 MySQL version: 5.1.52 I'm only use mysql_real_escape_string on the data, after htmlspecialchars and a trim, that's all the data cleaning on the variable. Yet, when I submit a single quote, the slash remains in the database. When running mysql_query I'm using "' . $var . '", although in the past this hasn't changed anything (could be due to the double quotes?). Any ideas? and please don't tell me about PDO/prepared statements, I'm aware of them and I have my reasons for doing it this way. Thanks!

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  • How do stop form posting to mysql if database contains a specific ID?

    - by user342391
    I have a form that I am using to post data to mysql. Before submitting the form I want to check the database and see if there are any fields in the column 'customerid' that equal 'userid' and if so not to post the form. Basically, I am trying to limit my users from posting more than once. Users will be able to login to my system and make ONE post. They will be able to delete and modify their post but are only limited to one post. How would I do this??? Code so far: <?php include '../login/dbc.php'; page_protect(); $userid = $_SESSION['user_id']; $sql="INSERT INTO content (customerid, weburl, title, description) VALUES ('$_POST[userid]','$_POST[webaddress]','$_POST[pagetitle]','$_POST[pagedescription]')"; if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } echo "1 record added"; ?>

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  • Secure method for linking to image files uploaded via custom CMS

    - by letseatfood
    How does one provide a direct URL for an image file (JPEG) that is secure? For example, if a PHP script writes uploaded images to directory http://www.somehost.com/images, is it okay to provide http://www.somehost.com/images/someimage.jpg as the URL, or is there a more secure way to do this? Should I look into something like the permalink feature that is a part of Wordpress? I am not sure if that is related. The main reason I ask is that I have a custom PHP/MySQL CMS for managing images. I would like for the client to be able to copy a link to the image they want and then include it in a TinyMCE text editor for inserting the image in their website. Thanks!

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