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  • Parameter error with Mysqli

    - by Morgan Green
    When I run this Query I recieve Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /home/morgan58/public_html/wow/includes/index/index_admin.php on line 188 SELECT * FROM characters WHERE id=5 Warning: mysqli_free_result() expects parameter 1 to be mysqli_result, boolean given in /home/morgan58/public_html/wow/includes/index/index_admin.php on line 194 The Query is running and it is strying to select the correct information, but for on the actual output it's giving me a fetch_array error; if anyone can see where the error lies it'd be much appreciated. Thank you. <?php $adminid= $admin->get_id(); $characterdb= 'characters'; $link = mysqli_connect("$server", "$user", "$pass", "$characterdb"); if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } $query = "SELECT * FROM characters WHERE id=$adminid"; $result = mysqli_query($link, $query); while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { echo $query; echo $row['name']; } mysqli_free_result($result); mysqli_close($link); ?>

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  • post values to external page explicitly PHP

    - by JPro
    Hi, I want to post values to a page through a hyperlink in another page. Is it possible to do that? Let's say I have a page results.php which has the starting line, <?php if(isset($_POST['posted_value'])) { echo "expected value"; // do something with the data } else { echo "no the value expected"; } If from another page say link.php I place a hyperlink like this: <a href="results.php?posted_value=1"> , will this be accepetd by the results page? If instead if I replace the above starting line with if(isset($_REQUEST['posted_value'])), will this work? I believe the above hyperlink evaluates to GET, but since the only visibility difference between GET and POST that is you can see parameters in the address bar with GET But, is there any other way to place a hyperlink which can post values to a page? or can we use jquery in the place of hyperlink to POST the values? Can anyone please suggest me something on this please? Thanks.

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  • Trouble defining method for Javascript class definition.

    - by btoverdrive
    I'm somewhat new to object oriented programming in Javascript and I'm trying to build a handler object and library for a list of items I get back from an API call. Ideally, I'd like the library functions to be members of the handler class. I'm having trouble getting my class method to work however. I defined as part of the class bcObject the method getModifiedDateTime, but when I try to echo the result of the objects call to this method, I get this error: Error on line 44 position 26: Expected ';' this.getModifiedDateTime: function(epochtime) { which leads me to believe that I simply have a syntax issue with my method definition but I can't figure out where. response( { "items": [ {"id":711,"name":"Shuttle","lastModifiedDate":"1268426336727"}, {"id":754,"name":"Formula1","lastModifiedDate":"1270121717721"} ], "extraListItemsAttr1":"blah", "extraListItemsAttr2":"blah2" }); function response(MyObject) { bcObject = new bcObject(MyObject); thing = bcObject.getModifiedDateTime(bcObject.videoItem[0].lastModifiedDate); SOSE.Echo(thing); } function bcObject(listObject) { // define class members this.responseList = {}; this.videoCount = 0; this.videoItem = []; this.responseListError = ""; // instantiate members this.responseList = listObject; this.videoCount = listObject.items.length; // populate videoItem array for (i=0;i<this.videoCount;i++) { this.videoItem[i] = listObject.items[i]; } this.getModifiedDateTime: function(epochtime) { var dateStringOutput = ""; var myDate = new Date(epochtime); dateStringOutput = myDate.toLocaleString(); return dateStringOutput; }; }

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  • PHP: Need a double check on an error in this small code

    - by Josh K
    I have this simple Select box that is suppose to store the selected value in a hidden input, which can then be used for POST (I am doing it this way to use data from disabled drop down menus) <body> <?php $Z = $_POST[hdn]; ?> <form id="form1" name="form1" method="post" action="test.php"> <select name="whatever" id="whatever" onchange="document.getElementById('hdn').value = this.value"> <option value="1">1Value</option> <option value="2">2Value</option> <option value="3">3Value</option> <option value="4">4Value</option> </select> <input type="hidden" name ='hdn' id="hdn" /> <input type="submit" id='submit' /> <?php echo "<p>".$Z."</p>"; ?> </form> </body> The echo call works for the last 3 options (2,3,4) but if I select the first one it doesnt output anything, and even if i change first one it still doesnt output anything. Can someone explain to me whats going on, I think it might be a syntax issue.

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  • How to get values from SQL query made by php?

    - by Ole Jak
    So I made a query like this global $connection; $query = "SELECT * FROM streams "; $streams_set = mysql_query($query, $connection); confirm_query($streams_set); in my DB there are filds ID, UID, SID, TIME (all INT type exept time) So I am triing to print query relult into form <form> <select class="multiselect" multiple="multiple" name="SIDs"> <?php global $connection; $query = "SELECT * FROM streams "; $streams_set = mysql_query($query, $connection); confirm_query($streams_set); $streams_count = mysql_num_rows($streams_set); for ($count=1; $count <= $streams_count; $count++) { echo "<option value=\"{$count}\""; echo ">{$count}</option>"; } ?> </select> <br/> <input type="submit" value="Submit Form"/> </form> How to print out as "option" "values" SID's from my sql query?

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  • Delete file using a link

    - by user329394
    Hi all, i want to have function like delete file from database by using link instead of button. how can i do that? do i need to use href/unlink or what? Can i do like popup confirmation wther yes or no. i know how to do that, but where should i put the code? this is the part how where system will display all filename and do direct upload. Beside each files, there will be a function for 'Remove': $qry = "SELECT * FROM table1 a, table2 b WHERE b.id = '".$rs[id]."' AND a.ptkid = '".$rs[id]."' "; $sql = get_records_sql($qry); foreach($sql as $rs){ ?> <?echo '<a href="download.php?f='.$rs->faillampiran.'">'. basename($rs->faillampiran).'</a>'; ?><td><?echo '<a href=""> [Remove]</a>';?></td><? ?><br> <? } ?> thankz all

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  • PHP rewrite an included file - is this a valid script?

    - by Poni
    Hi all! I've made this question: http://stackoverflow.com/questions/2921469/php-mutual-exclusion-mutex As said there, I want several sources to send their stats once in a while, and these stats will be showed at the website's main page. My problem is that I want this to be done in an atomic manner, so no update of the stats will overlap another one running in the background. Now, I came up with this solution and I want you PHP experts to judge it. stats.php <?php define("my_counter", 12); ?> index.php <?php include "stats.php"; echo constant("my_counter"); ?> update.php <?php $old_error_reporting = error_reporting(0); include "stats.php"; define("my_stats_template",' <?php define("my_counter", %d); ?> '); $fd = fopen("stats.php", "w+"); if($fd) { if (flock($fd, LOCK_EX)) { $my_counter = 0; try { $my_counter = constant("my_counter"); } catch(Exception $e) { } $my_counter++; $new_stats = sprintf(constant("my_stats_template"), $my_counter); echo "Counter should stand at $my_counter"; fwrite($fd, $new_stats); } flock($fd, LOCK_UN); fclose($fd); } error_reporting($old_error_reporting); ?> Several clients will call the "update.php" file once every 60sec each. The "index.php" is going to use the "stats.php" file all the time as you can see. What's your opinion?

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  • A stupid question about wordpress and php

    - by bubdada
    It may seem stupid question, but i've a serious problem... if you could check out orcik.net the thumbnail images does not appear. I figured out the reason but I don't know how to solve.. http://orcik.net/projects/thumb/orcikthumb.php?src=http://orcik.net/wp-content/uploads/2010/05/mac-safari-search-cache.png If you go to the above link you will get page not found error. However, if you go to the link below you'll get the thumbnail version of the image... http://orcik.net/projects/thumb/orcikthumb.php?src=/wp-content/uploads/2010/05/mac-safari-search-cache.png I'm using this piece of code on wordpress and the line appears like <a href="<?php the_permalink() ?>" rel="bookmark"> <img src="<?php bloginfo('template_directory'); ?>/includes/orcikthumb.php?src=<?php get_thumbnail($post->ID, 'full'); ?>&amp;h=<?php echo get_theme_mod($height); ?>&amp;w=<?php echo get_theme_mod($width); ?>&amp;zc=1" alt="<?php the_title(); ?>" /> </a> Thus, I believe I can't change the directory of image. But I could not figure out why I am getting page not found error. Is that might be CHMOD'es??? or something else?? Thanks

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  • How to write data option in jQuery.ajax() function when it include in a mysql_query?

    - by cj333
    I modify a php comment system. I want add it after every article witch are query from database. this is the php part <?php ... while($result = mysql_fetch_array($resultset)) { $article_title = $result['article_title']; ... ?> <form id="postform" class="postform"> <input type="hidden" name="title" id="title" value="<?=$article_title;?>" /> <input type="text" name="content" id="content" /> <input type="button" value="Submit" class="Submit" /> </form> ... <?php } ?> this is the ajax part. $.ajax({ type: "POST", url: "ajax_post.php", data: {title:$('#title').val(), content:$('#content').val() ajax_post.php echo $title; echo $content; How to modify the ajax data part that each article's comment can send each data to the ajax_post.php? thanks a lot.

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  • How to get aggregate days from PHP's DateTime::diff?

    - by razic
    $now = new DateTime('now'); $tomorrow = new DateTime('tomorrow'); $next_year = new DateTime('+1 year'); echo "<pre>"; print_r($now->diff($tomorrow)); print_r($now->diff($next_year)); echo "</pre>"; DateInterval Object ( [y] => 0 [m] => 0 [d] => 0 [h] => 10 [i] => 17 [s] => 14 [invert] => 0 [days] => 6015 ) DateInterval Object ( [y] => 1 [m] => 0 [d] => 0 [h] => 0 [i] => 0 [s] => 0 [invert] => 0 [days] => 6015 ) any ideas why 'days' shows 6015? why won't it show the total number of days? 1 year difference means nothing to me, since months have varying number of days.

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  • problem when uploading file

    - by Syom
    i have the form, and i want to upload two files. here is the script <form action="form.php" method="post" enctype="multipart/form-data" /> <input type="file" name="video" /> <input type="file" name="picture" > <input type="submit" class="input" value="?????" /> <input type="hidden" name="MAX_FILE_SIZE" value="100000000" /> </form> form.php: <? print_r($_FILES); $video_name = $_FILES["video"]["name"]; $image_name = $_FILES["picture"]["name"]; echo "video",$video_name; echo "image",$image_name; //returns Array ( ) videoimage ?> when i try to upload the file greater than 10MB, it doesn't happen. i try in many browsers. maybe i must change some field in php.ini? but i haven't permission to change them on the server. so what can i do? thanks

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  • Javascript variables are not working

    - by linkcool
    Hi, my problem is that my variables are not working in javascript. all variables need names without some character at the beginning, this is the stupid thing...Anyways, im trying to make a funtion that makes "select all checkboxes". It is not working so i looked at the page source/info and found out that the variables were not changing. this is my input: echo "<input onclick='checkAll(1);' type='checkbox' name='master'/><br/>"; My function: function checkAll(i) { for(var i=1; i < <?php echo $num; ?>; i++) { if(document.demo.master[i].checked == true) { document.demo.message[i].checked = true; } else { document.demo.message[i].checked = false; } } } so yes that's it. I can tell you that i also tried without the <i> in: checkAll("i") Thanks for the help.

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  • Not getting an array as result (calling a webservice by AJAX-JSON)

    - by Pasargad
    I'm trying to get the result of my web service as an array and then loop over the result to fetch all of the data; what I have done so far: In my web service when I return the result I use return json_encode($newFiles); and the result is as following: "[{\"path\":\"c:\\\\my_images\\\\123.jpg\",\"ID\":\"123\",\"FName\":\"John\",\"LName\":\"Brown\",\"dept\":\"Hr\"}]" tehn in my Web application I'm calling the rest web service by the following code in the RestService class: public function getNewImages($time) { $url = $this->rest_url['MyService'] . "?action=getAllNewPhotos&accessKey=" . $this->rest_key['MyService'] . "&lastcheck=" . $time; $ch = curl_init(); curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); $data = curl_exec($ch); if ($data) { return json_decode($data); } else { return null; } } and then in my controller I have the following code: public function getNewImgs($time="2011-11-03 14:35:08") { $newImgs = $this->restservice->getNewImages($time); echo json_encode$newImgs; } and I'm calling this `enter code here`controller method by AJAX: $("#searchNewImgManually").click(function(e) { e.preventDefault(); $.ajax({ type: "POST", async: true, datatype: "json", url: "<?PHP echo base_url("myProjectController/getNewImgs"); ?>", success: function(imgsResults) { alert(imgsResults[0]); } }); }); but instead of giving me the first object it is just giving me quotation mark (the first charachter of the result) " Why is that? I'm passing in JSON format and in AJAX I mentioned datatype as "JSON" ! Please let me know if you need more clarification! Thanks :)

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  • Lost in Nodester Installation

    - by jslamka
    I am trying to install my own version of Nodester. I have tried on Ubuntu 12.04 LTS and now with CentOS. I am not the most skilled Linux user (~2 months use) so I am at a loss at this point. The instructions are located at https://github.com/nodester/nodester/wiki/Install-nodester#wiki-a. They ask you to "export paths (to make npm work)" with the lines necessary to accomplish this. cd ~ echo -e "root = ~/.node_libraries\nmanroot = ~/local/share/man\nbinroot = ~/bin" > ~/.npmrc echo -e "export PATH=3d9c7cfd35d3628e0aa233dec9ce9a44d2231afcquot;\${PATH}:~/bin3d9c7cfd35d3628e0aa233dec9ce9a44d2231afcquot;;" >> ~/.bashrc source ~/.bashrc I can accomplish all of this until I get to the source ~/.bashrc line. When I run that, I get the following: [root@MYSERVER ~]# source ~/.bashrc -bash: /root/.bashrc: line 13: syntax error near unexpected token ';;' -bash: /root/.bashrc: line 13: 'export PATH=3d9c7cfd35d3628e0aa233dec9ce9a44d2231afcquot;${PATH}:~/bin3d9c7cfd35d3628e0aa233dec9ce9a44d2231afcquot;; I have tried changing the quot; to " and that didn't help. I tried changing quot; to colons and that didn't help. I also removed that and it didn't help (I am sure many of you at this point are probably wondering why I would even try those things). Does anyone have any insight as to what I need to do to get this to run properly?

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  • How to process a large post array in PHP where item names are all different and not known in advance

    - by Salnajjar
    I have a PHP page that queries a DB to populate a form for the user to modify the data and submit. The query returns a number of rows which contain 3 items: ImageID ImageName ImageDescription The PHP page titles each box in the form with a generic name and appends the ImageID to it. Ie: ImageID_03 ImageName_34 ImageDescription_22 As it's unknown which images are going to have been retrieved from the DB then I can't know in advance what the name of the form entries will be. The form deals with a large number of entries at the same time. My backend PHP form processor that gets the data just sees it as one big array: [imageid_2] => 2 [imagename_2] => _MG_0214 [imageid_10] => 10 [imagename_10] => _MG_0419 [imageid_39] => 39 [imagename_39] => _MG_0420 [imageid_22] => 22 [imagename_22] => Curly Fern [imagedescription_2] => Wibble [imagedescription_10] => Wobble [imagedescription_39] => Fred [imagedescription_22] => Sally I've tried to do an array walk on it to split it into 3 arrays which set places but am stuck: // define empty arrays $imageidarray = array(); $imagenamearray = array(); $imagedescriptionarray = array(); // our function to call when we walk through the posted items array function assignvars($entry, $key) { if (preg_match("/imageid/i", $key)) { array_push($imageidarray, $entry); } elseif (preg_match("/imagename/i", $key)) { // echo " ImageName: $entry"; } elseif (preg_match("/imagedescription/i", $key)) { // echo " ImageDescription: $entry"; } } array_walk($_POST, 'assignvars'); This fails with the error: array_push(): First argument should be an array in... Am I approaching this wrong?

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  • PHP class extends not working why and is this how to correctly extend a class?

    - by Matthew
    Hi so I'm trying to understand how inherteince works in PHP using object oriented programming. The main class is Computer, the class that is inheriting is Mouse. I'm extedning the Computer class with the mouse class. I use __construct in each class, when I istinate the class I use the pc type first and if it has mouse after. For some reason computer returns null? why is this? class Computer { protected $type = 'null'; public function __construct($type) { $this->type = $type; } public function computertype() { $this->type = strtoupper($this->type); return $this->type; } } class Mouse extends Computer { protected $hasmouse = 'null'; public function __construct($hasmouse){ $this->hasmouse = $hasmouse; } public function computermouse() { if($this->hasmouse == 'Y') { return 'This Computer has a mouse'; } } } $pc = new Computer('PC', 'Y'); echo $pc->computertype; echo $pc->computermouse;

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  • why my pagination link doesnt appear ?

    - by udaya
    This is my script which i have used to paginate ,,The datas are restricted to 4 but the pagination link doesn't appear <? require_once ('Pager/Pager.php'); $connection = mysql_connect( "localhost" , "root" , "" ); mysql_select_db( "ssit",$connection); $result=mysql_query("SELECT dFrindName FROM tbl_friendslist", $connection); $row = mysql_fetch_array($result); $totalItems = $row['total']; $pager_options = array( 'mode' => 'Sliding', // Sliding or Jumping mode. See below. 'perPage' => 4, // Total rows to show per page 'delta' => 4, // See below 'totalItems' => $totalItems, ); $pager = Pager::factory($pager_options); echo $pager->links; list($from, $to) = $pager->getOffsetByPageId(); $from = $from - 1; $perPage = $pager_options['perPage']; $result = mysql_query("SELECT * FROM tbl_friendslist LIMIT 5 , $perPage",$connection); while($row = mysql_fetch_array($result)) { echo $row['dFrindName'].'</br>'; } ?>

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  • Image error with wordpress and php

    - by bubdada
    It may seem stupid question, but i've a serious problem... if you could check out orcik.net the thumbnail images does not appear. I figured out the reason but I don't know how to solve.. http://orcik.net/projects/thumb/orcikthumb.php?src=http://orcik.net/wp-content/uploads/2010/05/mac-safari-search-cache.png If you go to the above link you will get page not found error. However, if you go to the link below you'll get the thumbnail version of the image... http://orcik.net/projects/thumb/orcikthumb.php?src=/wp-content/uploads/2010/05/mac-safari-search-cache.png I'm using this piece of code on wordpress and the line appears like <a href="<?php the_permalink() ?>" rel="bookmark"> <img src="<?php bloginfo('template_directory'); ?>/includes/orcikthumb.php?src=<?php get_thumbnail($post->ID, 'full'); ?>&amp;h=<?php echo get_theme_mod($height); ?>&amp;w=<?php echo get_theme_mod($width); ?>&amp;zc=1" alt="<?php the_title(); ?>" /> </a> Thus, I believe I can't change the directory of image. But I could not figure out why I am getting page not found error. Is that might be CHMOD'es??? or something else?? Thanks

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  • Array values changing unexpectedly

    - by Lizard
    I am using cakephp 1.2 and I have an array that appears to have a value change even though that variable is not being manipulated. Below is the code to that is causing me trouble. PLEASE NOTE - UPDATE Changing the variable name makes no difference to the outcome, The values get changed somewhere between the two print_r calls, and I can't see why the $this-find would do this . echo "Start of findCountByString()"; print_r($myArr); $test = $this->find('count', array( 'conditions' => $conditions, 'joins' => array('LEFT JOIN `articles_entities` AS ArticleEntity ON `ArticleEntity`.`article_id` = `Article`.`id`'), 'group' => 'Article.id' )); echo "End of findCountByString()"; print_r($myArr); I am getting the following output: Start of findCountByString() Array ( [0] => 4bdb1d96-c680-4c2c-aae7-104c39d70629 [1] => 4bdb1d6a-9e38-479d-9ad4-105c39d70629 [2] => 4bdb1b55-35f0-4d22-ab38-104e39d70629 [3] => 4bdb25f4-34d4-46ea-bcb6-104f39d70629 ) End of findCountByString() Array ( [0] => 4bdb1d96-c680-4c2c-aae7-104c39d70629 [1] => 4bdb1d6a-9e38-479d-9ad4-105c39d70629 [2] => 4bdb1b55-35f0-4d22-ab38-104e39d70629 [3] => '4bdb25f4-34d4-46ea-bcb6-104f39d70629' # This is now in inverted commas ) The the value in my array have changed, and I don't know why? Any suggestions?

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  • Header redirect and Session start() generating errors

    - by RPK
    I am using following code. I get errors: Warning: Cannot modify header information - headers already sent by (output started at /home/public_html/mc/cpanel/Source/verifylogin.php:11) Warning: session_start() [function.session-start]: Cannot send session cookie - headers already sent by (output started at /home/public_html/mv/cpanel/Source/verifylogin.php:11) Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at /home/public_html/mv/cpanel/Source/verifylogin.php:11) <?php error_reporting(E_ALL^ E_NOTICE); ob_start(); require("../Lib/dbaccess.php"); $inQuery = "SELECT mhuserid, mhusername FROM cpanelusers WHERE mhusername = '". $_POST['UserName'] ."' AND mhpassword = '". hash('sha512', $_POST['Password']) ."'"; try { $Result = dbaccess::GetRows($inQuery); echo $Result; $NumRows = mysql_num_rows($Result); if ($NumRows > 0) { header("Location: http://www.example.com/cpanel/mainwindow.php"); session_start(); } else { header("Location: http://www.example.com/cpanel/"); echo "Last login attempt failed."; exit; } } catch(exception $e) { } ob_clean(); ?>

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  • PHP session token can be used multipletimes?

    - by kornesh
    I got page A which is a normal HTML page and page which is an AJAX response page. And I want to prevent CSRF attacks by tokens. Lets say I use this method for an autocomplete form, is it possible to use same token multiple times (of course the session is only set one time) because i tired this method but the validation keep failing after the first suggestion (obviously the token has changed, somehow) page A <?php session_start(); $token = md5(uniqid(rand(), TRUE)); $_SESSION['token'] = $token; ?> <input id="token" value="<?php echo $token; ?>" type="hidden"></input> <input id="autocomplete" placeholder="Type something"></input> .... The form is autosubmitted every time theres a change using Jquery. page B <?php session_start(); if($_REQUEST['token'] == $_SESSION['token']){ echo 'Im working fine'; } ?>

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  • PHP: Trying to come up with a "prev" and "next" link

    - by fwaokda
    I'm displaying 10 records per page. The variables I have currently that I'm working with are.. $total = total number of records $page = whats the current page I'm displaying I placed this at the top of my page... if ( $_GET['page'] == '' ) { $page = 1; } //if no page is specified set it to `1` else { $page = ($_GET['page']); } // if page is specified set it Here are my two links... if ( $page != 1 ) { echo '<div style="float:left" ><a href="index.php?page='. ( $page - 1 ) .'" rev="prev" >Prev</a></div>'; } if ( !( ( $total / ( 10 * $page ) ) < $page ) ) { echo '<div style="float:right" ><a href="index.php?page='. ( $page + 1 ) .'" rev="next" >Next</a></div>'; } Now I guess (unless I'm not thinking of something) that I can display the "Prev" link every time except when the page is '1'. How can make it where the "Next" link doesn't show on the last page though?

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  • How to add sub category on my codeigniter php application ?

    - by sagarmatha
    Hello friends I have a view echo "<label for='parent'>Category</label><br/> "; echo form_dropdown('category_id', $categories). "<p>"; controller function create(){ if($this->input->post('name')){ $this->MProducts->addProduct(); $this->session->set_flashdata('message', 'Products Created'); redirect('admin/products/index', 'refresh'); }else{ $data['title'] = "Create Product"; $data['main'] = 'admin_product_create'; $data['categories']= $this->MCats->getTopCategories(); $this->load->vars($data); $this->load->view('dashboard'); } } and the model is function getTopcategories(){ $data = array(); $data[0] = 'root'; $this->db->where('parentid',0); $Q = $this->db->get('categories'); if($Q->num_rows() > 0){ foreach($Q->result_array() as $row){ $data[$row['id']] = $row['name']; } } $Q->free_result(); return $data; } Basically, what i want is we get sub categories when we click categories and 'selected' subcategory id goes to database when we create products, So that we can list products by sub categories. Please help me how do we do that ?

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  • wp+sql+image not goin in the folder

    - by happy
    this is my code for uploading image in database but image are going to the desird forlder...but when i m tryin to retrieve the images to diaplay,,they are not displayed..anyone help me...... $category=$_POST['category']; $uploadDir = 'D:/xampp/htdocs/js/wordpress/wp-content/plugins/img/imagess/ '; $fileName = $_FILES['Photo']['name']; $tmpName = $_FILES['Photo']['tmp_name']; $fileSize = $_FILES['Photo']['size']; $fileType = $_FILES['Photo']['type']; $filePath = $uploadDir . $fileName; $result = move_uploaded_file($tmpName,$filePath); if (!$result) { echo "Error uploading file"; exit; } if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); $filePath = addslashes($filePath); } global $wpdb; //$insert=$wpdb->insert('images',array('image_name'=>$filePath,'cat_name'=>$category),array('%b','%s')); $insert=$wpdb->insert('images',array('image_name'=>$filePath,'cat_name'=>$category)); $wpdb->insert('categories',array('cat_name'=>$category)); echo "Successfully Submitted";

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  • Textbox auto generate by php and html

    - by user2892997
    i have few field of data such as product , amount and barcode.The system just show 1 row of data insert form that contain the 3 field.when i completed insert the first row, then second row of textbox will auto generate, i can do it by microsoft access, can i do so for php ? <?php $i=0; ?> <form method="post" action=""> <input type="text" name="<?php echo $i; ?>" /> </form> <?php if(isset($_POST[$i])){ $i++; ?> <form method="post" action=""> <input type="text" name="<?php echo $i; ?>" /> </form> <?php }?> it work for the first and second textbox, but how can i continue to create more textbox accordingly?

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