PHP - Query String Causing Problem
i want to change this code :
$result = mysql_query("select count(*) from 3gp where category='Bollywood'"); with this $result = mysql_query("select count(*) from 3gp where category='$category'"); but i am getting error message "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/funxy/public_html/salmobile.info/3gp-videos.php on line 151". I am calling the page using http://abc.com/3gp-videos.php?category=Bollywood Please help. Code: [Select] <?php $db="funxy_db"; $conn = mysql_connect("localhost","funxy_saba","myasco001"); @mysql_select_db($db) or die( "Unable to select database, Please contact your administrator"); class Pager { function getPagerData($numHits, $limit, $page) { $numHits = (int) $numHits; $limit = max((int) $limit, 1); $page = (int) $page; $numPages = ceil($numHits / $limit); $page = max($page, 1); $page = min($page, $numPages); $offset = ($page - 1) * $limit; $ret = new stdClass; $ret->offset = $offset; $ret->limit = $limit; $ret->numPages = $numPages; $ret->page = $page; return $ret; } } // get the pager input values $page = $_GET['page']; $limit = 3; $result = mysql_query("select count(*) from 3gp where category='Bollywood'"); $total = mysql_result($result, 0, 0); // work out the pager values $pager = Pager::getPagerData($total, $limit, $page); $offset = $pager->offset; $limit = $pager->limit; $page = $pager->page; // use pager values to fetch data $query = "select * from 3gp where category='Bollywood' order by id DESC limit $offset, $limit"; $result = mysql_query($query); // use $result here to output page content //my addition //grab all the content //Custom Table Stsrt// $cols = 4; //number of coloms $i =1; echo "<table border=\"0\" cellpadding=\"2\" cellspacing=\"2\" width=\"100%\" id=\"table1\" bordercolor=\"#FFFFFF\" bgcolor=\"#FFFFFF\">" ."<tr>"; while($r=mysql_fetch_array($result)) { //the format is $variable = $r["nameofmysqlcolumn"]; //modify these to match your mysql table columns $id=$r["id"]; $name=$r["name"]; $views=$r["views"]; $url=$r["url"]; $image=$r["image"]; $category=$r["category"]; //display the row $mybox = " <br> <a href='http://salmobile.info/3gp-videos-download.php?id=$id' class=\"classd\"><b><u> $name</u></b></a> <BR> <span class=\"text\"><img src=\"$image\" width=\"60\" height=\"60\"></span> <BR> <span class=\"text2\">Download This 3GP Video</span> <br>"; if (is_int($i / $cols)){ echo "<td width='336' align='center' style=\"border-style: dotted; border-width: 0\">$mybox</td></tr><tr>"; }else{ echo "<td width='336' align='center' style=\"border-style: dotted; border-width: 0\">$mybox</td>"; } if ( $i / $cols == 3) echo "<td colspan='3' align=\"center\"> </td></tr><tr>"; if ( $i / $cols == 8) echo "<td colspan='3' align=\"left\"> </td></tr><tr>"; $i++; //end if }//end while echo "</tr></table>"; //Custom Table End// //ends my addition // output paging system (could also do it before we output the page content) if ($page == 1) // this is the first page - there is no previous page echo "<font color=\"#FFB300\">Previous</font>"; else // not the first page, link to the previous page echo "<a href=\"http://salmobile.info/index-" . ($page - 1) . ".html\" id=\"navigationURL\"><font color=\"#FFB300\">Previous</font></a>"; for ($i = 1; $i <= $pager->numPages; $i++) { echo " <font color=\"#FFFFFF\">|</font> "; if ($i == $pager->page) echo "<font color=\"#FFB300\"> $i</font>"; else echo "<a href=\"http://salmobile.info/index-$i.html\" id=\"navigationURL\"> <font color=\"#FFB300\">$i</font></a>"; } if ($page == $pager->numPages) // this is the last page - there is no next page echo "Next"; else // not the last page, link to the next page echo " <a href=\"http://salmobile.info/index-" . ($page + 1) . ".html\" id=\"navigationURL\"><font color=\"#FFB300\">Next</font></a>"; ?> Similar Tutorials
this is what i want in the url
when this link is clicked
but i keep getting this error: appreciate any help $params = array(); $sqlFilters = ""; if(isset($filtertown)) { $sqlFilters .= " AND town =:town"; $params["town"] = $filtertown; } if(isset($filterinstagram) && !empty($filterinstagram)) { $sqlFilters .= " AND instagram >=:instagram"; $params["instagram"] = $filterinstagram; } if(isset($filtertwitter) && !empty($filtertwitter)) { $sqlFilters .= " AND twitter >=:twitter"; $params["twitter"] = $filtertwitter; } if(isset($filterfacebook) && !empty($filterfacebook)) { $sqlFilters .= " AND facebook >=:facebook"; $params["facebook"] = $filterfacebook; } if(isset($filterpinterest) && !empty($filterpinterest)) { $sqlFilters .= " AND pinterest >=:pinterest"; $params["pinterest"] = $filterpinterest; } if(isset($filteryoutube) && !empty($filteryoutube)) { $sqlFilters .= " AND youtube >=:youtube"; $params["youtube"] = $filteryoutube; } if (isset($userstatus)) { if ($userstatus == "active") { $sqlFilters .=" AND profilecomplete = 'yes'";} if ($userstatus == "incomplete") { $sqlFilters .=" AND profilecomplete IS NULL";} if ($userstatus == "block") { $sqlFilters .=" AND status = 'block'";} if ($userstatus == "cancel") { $sqlFilters .=" AND status = 'cancel'";} if ($userstatus == "showall") { $sqlFilters .=" AND status IS NULL";} } if (isset($search)) { if (isset($searchfirstname)) { $sqlFilters .= " AND firstname LIKE :searchfirstname"; $params["searchfirstname"] = $searchfirstname; } if (isset($searchsurname)) { $sqlFilters .= " AND surname LIKE :searchsurname"; $params["searchsurname"] = $searchsurname; } } if (isset($sort)) { $sqlStart = "SELECT * FROM users WHERE usertype = 'influencer'"; $sqlEnd = " ORDER BY `$col` $order LIMIT $offset, $rowsPerPage;"; $sql = $sqlStart.$sqlFilters.$sqlEnd; $sqlfiltercount = $sqlStart.$sqlFilters; $result = $pdo->prepare($sql); $result->execute($params); $resultfiltercount = $pdo->prepare($sqlfiltercount); $resultfiltercount->execute($params); $num_users = $result->rowCount(); $num_usersfiltered = $resultfiltercount->rowCount(); } else { $sqlStart = "SELECT * FROM users WHERE usertype = 'influencer'"; $sqlEnd = " LIMIT $offset, $rowsPerPage;"; $sql = $sqlStart.$sqlFilters.$sqlEnd; // sqlfiltercount removes the LIMIT to show ALL users on this userstatus filter $sqlfiltercount = $sqlStart.$sqlFilters; $result = $pdo->prepare($sql); $result->execute($params); $num_users = $result->rowCount(); echo "$sqlfiltercount"; $resultfiltercount = $pdo->query($sqlfiltercount); $num_usersfiltered = $resultfiltercount->rowCount(); } Hi there. I am writing a tool that displays a list of users. The page has pagination, set at 99 per page. This script combines the various filters added by the administrator, but if they select Active, Blocked or a few others, it bypasses that and runs the bottom {} brackets of query. You have the $num_users which is the total on screen at the time, plus the secondary count of TOTAL users in the system. This count is then show by 'users' on the page.
This runs fine on my local machine running php7.2, but on the live 7.2 server, it's showing nothing. No errors. If I comment out the $resultfiltercount lines, the page loads, but of course no count. Why might it be getting upset at that, and why would it not throw errors on screen at me? If I hard code the To address such as a@acme.com into my email function, no problems. But if I try using a variable I get an error. The error doesn't give me any valuable information and is generated from some old code that I never wrote. If I place the same variable such as "a@acme.com" into any other field such as content, subject I get the value for the variable so I know that it isn't null. I have tried all the methods of converting to a string and nothing changes. Any ideas? I'm going to create a web page with <iframe> tag & want that the src of <iframe> change as I've made a few attempts at this, but I'm likely not making much sense.... I'm working on how to phrase the question. I had a MySQL database where I used to combine user first AND last names in a single string... I called that string "Member" So if I filtered: www.mysite.com/members.php?Member=Peter It would find and display all Peters: Peter Rabbit and Peter Griffith and Pope Peter. This is what I want. I'm now keeping each name (first and last) in separate strings (as separate variables?), and I am combining them in an array to display on the website: Code: [Select] $First = $First; $Last = $Last; $member = array($Last, $First,); foreach ($Member as $key => $v ) if (!$v) unset ($Member[$key]); $Member = implode(', ', $Member); echo $Member; Now I've lost the ability to search all the members for Peter. So if I filtered: www.mysite.com/members.php?First=Peter ...it would work.... for Peter Rabbit and Peter Griffith, but not Pope Peter I cannot query the entire array? I can no longer check members (first and last names) for Wayne? www.mysite.com/members.php?Member=Peter I'm betting I can, but there's a lot more to it... and 8 hours of Googling haven't helped much. Thanks for listening. ~Wayne Hi there, im trying to have a form show up when user clicks "add joke". I need the variable to be retrieved from the url query string. I cant get the form to show up. I think its either an issue with the GET function at the top or the link down at the bottom. Please help! <?php // If the user wants to add a joke $_GET['addjoke'] = $addjoke; if (isset($addjoke)): ?> <FORM ACTION="<?php echo($PHP_SELF); ?>" METHOD=POST> <P>Type your joke he <BR> <TEXTAREA NAME="joketext" ROWS=10 COLS=40 WRAP> </TEXTAREA><BR> <INPUT TYPE=SUBMIT NAME="submitjoke" VALUE="SUBMIT"> </FORM> <?php else: // Connect to the database server $dbcnx = @mysql_connect("servername", "username", "password"); if (!$dbcnx) { echo( "<P>Unable to connect to the " . "database server at this time.</P>" ); exit(); } // Select the jokes database if (! @mysql_select_db("jhodara2") ) { echo( "<P>Unable to locate the joke " . "database at this time.</P>" ); exit(); } // If a joke has been submitted, // add it to the database. $joketext = $_POST['joketext']; $submitjoke = $_POST['submitjoke']; if ("SUBMIT" == $submitjoke) { $sql = "INSERT INTO jokes SET " . "JokeText='$joketext', " . "JokeDate=CURDATE()"; if (mysql_query($sql)) { echo("<P>Your joke has been added.</P>"); } else { echo("<P>Error adding submitted joke: " . mysql_error() . "</P>"); } } echo("<P> Here are all the jokes " . "in our database: </P>"); // Request the text of all the jokes $result = mysql_query( "SELECT JokeText FROM jokes"); if (!$result) { echo("<P>Error performing query: " . mysql_error() . "</P>"); exit(); } // Display the text of each joke in a paragraph while ( $row = mysql_fetch_array($result) ) { echo("<P>" . $row["JokeText"] . "</P>"); } // When clicked, this link will load this page // with the joke submission form displayed. echo("<P><A HREF='$PHP_SELF?addjoke=1'>Add a Joke!</A></P>"); endif; ?> see the problem live at http://www.freewaycreative.com/insert2.php Greetings, I'm looking for a way to pass a query string (from page1) as part of a query string (to page2) as a single key=>value pair. The idea is the use the query string to return the user to the previous page after the action has been completed. query results[page1]->view record/action selection[page2]->back to results[page1] I'm sure someone has been down this path before. P.S. the script is all contained within one file, thus the filename.ext is already known. Thanks I have seen forums highlight what we searched on google to get to their site. is there a script that would allow me do the same??? Hello everyone. I have a small problem. I might receive a ?aff=## or not on the end of my url when I get a visitor to my website. This depends on if they are sent from a affiliate website or not. If they are it shows fine on the home page but I loss it if they go to another page on my website. I need to keep this ?aff=## information while they look at the other pages. How would I capture and pass this information to my other pages as they surf my site? I tried this to no avail. $aff=$_GET['aff']; I have no clue no adding it back to the next page they go to. Any ideas would be helpful.. Is it possible to add a query string for example some_var=jk84 to any sort of link be it, http://www.website.com , http://www.website.com/?some_id=4, or http://www.website.com/post=45&category=9 or http://www.website.com/somepost/ ? While adding that extra query string how can I make sure I'm not affecting the website's content or causing some script error? Hi guys,
I am using this code to open and close a pop up window, but as soon as i click the close button this
http://localhost/popup.php?random=&button=
automatically adds in the url, Please tell me what is wrong with the script
<script type="text/javascript"> $(document).ready(function(){ $('a.popup-window').click(function(){ var popupBox = $(this).attr('href'); $(popupBox).fadeIn(400); var popMargTop = ($(popupBox).height() + 24)/2; var popMargLeft = ($(popupBox).width() + 24)/2; $(popupBox).css({ 'margin-top' : -popMargTop, 'margin-left' : -popMargLeft }); $('body').append('<div id="mask"></div>'); $('#mask').fadeIn(400); return false; }); $('button.close,#mask').live('click', function(){ $('#mask,.popupInfo').fadeOut(400,function(){ $('#mask').remove(); }); return false; }); }); $(document).keyup(function(e){ if(e.keyCode ==27){ $('#mask,.popupInfo, #popup-box').fadeOut(400); return false; } }); </script> </head> <body> <a href="#popup-box" class="popup-window">Click</a> <div id="popup-box" class="popupInfo"> <form> <label>ANYTHING</label></br> <input type="text" name="random"/></br> <button type="submit" name="button" class ="close">close</button> </form> </div> </body> </html> Edited by chauhanRohit, 27 June 2014 - 09:46 AM. Hi, My issue here is that I cant get my query string variables to ONLY feed into if/else statement on my secondary page. I include my secondary page (fine.php) from my index page. My query string variables keep being fed back into my original if/ese statement on index.php. here is the if/else on index.php (these links work fine): <?php if($_SERVER['QUERY_STRING']=='/index.php' || $_SERVER['QUERY_STRING']=='') { include 'port.php'; } elseif (isset($_GET['pos'])){ include 'pos.php'; } elseif (isset($_GET['web'])){ include 'web.php'; } elseif (isset($_GET['fine'])){ include 'fine.php'; } else {include '404.php';} here is the if/else on my secondary page (fine.php). These links are supposed to alert the if/else in the next table cell. However, they instead alert the if/else in index.php. <td><br/> <a href="?backset"><img src="fine/thumbs/x-backset.jpg" border="0"></a><br/><br/> <a href="?backside"><img src="fine/thumbs/x-backside.jpg" border="0"></a><br/><br/> <a href="?bannerprint"><img src="fine/thumbs/x-bannerprint.jpg" border="0"></a><br/><br/> <a href="?chopu"><img src="fine/thumbs/x-chopu.jpg" border="0"></a><br/><br/> </td> <td><br/> <div id="DivPiece" align="left"> <?PHP if (isset($_GET['backset'])){ include 'fine/backset.php'; } elseif (isset($_GET['backside'])){ include 'fine/backside.php'; } elseif (isset($_GET['bannerprint'])){ include 'fine/bannerprint.php'; } elseif (isset($_GET['chopu'])){ include 'fine/chopu.php'; } ?> </div> </td> How can I get the links on the secondary page to only alert the if/else statement on that page, and BLOCK the if/else statement on index.php from seeing them? I still want to use the query string though. Thanks! [/quote] Okay so i have a URL like this..
https://www.mysite.com/board/profile.php?user=1I want the URL to be like this.. https://www.mysite.com/user/1.htaccess rewrites have always confused me. I want the page to refresh on with the query string, but this isn't working for me... Code: [Select] header("Location: " . $_SERVER['php_self'] . "?" . $_SERVER['query_string'] ); Hi All Thanks in advance for your help. I want to have to following query string Type=myparam&Username=dazd&Password=nk98830&id=0&Cols_Returned=numfrom,sentdata But my code returns the following Type=myparam&Username=dazd&Password=nk98830&id=0&Cols_Returned=%2F%22numfrom%2F%22%2C%2F%22sentdata%2F%22 Below is the code: $data= array( "Type"=> "myparam", "Username" => "dazd", "Password" => "nk98830", "id" => "0", "Cols_Returned" => '/"numfrom/",/"sentdata/"' ) ; //This contains data that you will send to the server. $data = http_build_query($data); //builds the post string ready for posting echo "The Query String is "; echo $data; Regards Hey all, When someone clicks a link, link looks like this: http://site/homes/edit/1?image=true Now on the same page, I have another link, which posts to the edit method as well, but no query string passed: http://site/homes/edit/1 I want to check if image key is set to true and basd on that, determine which view to render for the user, that is, whether to display the edit form to edit post or the image form to edit image. But by using the get array to try to grab query string: Code: [Select] if($_GET["image"]=="true"){ I get the following error: Message: Undefined index: image It appears that $_GET is not successfully grabbing the parameter. Thanks for response. I have this form
<form action="" method="get" name="size"> Størrelse <hr><label><input type="radio" value="9" name="size" onchange="this.form.submit()">92 (2år)</label> <label><input type="checkbox" value="147" name="size" onchange="this.form.submit()">68 (6-9 mdr.) </label><label><input type="checkbox" value="150" name="size" onchange="this.form.submit()">86 (18-24 mdr.)</label> <input type="checkbox" value="149" name="size" onchange="this.form.submit()">80 (12-18 mdr.)</label> <label><input type="checkbox" value="148" name="size" onchange="this.form.submit()">74 (9-12 mdr.)</label> </form>I want to make it so that when the user clicks one checkbox for example size 68 and then checks size 80 I want both sizes to be in the URL query string. But every time the user clicks a new checkbox, the query string changes accordingly. Maybe with a separator like ?size=4:23 and then I will handle it with the DB stuff. Please tell me how to add multiple elements in the query string with this form, thank you! Edited by Stefany93, 21 May 2014 - 08:03 AM. Hi, how do I store as a string (in a variable) a mysql query b/c what I'm doing below outputs Resource id in client browser: Code: [Select] <?php //database connection set up etc $show=mysql_query("SELECT file_Name FROM xdocument WHERE doc_id=95"); print $show; ?> Any help much appreciated, thanks. Thanks in advance for help. I have a PHP script that has a function called traverseXMLNodes and it passes an xml string. I want anyone with a username and password to be able to call this function from his website or application or programming. In short i want my site to act as an API(Application Programming Interface). I am really stuck as to how to go about this. There are other functions i would like to call as well so traverseXMLNodes is not the only function i want to my "API" users to access. Below is the code I am using. the filename for my php i called remote.php. I am using the following query string to test and no results are returned. http://localhost/testfolder/remote.php?xmlstring=20 Code: [Select] <?php $allowedfuncs = array('xmlstring','remotefunction2'); if(in_array($_get['xmlstring'], $allowedfuncs)) { $func = $_get['xmlstring']; $output = traverseXMLNodes($func); // This calls $_get['funccall']() echo $output; } else { exit(); } function traverseXMLNodes ($func) { echo "The XML string is"; echo "<br/>"; echo $func; echo "<br/>"; } /* More functions go here */ ?> Hi, When I run the following code... <?php if ($_GET['view']) { echo 'test'; } .. and then navigate to index.php?view=1 the above code outputs 'test', as intended. The same goes for any non-zero value after the ?view= However, navigating to index.php?view=0 does not work as intended because it assumes that the 0 is FALSE. So if I modify the code, like this: <?php if ($_GET['view'] || $_GET['view'] == 0) { echo 'test'; } ... it outputs 'test' even when view is not set, i.e. when I navigate to index.php without the ?view=. Is there an easy way to force the value to behave as an integer instead of boolean? I have tried putting (int) before, as well as the intval() function, but neither seem to work. Thanks. |