|
PHP - Query String Problem Is Whooping My Behind..help Please!
this is what i want in the url
when this link is clicked
but i keep getting this error: appreciate any help Similar Tutorialsi 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>"; ?> 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 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 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 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? 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.. 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){
$('#mask,.popupInfo, #popup-box').fadeOut(400);
return false;
}
});
</script>
</head>
<body>
<a href="#popup-box" class="popup-window">Click</a>
<div id="popup-box" class="popupInfo">
<form>
<label>ANYTHING</label></br>
<input type="text" name="random"/></br>
<button type="submit" name="button" class ="close">close</button>
</form>
</div>
</body>
</html>
Edited by chauhanRohit, 27 June 2014 - 09:46 AM. Stumped on this. I copied the syntax of this from another site I'd done (where it works) Then I changed the query, of course. But I'm getting an error, that I'm not sure I understand: The page prints: 451made it to here Warning: mysql_query() expects parameter 1 to be string, resource given in /home/workshop/public_html/register5b.php on line 20 Resource id #4 Warning: mysql_num_rows() expects parameter 1 to be resource, null given in /home/workshop/public_html/register5b.php on line 23 Query failed: And the code is: Code: [Select] <?php if ((isset($_POST['rid'])) && (is_numeric($_POST['rid']))) { $rid = $_POST['rid']; print_r($rid); echo "made it to here"; // connect to database include("x.php"); $query = mysql_query("SELECT tbl_registration.*, tbl_workshops.* FROM tbl_registration_workshop LEFT JOIN tbl_registration ON tbl_registration_workshop.workshop_id = tbl_registration.reg_id LEFT JOIN tbl_workshops ON tbl_registration_workshop.workshop_id = tbl_workshops.workshop_id WHERE tbl_registration_workshop.registration_id = $rid"); $result = mysql_query($query); echo $query; if (mysql_num_rows($result) == 1){ while ($row = mysql_fetch_array($result)) { $workshop = $row['workshop_id']; ?> <!-- from here on, I just print out fields from the data I pull. --> <?php } } }die("Query failed: " . mysql_error()); ?> Hi, When I run the following code... <?php if ($_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'] || $_GET['view'] == 0) { echo 'test'; } ... it outputs 'test' even when view is not set, i.e. when I navigate to index.php without the ?view=. Is there an easy way to force the value to behave as an integer instead of boolean? I have tried putting (int) before, as well as the intval() function, but neither seem to work. Thanks. Hi, My issue here is that I cant get my query string variables to ONLY feed into if/else statement on my secondary page. I include my secondary page (fine.php) from my index page. My query string variables keep being fed back into my original if/ese statement on index.php. here is the if/else on index.php (these links work fine): <?php if($_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] 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 $_GET is not successfully grabbing the parameter. Thanks for response. Hello!
I'm hoping this makes sense.
Basically I have this code:
What it does
1) Looks at the XML
2) Looks for the PIN number in the URL, then finds it in the XML
3) Returns values for the PIN number (which is a profile) allowing me to do echo statements like name, tools, description, etc.
<?php
if (!empty($_GET['pin']))
{
$feedURL = 'http://www.inveroak.co.uk/readerimages/livepanel/91255.xml';
$showOpsWithNoPics = false;
try
{
$xml = simplexml_load_file($feedURL . '?' . time());
$readers = $xml->xpath('/ReaderDetails/Reader[Pin="' . $_GET['pin'] . '"]');
foreach ($readers as $reader)
{
if ($reader->Picture == 'None' && $showOpsWithNoPics == false)
{
$picture = '';
}
elseif ($reader->Picture == 'None' && $showOpsWithNoPics == true)
{
$picture = 'images/defaultpic.jpg';
}
else
{
$picture = $reader->Picture;
}
if ($picture != '')
{
$name = $reader->Name;
$pin = $reader->Pin;
$status = $reader->Status;
$description = $reader->Description;
$arr_skills = '';
$arr_subjects = '';
foreach ($reader->Categories->Category as $Category)
{
foreach ($Category->Skill as $Skill)
{
switch ($Category['name'])
{
case 'Skills':
$arr_skills .= "<li>$Skill</li>";
break;
case 'Subjects':
$arr_subjects .= "<li>$Skill</li>";
break;
}
}
}
$arr_shift = '';
$hourWidth = 20;
$PreviousDate = "";
foreach ($reader->Rota->Shift as $Shift)
{
$Date = $Shift['Date'];
$Day = date("l", strtotime($Date));
$Start = $Shift['Start'];
$Stop = $Shift['Stop'];
if($Stop == '23:59:59' || $Stop == '00:00:00' ){$Stop = '24:00:00';}
$Left = date("H",strtotime($Date.' '.$Start)) * $hourWidth;
$Width = (round(abs(strtotime($Date.' '.$Stop) - strtotime($Date.' '.$Start)) / 60,2)/60)*$hourWidth;
$compare = strcmp ($PreviousDate, $Date);
if ($compare != 0)
{
if ($PreviousDate != '')
{
//this is not the first loop so close off the previous dayrow and hours divs
$arr_shift.= '</div></div>';
}
//create a new day row
$arr_shift .= '<div class="dayrow">';
$arr_shift .= '<div class="day">'.$Day.'</div>';
$arr_shift .= '<div class="hours">';
}
//add the shift object
$arr_shift .= '<div class="shift" title="'. substr($Start,0,5).'-'.substr($Stop,0,5) .'" style="left:'.$Left.'px; width:'.$Width.'px; opacity: 1;"></div>';
//set previouse date to the current shift's date so that we can check if we need to create a new day row next loop
$PreviousDate = $Date;
}
//now we have finshed looping the shifts, either close the last div or display a message to show
//we have no schedule for the op
if ($Date != "")
{
$arr_shift.= '</div></div>';
}
else
{
$arr_shift.= 'We are sorry but they have not scheduled their hours this week. Email for further details.';
}
}
}
}
catch (Exception $e)
{
echo 'cannot display operator profile';
}
}
?>
What I am want to achieve
1) Looks at the XML
2) Looks for a set PIN number in the CODE, then finds it in the XML
3) Returns values for the PIN number (which is a profile).
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, 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. Hi, i have a query string id=5&start=0 that i want to remove from the url and return the url without it. I am searching a cleaner way for doing it. Here is an example :
Before : http://www.example.com/?a=3&id=5&start=0 After : http://www.example.com/?a=3 Before : http://www.example.com/?id=5&start=0 After : http://www.example.com/ Before : http://www.example.com/?alpha=1&id=5&start=0&beta=2 After : http://www.example.com/?alpha=1&beta=2Thank you Edited by Dareros, 11 August 2014 - 04:06 PM. 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. hey @requinix, I know you and I have talked about this before but I just want to clarify this issue. Sadly, I have produced very few websites that have made use of dynamics in the form of database querying. But, the next project that I'm looking at will require it. The last thing I did with a query string involved the recording of downloads of file on a website. and on the page, I had this code: The archive is located he <a href="dl.php?f=archive.zip">archive.zip</a> and the ''dl'' page had this code:
if(!$_GET['f']) error('Missing parameter!');
$stmt = $conn->prepare("INSERT INTO tblDownloads (ip, file, date, time)
VALUES (?, ?, ?, ?)");
$stmt->bind_param("ssss", $ip, $file, $date, $time);
$stmt->execute();
$stmt->close();
$conn->close();
header("Location: archive.zip");
exit;
and all the variables in the above ''dl'' page are declared in a file that is required by it, called ""conn.php"". My question for anyone here is: I am going to be producing a website similar to the testing site called ""jsFiddle"": https://jsfiddle.net/ . And it is slated to have a huge amount of content on it. for example, a main page might have 100 links on it whereby, if a user clicks on any of them, it should take them to another page (or load async content) that illustrates and example of any given issue being asked about. So, I know we've talked about $_GET before Requinix, and how most websites use it to produce dynamic content. And we've also talked about the concept of markdown. But once again, can someone here remind me....what's the best way to go about doing this? Storing the actual ""answer to the issue"" content in a backend database and pulling/displaying it appropriately, kind of like what I am showing I did with the file in the above code? thanks. Adam Edited September 6, 2020 by ajetrumpet |
|||||||