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PHP - Executing A Query Without Page Reloading
hey guys im sure this is possible im wondering how a user can execute a mysql query by a click of a button but without the page reloading...thanks guys
Similar TutorialsWhat is the best way to display the next record on a web page, from a list of records, without executing the query over and over again each time you display the next record? For example, imagine my user has a list of messages displayed on a web page, and the user filters that list to display all messages with the title "Test", and let's say that returns 12 records. He then clicks on record 1 to read that message. Once he is done reading it, he then wants to read msg 2, and then msg 3 etc. He could hit the Back button in his browser and click on the next record from the list, but it would be more useful to just click on a link that says "Next" which displays the next message in the filtered list of messages (without having to go back to the list). A lot of web sites do this sort of thing, but what is the best way to do it? Do you have to run the original query again each time a new message is loaded, so you know which message is next, or can you somehow store the query to save the server from having to run the same query over and over again? If a user is viewing a list of pictures for example, they could be flicking through the pictures quite quickly, and I would not like to have the server running the same query over and over again every few seconds when it doesn't need to. Advice? (Thanks.) Hi, I know this has to be possible but I have been unable to get this to work properly. I have a page with two forms. The first form is a list of sources associated with a subject, with checkboxes, so that it is possible to remove these associations, like so: Code: [Select] $qs = "SELECT s.source_id, s.source_name from source s, source_subject ss, subject sb where s.source_id = ss.source_id and sb.subject_id = ss.subject_id and sb.subject_id = $subject_id order by source_name"; $rs = mysqli_query($dbc, $qs) or die ('Error querying database'); while($row = mysqli_fetch_array($rs)) { echo '<input type="checkbox" value="' . $row['source_id'] . '" name="markdelete[]">' . $row['source_name'] . '<br />' ; } The second form is a list of sources that are not associated with this subject, with checkboxes enabling the addition of more sources to this subject, like so: Code: [Select] $r = "SELECT source_id, source_name FROM source WHERE source_id NOT IN (select s.source_id from source s, source_subject ss where s.source_id = ss.source_id and ss.subject_id = '$subject_id')"; $r1 = mysqli_query($dbc, $r) or die ('Update Error: '.mysqli_error($dbc)); while ($row = mysqli_fetch_array($r1)) { echo '<input type="checkbox" id="source_id" name="source_id[]" ' ; echo 'value="'. $row['source_id'] .'"'; echo '> ' . $row['source_name'] . '<br />'; } The first form works fine... when the "remove" submit button is clicked, it successfully removes any selected sources and then shows the remaining sources still associated, and then automatically populates this source in the below list (form 2) as one that is not selected. On the second form, when adding a new source to the subject, it successfully enters the data into the database, and removes this source from the list of unselected sources, but it does not seem to re-populate in the first form. The data is correct on the tables, but the page needs to reload the first query... how do I get it to do this? I'd like to be able to make this page editable, and re-editable (in case someone makes a mistake) without having to go back and reload the entire page. Note, both forms call the page itself; not sure if this is part of it. Using this: Code: [Select] <form method="POST" action="<?php echo $_SERVER['PHP_SELF'].'?subject_id='.$subject_id ; ?>"> Does anyone have any ideas? Hello, I have links at above of the page. if i click a link a new page will appear. The menu links above are same for all pages. Which is better reloading the whole page with all menu links or keeping the above menu link permanent/unchange reloading bottom division with ajax? I mean only one page will contain the menu.Is it better? Or every page will contain the menu above.Is it better then the first one? I say many web site they have link menus above if i click a link from above it seems the whole page reloads or chages. For example:http://www.sitepoint.com/ Thank you. I have the following query where $userid is about 100,000 userids separated by commas. Code: [Select] <? $getpostid=mysql_query("SELECT DISTINCT post.username,post.threadid,thread.threadid,thread.forumid from post,thread WHERE post.username='$username' AND post.threadid=thread.threadid AND post.userid IN ($userid) AND thread.forumid IN ($forumids)") or die (mysql_error()); ?>can i just print all the usernames from above result separated by ';' without looping it multiple times? Could someone help me out how to do above in an efficient manner? Hi, In some cases I need to execute a default query , so I am storing the SELECT statement in a variable like this and executing it: Code: [Select] $default_query="SELECT * from user where userid='$userid'"; $user_query=mysql_query($default_query); The above code returns an empty results. But if I execute it without the variable it works fine. Code: [Select] $user_query=mysql_query("SELECT * from user where userid='$userid'"); Am I syntatically wrong somewhere? Im trying to execute an update sql query, but its not working. IDK if I got the right query tho what i want to do is, when the form is submitted, subtract 15 from the value 'creditleft'. So say the value is 20 , i want to subtract 15 and get 5. Here is the sql query im using; $query2="update userdata set creditleft = - 15 where username = '".$_SESSION['login_name']."'"; mysql_query($query2, $link); I dont get any errors, just no output I am facing problem to execute query by assigning NULL value to a variable and then executing query.In MySQL DB four fields Mobile,landline, pincode,dob are set as integer and date(for dob) respectively.The default is set as NULL and NULL option is selected as yes.All these fields are not mandatory.The problem is that when I edit the form my keeping the value as empty in DB these are saved as 0, 0 , 0 & 0000-00-00 inspite of Null. I have tried everything but still the defect persist. Please help me to come out of the problem The code, I have used: <?php //require_once 'includes/config.php'; $dbusername = $_POST['email']; $dbfirstname = $_POST['first_name']; $dblastname = $_POST['last_name']; //$dbmobile_number = $_POST['mobile']; if (isset($_POST['mobile'])) { $dbmobile_number = $_POST['mobile']; } else { $dbmobile_number = "NULL"; } $dblandline_number = $_POST['landline']; $dbdob = $_POST['dob']; if(isset($_POST['is_email'])) { $dbSubscribe_Email_Alert = '1'; } else { $dbSubscribe_Email_Alert = '0'; } if(isset($_POST['is_sms'])) { $dbSubscribe_SMS = 0; } else { $dbSubscribe_SMS = 0; } $dbAddress_firstname = $_POST['shipping_first_name']; $dbAddress_lastname = $_POST['shipping_last_name']; $dbAddress = $_POST['shipping_address']; $dbcity = $_POST['shipping_city']; $dbpincode = $_POST['shipping_pincode']; $dbstate = $_POST['shipping_state']; $dbcountry = $_POST['shipping_country']; echo "Welcome".$dbusername; //if($_POST['btnSave']) //if ($_POST['btnSave']) //{ //echo "Inside query loop"; $connect = mysql_connect("localhost","root","") or die("Couldn't connect!"); mysql_select_db("salebees") or die ("Couldn't find DB"); //$query = mysql_query("SELECT * FROM users WHERE username='$username'"); $query = mysql_query("update users set firstname = '$dbfirstname', lastname = '$dblastname', mobile_number = '$dbmobile_number', landline_number = '$dblandline_number', dob = '$dbdob', Subscribe_Email_Alert = '$dbSubscribe_Email_Alert', Subscribe_SMS = '$dbSubscribe_SMS', Address_firstname = '$dbAddress_firstname', Address_lastname = '$dbAddress_lastname', Address = '$dbAddress', city = '$dbcity', pincode = '$dbpincode', state = '$dbstate', country = '$dbcountry' where username = '$dbusername' "); header("location:my_account.php"); //} //else //{ //die(); //} ?> I have a php generated image: (number.txt contains a number 1 or 2) pic.php Code: [Select] <?php // Set the content-type header('Content-type: image/bmp'); // Create the image $im = imagecreatetruecolor(80, 80); $count_my_page = ("number.txt"); $hits = file($count_my_page); if ($hits[0] == 1) { $im = imagecreatefrompng("image1.png"); $hits[0] ++; $fp = fopen($count_my_page , "w"); fputs($fp , "$hits[0]"); fclose($fp); }else{ $im = imagecreatefrompng("image2.png"); $hits[0] --; $fp = fopen($count_my_page , "w"); fputs($fp , "$hits[0]"); fclose($fp); } // Create some colors $white = imagecolorallocate($im, 255, 255, 255); $black = imagecolorallocate($im, 1, 1, 1); // Using imagepng() results in clearer text compared with imagejpeg() imagepng($im); imagedestroy($im); ?> And I have another page index.html Code: [Select] <img src="pic.php"> <img src="pic.php"> Basically, I need one of the images on the index page to be image1 and then the second image to be image2, but it must be contained within the php file because I am unable to edit index.html. Is there anyway to do this? The method I have now where it reads a file to see if it's value is 1 or 2 doesn't seem to compute each time the image is generated. I have tried header("Cache-Control: no-cache"); header("Pragma: no-cache"); in the image code to force it not to cache but to no avail. Can anyone solve this? Hi everybody I am a PHP rookie for now, and I want to reload the same dynamic page - it's content in several languages - by clicking little flags which are located at some point on the page - AND ITS NOT WORKING. Clicking on the flag triggers a PHP script which basically says Code: [Select] ]$_SESSION['lang'] = <Some_language>; require ('webpage.php'); where webpage.php is the dynamic page, in different languages, with it's content drawn out from a database (which contains the content in all languages I chose to display, blah, blah) Funny thing - when I look at the page html (View Page Source) - it seems OK (identical from one page to the next) ! But WHAT is displayed after the first attempt to change the language is not. Is something fundamentally wrong with my approach ? Mike I'm trying to show the same data, but updated right away. For example. I want to update my coords on a map and refresh a div to show the new data, but as the code now, it keeps the same data until I reload the page. Here is the code I have now. if ($north) { $ylocation = $users['y'] + 1; if ($ylocation > 5) { $ylocation = 0; } $locationyupdate = ("UPDATE players SET y = '$ylocation' WHERE name='$users[name]'"); mysql_query($locationyupdate) or die("could not register");?> <script type="text/javascript"> $('#npc').load('npc.php'); $('#description').load('description.php'); </script><?} The update code is before the reload script for the two div's. The data DOES change in the database, but the two div's won't display the new data until it is refreshed again. Do I need to reactivate fetch to get the new data? What is the best function to use to ensure an order is not re-entered when the user selects reload page? I can link code if you'd like but I'm only looking for a direction to go in and search google with. 'reload current page inserts extra' that google search hasn't been fruitful. This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=305930.0 This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=314750.0 Looking to INSERT a row in a data table and keeping the User on the current page. Will it work like what I have below, or do I need to send it to its own page then return the User back to where he started? What I have below isn't inserting a row. No errors showing.
function bookmark_add () {
if (isset($_POST['submit'])) {
include("/home2/csi/public_html/resources/con.php");
$query = "INSERT INTO a_player_bookmark (bookmark)
VALUES ('1')";
$results = mysqli_query($con,$query);
echo mysqli_error($con);
}
$output = '<form name="bookmark_add" method="post" action="" class="bookmark-plus">';
$output .= '<span><input type="submit" name="Bookmark"></span>';
$output .= '</form>';
return $output;
}
Hello all, I have been working with PHP for almost a year now, so somewhat still new to it all. OK, this may be an Apache issue, but this involves a php script I wrote that tries to execute on pages that do NOT call it (via the include command). So, here is a little info on how I have set things up. I use WAMP Server to test php and MySQL code for myself (for fun and for practice) and for clients and their web sites. I run WAMP Server off of my laptop, so I can code on the go and because it is on my laptop, I decided to create a PHP & MySQL user authentication setup to keep out unwanted visitors, but allow trusted friends and fellow developers to view my work on client web sites. Let's say the authentication script is called "auth.php" and for all of the pages I want private, I include that script at the top of those pages, if the visitor is not logged in, redirect them to the login page, otherwise, let them view the page (provided their permission level is adequate). Now, for the client web sites that I am working on, I do not call auth.php anywhere on those pages. Now, the problem I notice is that when I visited a client's web site that I am working on (on my WAMP Server), I reviewed my Apache error log and it gave me a PHP error stating the following: Code: [Select] Maximum execution time of 60 seconds exceeded in E:\\wamp\\www\\assets\\auth.php on line 3, referer: http://192.168.1.4/Web_Projects/NCCCLP/home.php Of course I am able to view the client web site pages without having to login, however, the auth.php script tries to run nonetheless. So my question is, does anyone know why this is happening and if and how I can stop it? This is not a serious (urgent) issue, but it does clog up my error logs and so forth and I prefer to have as few errors as possible. Plus, I would just like to know why and how it is running without being called. Any and all help is greatly appreciated and if I need to provide anymore details or be more specific, please let me know. Also, if an issue like this has already been solved and I missed that thread, please let me know and have my apologies. Thanks in advance. This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=326913.0 I am delete records using $_POST on the current page using isset() function. The records are deleting and the isset function is working but the corresponding record still appears in the list of records until the page is reloaded again. I have a session to check when the users login and fetch the details from their account but when they edit their avatar I want it to update the value right away. For it to change the value for the user they need to log out and back in, destroying the original session and recreating a new one on that account. Is there a way to have it always getting the data I need from the database while they're logged in? Because I don't want it to only get the orignial data in case they change a setting so they don't have to log out and then back in to change the setting. I hope you understand So, I just don't know where to post this question. I've written a quiz program where 50 English vocab words are pulled from a dB and displayed on one page with the order scrambled. My students then type the Spanish translation for each. When they submit the form the next page grades their work. They have to get a 90 or better for the grade to be automatically recorded, so I want them to click the back button and go back and make corrections. The problem is when they use a Chrome browser and click the back button the program re-sorts the order of questions. It leaves the order of their answers the same. So what they answered for number 1 is still in the field for number 1, but the questions (the English vocab word) is different. Chrome does this consistently so I thought this was a Chrome issue only. Then one of my students said the same thing happened to her when using Safari. So here is my question, how can I stop Chrome (or any other browser) from re-running my server-side PHP scripts (re-sorting the vocab array) when clicking the back button?
Here is the code for students taking the quiz:
<form method="post" action="portal.php?load=vquiz_grade" style="margin-left: 50px; line-height: 30px;">
$max = $count-1;
?>
//print "<br>";
<input type="hidden" name="max" value="<?print $max?>"> And here is the code showing them what they missed.
<?
// Total grade
print "You missed ".$wcount." words.<br>";
print "<table>";
// Grade response and set color
if ($wrong == 1){ if ($grade >= 90){
$query = "SELECT * FROM inno_vocab_assignments WHERE key_code = '$key_code' AND uname = '$secure_uname' AND teacher = '$teacher'";
// Prevent student from changing vocab list so as to do an easier list
// Insert grade
}
Thank you!
I have a form where people have to fill in some fields and send an email. When there is an error the user has to go back to the form with: Code: [Select] if (!isset($_POST['checkbox'])) { $msgToUser = '<br /><br /><font color="#FF0000">You did not add any recipients!<br><a href="javascript: history.go(-1)">Go Back</a></font>'; include_once 'msgToUser.php'; exit(); } I want to save the data that has been filled in the textarea so he doesn't have to type eveything again. I tried several options but the textarea keeps getting blank. Code: [Select] <?php $message = isset($_POST['message']) ? $_POST['message'] : ''; $fhtml = "<p>Welcome <b>'$logOptions_username'</b> write your email here. <a href=\"#\" onclick=\"return false\" onmousedown=\"javascript:toggleViewFlags('country_flags');\">Add Recipients</a> <br> </p><input type=hidden name=post value=yes><p> Subject:<br> <input type=text name=name size=100> </p> <p> Message:<br> <textarea name=message rows=10 cols=75>$message</textarea> </p> <p> <input type=submit name=submit value=\"Send\"> $sendMsg </p> "; echo $fhtml; ?> This code is apparently not working. Marco |
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