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PHP - Updating A Query On One Page, From Another?
Hey Guys,
Ive been searching for a while, but havent really come up with any answers. At the minute i have a page where i can add a new topic to a database, but i want to be able to update the query on another page if that makes sense So as an example say i have this below thats just been inserted into the database with an ID of 2. <?php // Details $id = $_POST['id']; $title = $_POST['title']; $detail = $_POST['detail']; mysql_connect("localhost", "web148", "123") or die(mysql_error()); mysql_select_db("web148") or die(mysql_error()); mysql_query("SET names 'utf8'"); mysql_query("INSERT into sunnyargues(ID, title, body) VALUES ('2', ''超级马里奥意大利再遭种族歧视', '当被问及在友谊赛上的不友好一幕会不会重演时，曼奇尼说，“有这种可能性。” “我当然不希望这样的事情发生，但是谁也不能保证。”'"); echo "<h1>Article Updated!</h1>"; ?> On the sunny argues page i currently have, where it only selects ID 1: <?php mysql_connect("localhost", "web148", "123") or die(mysql_error()); mysql_select_db("web148") or die(mysql_error()); mysql_query("SET names 'utf8'"); $result = mysql_query("SELECT body FROM sunnyargues WHERE ID='1'"); $row = mysql_fetch_array( $result ); echo nl2br($row['body']); ?> Is there a way i can update this query with the new ID of the 2nd article once its been inserted into the database. I hope that makes sense Similar Tutorialswhy is it not updating? <?php $_POST['email']; $_POST['submit']; if (isset($_POST['submit'])) { /*$sql =" UPDATE class_members SET email ='".$_POST['email']."', WHERE S_UID ='".$_POST['mem_id']."' ";*/ $result = mysql_query("UPDATE class_members SET email ='".$_POST['email']."', WHERE mem_id='".$_POST['mem_id']."'") or die(mysql_error()); } ?> Hi all, for my website ive got a fuel station which ive coded so users can fill up there there car fuel tanks, but the code all works part from one part of it. The code: The part of the code which isnt working: <?php $usermoney = $fetch->money; $costs2 = $fetch_fuelstation->price * $fueldiff; if ($costs2 > $usermoney){ echo "You dont have enouth money."; } elseif ($costs2 <= $usermoney){ $nmoney = $usermoney - $costs2; mysql_query("UPDATE users SET money='$nmoney' WHERE username='$username'") or die (mysql_error()); } ?> Ive got a varibal called $usermoney which selects the users current money from the database, but what im trying todo is take the fuel price away from there current money, but its not subtracting the money as it should. Anyone see why its not taking the money? Thanks. This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=334445.0 Hi there. I'm totally new (about a week!) with php and mysql and am encountering a problem that perhaps someone can help me with?
I've looked through to see if a similar problem has appeared or been solved, but without success, so apologies if I am repeating something.
In php I am trying to update 7 fields from a form from which a user has edited/modified any of the fields in a chosen record (except id).
Here is the code:
$id=$_GET['id']; I have a form that allows my client to update some products. Now the products are simple just basic info and 1 picture. I have set this up so they can edit the products and change the information, having done this many times in the past, but now hit a puzzling block that I am baffled. The client when editing is presented with the form with the information pulled from the database and the form fields loaded with that data ready to edit. The image can either be left alone or they can choose to upload a new image. They are shown the image they currently have stored in the database. The problem I have is EVEN if they decide not to upload an image and change other information, when the submit the form it must be sending a blank value for the image somewhere as it is updating the database and removing the image reference as if it has been removed. I have an if/else statement based on the form to perform 2 different queries for the update in mysql. Here is the code for the form update, as you can see the image should not update?? Please help?? if ($_SERVER['REQUEST_METHOD'] =='POST') { //This stops SQL Injection in POST vars foreach ($_POST as $key => $value) { $_POST[$key] = mysql_real_escape_string($value); } // **************************** THIS IS FOR NO NEW IMAGE ******************************** if ($_SERVER['REQUEST_METHOD'] =='POST' && empty($_FILES['product_image']['name'])) { # setup SQL statement for no new image $SQL = " UPDATE products SET product_title = '{$_POST['product_title']}', product_description = '{$_POST['product_description']}', standard_price = '{$_POST['standard_price']}', deluxe_price = '{$_POST['deluxe_price']}' WHERE product_id = '{$_REQUEST['product_id']}' "; } // **************************** THIS IS FOR A NEW IMAGE ******************************** else { // Check the image type is a jpeg or gif for the image. if (($_FILES['product_image']['type'] != "image/gif") && ($_FILES['product_image']['type'] != "image/jpeg") && ($_FILES['product_image']['type'] != "image/pjpeg")) { echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\">You have chosen not to upload a <b>Product Image</b>.<BR></SPAN>" ; } elseif ($_FILES['product_image']['size'] > 100000) { echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\">The file size is bigger than 300kb.<BR></SPAN>" ; } else { move_uploaded_file($_FILES['product_image']['tmp_name'], "/httpdocs/product_images/".$_FILES['product_image']['name']) ; echo "<FONT FACE=\"Verdana\"><SPAN CLASS=\"content\"><B>Your front image has successfully uploaded.</B><BR></SPAN>" ; } } # setup SQL statement for update $SQL = " UPDATE products SET product_title = '{$_POST['product_title']}', product_description = '{$_POST['product_description']}', standard_price = '{$_POST['standard_price']}', deluxe_price = '{$_POST['deluxe_price']}', product_image = '{$_FILES['product_image']['name']}' WHERE product_id = '{$_REQUEST['product_id']}' "; } #execute SQL statement $result = mysql_db_query( *****,"$SQL",$connection ); # check for error if (!$result) { echo("ERROR: " . mysql_error() . "\n$SQL\n"); } Hello all, I'm relatively new to all of this, but making progress.... I figure this has come up before, but couldn't fin anything by search or browsing. I have an sql table that is cron updated every two minutes from externally generated data in table X columns a,b,c,... . As part of more complicated site, I have one body element page (selected by a tab in the header) that does various sql queries on the data in table X and displays the data. I need this to rerun the queries and update the displayed results every 2 minutes or so as well. However, the standard solutions do not seem to be working: <META HTTP-EQUIV="refresh" CONTENT="15"> A refresh button refreshes the entire site, and not just this element. <FORM> <INPUT TYPE="button" onClick="history.go(0)" VALUE="Force Flight Data Refresh"> </FORM> I hope it is clear. Any ideas of how I could get this to work? Thanks, Kalle Here is a snippet of the code on that page for what it's worth... </head> <body> <div id="header"> <center><h2><font color="red">TRAFFIC INFORMATION</font></h2> </div> <div id="navigation"> </div> <div id="content"> <!-- refresh button --> <FORM> <INPUT TYPE="button" onClick="history.go(0)" VALUE="Force Flight Data Refresh"> </FORM> </center> <?php $IDS->db_build($db); $IDS->db_query($db,$res,"SELECT * FROM `Pilots` WHERE `dest`='KORD'"); echo "<h5>Arrivals to O'Ha </h5>"; while ($arr_row = mysql_fetch_assoc($res)) { echo "<font color=yellow>".$arr_row['callsign']."</font> using route: ". $arr_row['route']."<br>"; } //end queries //close connect to sql database mysql_close($con) </div> <div id="footer"> </div> </body> </html> I have the following PHP script to update two time/date fields in the database. When i run this the fields are not updated. Can anyone see where i m going wrong. <?php $con = mysql_connect("localhost","dbname","dbpassword"); if (!$con) { die('Could not connect: ' . mysql_error()); } echo 'Connected successfully'; mysql_select_db("my_db", $con); mysql_query("UPDATE msm_content SET created = '2011-01-02 00:00:00', modified = '2011-01-01 00:00:00'"); echo 'Query Updated successfully'; mysql_close($con); ?> Your guidance is much appreciated. Hi all, I'm currently working on an admin page and I'm trying to figure out how I can make an update query where all information will be updated. The code below is where I extract all the Users and their status from the database and now I when the statuses are updated in the text fields they need to be send to the database. Does anyone know how this could be done? Thnx Ryflex <?php require_once('auth.php'); require_once('config.php'); $global_dbh = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD) or die("Could not connect to database"); mysql_select_db(DB_DATABASE, $global_dbh) or die("Could not select database"); $user_query = "SELECT * FROM members"; $user_result = mysql_query($user_query); ?> <form ID="gotoresource" NAME="gotoresource" METHOD="POST" ACTION="admin_exec.php"> <table width="500" border="1" align="center" cellpadding="2" cellspacing="0"> <tr> <td><b>ID</b></td> <td><b>User</b></td> <td><b>Status</b></td> </tr> <?php while($row = mysql_fetch_assoc($user_result)) { echo "<tr>"; echo "<td>"; echo $row['member_id']; echo "</td>"; echo "<td>"; echo $row['login']; echo "</td>"; ?> <td> <input type="text" name="status" value="<?php echo $row['status'];?>" /> </td> <?php echo "</tr>"; } ?> <input type="image" src="/images/button.gif" alt="Submit button"> </form> <html> <head> <title>AdminPage</title> <link href="loginmodule.css" rel="stylesheet" type="text/css" /> </head> <body> </body> </html> I used to be good at this but I changed servers and everything is different... Heres my code so far: Code: [Select] <?php $rated=$_REQUEST['rated']; echo $rated; $rating=$_REQUEST['rating']; echo $rating; // Make a MySQL Connection mysql_connect("localhost", "********", "********") or die(mysql_error()); mysql_select_db("*********") or die(mysql_error()); $result = mysql_query("SELECT * FROM main WHERE username = '$rated'") or die(mysql_error()); $row = mysql_fetch_array( $result ); $votes = $db_field['$rating']; $newvotes = $votes + 1; echo $newvotes; mysql_query("UPDATE main SET $rating = '$newvotes' WHERE username = '$rated'"); ?> Whats going on here is the colomb that I want to update comes as a variable $rated (That works) and then the database selects the row to update with $username (That works) and gets the variable $newvotes by taking the original value of the data its about to update and add 1 to it (That works) Then it updates the field to $newvotes.... I don't know why the update won't go through... there are no errors.... friends whenever i include this coding to my page, everything goes blank. how could i fix this? Code: [Select] $time_count = 7*24*60*60; // count the 7 days Ads will be deleted $match_time = time()-$time_count; $query = "SELECT id FROM classifieds WHERE time<'".$match_time."' ORDER BY RAND() LIMIT 1"; $result = mysql_query($query); while($ad_id = mysql_fetch_array($result)) $delete =$ad_id[0]; mysql_query("DELETE FROM classifieds WHERE id='$delete'"); I am trying to clean up some code in my pages and would like some advice on good practice for the structure. Initially the code was generated by Dreamweaver and everything relating to the queries were place at the very top of the page, outside any of the HTML. A lot of the coding help I have had from this forum has put the queries etc in with the actual output php. Is there any right or wrong way to do this? I assume both ways have their pros and cons and both generate the same end result but I want to start getting into good habits before it is too late!! Thanks in advance Steve Hello all. I'm using this code to go through the database and output certain user-defined numbers Code: [Select] <?php mysql_connect ("pdb1.awardspace.com", "anastasov_db","moscow1945") or die (mysql_error()); mysql_select_db ("anastasov_db"); $term = $_POST['term']; $sql = mysql_query("select * FROM countries WHERE cocode = '$term'"); while ($row = mysql_fetch_array($sql)){ echo '<br/> Code: '.$row['cocode']; echo '<br/> Country: '.$row['coname']; echo '<br/><br/>'; } ?> Now how would I go about making a page that appears if the script can't find any results in the database? Thank you in advance I have an MS Access Database that resides on the server that I will be hosting the pages I'm building. In my MS Access Database I have a query called "CheckedInEquipment" that I would like to have constantly display on a PHP page. It needs to be 'fluid', so the data is always up to date on the web page. (This is all in a sandbox XAMPP environment). Has anyone done this, can you lead me to examples or push me in the right direction? Carl OK, here it is. I have been trying to do this myself, but it has been driving me insane and I turn to professionals here for help.
I am a basic web developer for a company that I work for in other capacities. I have a reasonable understanding of HTML and that is about where my expertise ends. I am not typically a programmer, just a simple (extremely) part time designer that uses Muse and Dreamweaver when necessary.
However, recently my company has asked me to accomplish a task for their website. In plain English, they want a large database that exists currently as a CVS file made into a searchable web page. It is 21 columns by approximately 6,700 rows.
To explain what I need a little more technically, here are my ideas and where I have gotten to so far:
1. The company uses godaddy, into which I *believe* I have successfully imported the spreadsheet. I believe it is successful because through godaddy's SQL Control Panel (phpMyAdmin console), I can do the EXACT searches that the company needs, and it spits out the EXACT results that I need.
2. The end result needs to be a .php that I can upload to the website's root folder that can be then inserted into premade pages using:
<iframe src="SMQ.php" scrolling="yes" width="950" height="800"></iframe>3. On the .php page, I need to have a way to log in to the SQL server and a simple search box built in that will allow the user to input a very simple search string consisting of no more than 4 numbers and 3 letters at a time. No buttons, no check boxes, just a search box. 3. This query then needs to be output as a nice data table, similar to this: This in fact is a screenshot of a search I performed out of my SQL database, in phpMyAdmin using the column "Scott" for the search, and the number 226 as the search term. All column names are visible with the exception of the first column, entitled LINEID, made to be the key, and the output should not include the key but have everything else as above. 4. I can see what the simple line of php is that performed this task: SELECT * FROM `SMQSQL` WHERE `Scott` = '226' ORDER BY `LINEID` ASCbut I can't figure out how the hell to get this incorporated to a .php search. To sum it up, I need a .php page written that can connect to a SQL database, perform a data based search, and spit out a clean table when it is done. I had accomplished this in the past using an import into google docs and using it to perform a search and result display via the following code built into a php called SMQ.php:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Example of Google Spreadsheet Data Visualisation</title>
</head>
<body>
<form id="form1" method="post" action ="<?php echo $_SERVER['PHP_SELF']; ?>"> <label>
<input id="search" name="search" type="text" />
</label>
<label>
<input id="Scott #" name="Scott #" type="submit" value="Scott #" />
</label>
<img src="loading.gif" width="16" height="11" />
</form>
<p>
<?php
$search= $_REQUEST['search'];
if ($search > ''){ $search = $search;} else { $search = '';}
?>
<script type="text/javascript" src="http://www.google.com/jsapi"></script>
<script type="text/javascript">
google.load('visualization', '1', {packages: ['table']});
</script>
<script type="text/javascript">
var visualization;
function drawVisualization() {
var query = new google.visualization.Query(
'https://docs.google.com/spreadsheet/ccc?key=0AronCwm9QPefdGpIUllscGgtLUJod2pOazc0bjU0cUE&usp=sharing');
query.setQuery('SELECT A, B, C, D, E, F, G, H, I, J, K, L, M, N, O ,P ,Q ,R ,S ,T WHERE (A) LIKE ("<?php echo $search; ?>") order by A asc label A "Scott #", B "Den", C "Color", D "Cond", E "40", F "60", G "70", H "70J", I "75", J "75J", K "80", L "80J", M "85", N "85J", O "90", P "90J", Q "95", R "95J", S "98", T "98J"');
query.send(handleQueryResponse);
}
function handleQueryResponse(response) {
if (response.isError()) {
alert('Error in query: ' + response.getMessage() + '' + response.getDetailedMessage());
return;
}
var data = response.getDataTable();
visualization = new google.visualization.Table(document.getElementById('table'));
visualization.draw(data, { page: 'enable', page: 16, pageSize: 16, legend: 'bottom'});
}
google.setOnLoadCallback(drawVisualization);
</script>
<div id="table"></div>
</div>
</body>
</html>
But as you can see, this may not be the most secure thing in the world, plus we want to be able to expand it in the future and not be so simplistic, hence the need to switch to SQL.
Please let me know right away by contacting me at disead@gmail.com if this is something YOU might be able to help with. I'm sure for an experienced programmer, once you have the details from me that you need, it would take maybe 10 minutes to write. I don't have much, but I can pay a little bit for this one time job. If it ends up working out, I may be able to pay more down the line for more advanced options such as being able to do drop-down searches based on the column titled "ISSUE", as well as more things down the line as it grows.
Thank you so much, I hope to hear from someone soon!!!
hi there alll newbie here lol just needed help with the code for the next page of results ? IE i have say 30 results returned from a sql query but only want to show ten at a time i no how to limit the result with the LIMIT 10 but how do i get it to put link show the other ten and so on. the code i have is here at the bottom of page http://www.phpfreaks.com/forums/index.php/topic,307971.0.html thanks in advance for any help kaine 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 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 ajetrumpetHi, I'm making an attempt to build a small cms based site. I have read in a lot of places it is best to use the id of a table row to fetch contents, but being as I am using url rewrite in frontend, it won't be showing query strings, so I thought I would ask if anyone saw any potential problems if I did the following. Page table looks like this: Pages - id - name - content In backend, the user has a form to name the page and use ckeditor to add rich content. I will use php to ensure that the name is alphanumeric and use strtolower & str_replace to ensure that My First HTML Page is sent to mysql as my-first-html-page and also check that a row with this name does not already exist. So on front end this page url will be mywebsite.com/my-first-html-page So to output the content I strip out the domain & / so I'm left with the name as entered in the db. $url = $_SERVER['REQUEST_URI']; $url = str_replace ('/','',$url); Then in my function to output content I will use WHERE name = $url What 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.) |
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