PHP - Onchange To Update A Variable And Pass It Through To A Query?
How can I make my on change in the form pass through the year thats been selected, then use that in the second query?
Code: [Select] <select name="mySelect" onchange=""> <?php $result= mysql_query('SELECT DISTINCT Year FROM MonthlySales'); ?> <?php while($row= mysql_fetch_assoc($result)) { ?> <option value="<?php echo htmlspecialchars($row['Year']);?>"> <?php echo htmlspecialchars($row['Year']); ?> </option> <?php } ?> </select> <?php $query="SELECT * FROM MonthlySales WHERE Year = PASS YEAR SELECTED VARIABLE TO THIS BIT HERE"; $results = mysql_query ( $query, $conn); Similar TutorialsI am working on a room availability calendar that has links for each day. When you click on a link it passes the day month and year in the URL to another page like this: echo "<td class=\"today\"> <a href=\"status.php?month=$month&day=$day_num&year=$year\">$day_num</a> </td>\n"; I can see that the correct information is being passed through the URL like this: Code: [Select] .../status.php?month=12&day=9&year=2010 Then the information is supposed to be passed to the MySQL query, but here is my question: How do I do this? I have a DB table set up, but the query is currently returning a blank page. Here is my current query: // connects to server and selects database. include ("../includes/dbconnect.inc.php"); // table name $table_name = "availability"; // query database for events $result = mysql_query ("SELECT id FROM $table_name WHERE month=$month AND year=$year AND day=$day_num LIMIT 1") or die(); if (mysql_num_rows($result) > 0 ) { while($row = mysql_fetch_array($result)) { extract($row); echo "<h1>Current availability for ".$row['month'] . "/" . $row['day'] . "/" . $row['year'] . "</h1>"; echo " <ul>"; echo " <li>Earth Room: " . $row['earth_room'] . "</li>"; echo " <li>Air Room: " . $row['air_room'] . "</li>"; echo " <li>Fire Room: " . $row['fire_room'] . "</li>"; echo " <li>Water Room: " . $row['water_room'] . "</li>"; echo " </ul>"; } } else { echo " <ul>"; echo " <li>Currently no reservations.</li>"; echo " </ul>"; } Any help is appreciated. Thanks, kaiman Hi This does not make sense to me, I hope someone can tell me why this query don't work: Code: [Select] safe_query("UPDATE ".PREFIX."cup_challenges SET reply_date='$postdate', $select_map $select_date challenged_info='".$_POST['info']."', status='2' WHERE chalID='".$_GET['challID']."'");; and this query does work: Code: [Select] safe_query("UPDATE ".PREFIX."cup_challenges SET reply_date='$postdate', map1='".$_POST['map1']."', map2='".$_POST['map2']."', map3='".$_POST['map3']."', date1='".$_POST['date1']."', date2='".$_POST['date2']."', date3='".$_POST['date3']."', date4='".$_POST['date4']."', challenged_info='".$_POST['info']."', status='2' WHERE chalID='".$_GET['challID']."'"); Variables in $select_map and $select_date makes the query fail but when I copy/paste this (bold) in query it works: $select_map = map1='".$_POST['map1']."', map2='".$_POST['map2']."', map3='".$_POST['map3']."', $select date = date1='".$_POST['date1']."', date2='".$_POST['date2']."', date3='".$_POST['date3']."', date4='".$_POST['date4']."', am I using the variable in query incorrectly? 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(); //} ?> Hey phpFreaks, im having some troubles getting my script to work correctly and im also not sure if this issue is in the right section of the forum. but heres what i have going on. I have a query result that displays a list of images with a checkbox and a couple buttons for edit and delete. everything works fine other than the checkbox stuff. I had it working when i was using a submit button, but i wanted to get rid of that button cuz it was only dealing with the checkboxes. so whats going on when using the checkboxes is that when checked or uncheck it would submit the form. It works to submit but its not submitting any data to the database. heres what i got for code for this section of checkbox. Code: [Select] <?php echo '<form method="post">'; if(isset($_POST['submit'])){ foreach($_POST['id'] as $id){ $value = (isset($_POST['location'][$id]) && $_POST['location'][$id]=="0" ? '0' : '1'); $insert = mysql_query("UPDATE items SET location='$value' WHERE id='$id'") or die('Insert Error: '.mysql_error()); } } $result = mysql_query("SELECT * FROM items") or die("Query Failed: ".mysql_error()); $counter = 0; echo '<div class="specialcontainer">'; while($row = mysql_fetch_array($result)){ list($id, $item_info, $item_img, $price, $sale, $location) = $row; if($location == '0'){ $set_checked = 'checked="checked"'; }else{ $set_checked = ''; } if($counter % 5==0) { echo '</div>'; echo '<div class="specialcontainer">'; } echo '<div class="special"><img src="../images/items/'.$item_img.'" width="130" /><br />'.$item_info.'<br />$'.$price.'<br />$'.$sale.'<br />Slide Show: <input type="checkbox" id='.$id.' value="0" name="location['.$id.']" '.$set_checked.' onchange="this.form.submit()"/><br /><input type="button" value="Edit" name="edit" id="'.$id.'" onclick="window.location.href=\'specials.php?action=edit&id='.$id.'\'"><input type="button" value="Delete" name="Delete" id="'.$id.'" onclick="window.location.href=\'specials.php?action=delete&id='.$id.'\'"><input type="hidden" name="id[]" value='.$id.' /></div>'; $counter++; } echo '</div>'; echo '</form>'; ?> Hello all,
Based on the suggestion of you wonderful folks here, I went away for a few days (to learn about PDO and Prepared Statements) in order to replace the MySQLi commands in my code. That's gone pretty well thus far...with me having learnt and successfully replaced most of my "bad" code with elegant, SQL-Injection-proof code (or so I hope).
The one-and-only problem I'm having (for now at least) is that I'm having trouble understanding how to execute an UPDATE query within the resultset of a SELECT query (using PDO and prepared statements, of course).
Let me explain (my scenario), and since a picture speaks a thousand words I've also inlcuded a screenshot to show you guys my setup:
In my table I have two columns (which are essentially flags i.e. Y/N), one for "items alreay purchased" and the other for "items to be purchased later". The first flag, if/when set ON (Y) will highlight row(s) in red...and the second flag will highlight row(s) in blue (when set ON).
I initially had four buttons, two each for setting the flags/columns to "Y", and another two to reverse the columns/flags to "N". That was when I had my delete functionality as a separate operation on a separate tab/list item, and that was fine.
Now that I've realized I can include both operations (update and delete) on just the one tab, I've also figured it would be better to pare down those four buttons (into just two), and set them up as a toggle feature i.e. if the value is currently "Y" then the button will set it to "N", and vice versa.
So, looking at my attached picture, if a person selects (using the checkboxes) the first four rows and clicks the first button (labeled "Toggle selected items as Purchased/Not Purchased") then the following must happen:
1. The purchased_flag for rows # 2 and 4 must be switched OFF (set to N)...so they will no longer be highlighted in red.
2. The purchased_flag for row # 3 must be switched ON (set to Y)...so that row will now be highlighted in red.
3. Nothing must be done to rows # 1 and 5 since: a) row 5 was not selected/checked to begin with, and b) row # 1 has its purchase_later_flag set ON (to Y), so it must be skipped over.
Looking at my code below, I'm guessing (and here's where I need the help) that there's something wrong in the code within the section that says "/*** loop through the results/collection of checked items ***/". I've probably made it more complex than it should be, and that's due to the fact that I have no idea what I'm doing (or rather, how I should be doing it), and this has driven me insane for the last 2 days...which prompted me to "throw in the towel" and seek the help of you very helpful and intellegent folks. BTW, I am a newbie at this, so if I could be provided the exact code, that would be most wonderful, and much highly appreciated.
Thanks to you folks, I'm feeling real good (with a great sense of achievement) after having come here and got the great advice to learn PDO and prepared statements.
Just this one nasty little hurdle is stopping me from getting to "end-of-job" on my very first WebApp. BTW, sorry about the long post...this is the best/only way I could clearly explaing my situation.
Cheers guys!
case "update-delete": if(isset($_POST['highlight-purchased'])) { // ****** Setup customized query to obtain only items that are checked ****** $sql = "SELECT * FROM shoplist WHERE"; for($i=0; $i < count($_POST['checkboxes']); $i++) { $sql=$sql . " idnumber=" . $_POST['checkboxes'][$i] . " or"; } $sql= rtrim($sql, "or"); $statement = $conn->prepare($sql); $statement->execute(); // *** fetch results for all checked items (1st query) *** // $result = $statement->fetchAll(); $statement->closeCursor(); // Setup query that will change the purchased flag to "N", if it's currently set to "Y" $sqlSetToN = "UPDATE shoplist SET purchased = 'N' WHERE purchased = 'Y'"; // Setup query that will change the purchased flag to "Y", if it's currently set to "N", "", or NULL $sqlSetToY = "UPDATE shoplist SET purchased = 'Y' WHERE purchased = 'N' OR purchased = '' OR purchased IS NULL"; $statementSetToN = $conn->prepare($sqlSetToN); $statementSetToY = $conn->prepare($sqlSetToY); /*** loop through the results/collection of checked items ***/ foreach($result as $row) { if ($row["purchased"] != "Y") { // *** fetch one row at a time pertaining to the 2nd query *** // $resultSetToY = $statementSetToY->fetch(); foreach($resultSetToY as $row) { $statementSetToY->execute(); } } else { // *** fetch one row at a time pertaining to the 2nd query *** // $resultSetToN = $statementSetToN->fetch(); foreach($resultSetToN as $row) { $statementSetToN->execute(); } } } break; }CRUD Queston.png 20.68KB 0 downloads I created this code to upload a member's main picture on his member page on website. I'll only include the query part of the code since that's what is relevant to my problem. The idea is basically to upload a new picture onto the database if no picture already exists for that member and display the picture on the page. If a picture already exists, then the script replaces the old picture with the new one upon upload. But for whatever reason I don't understand, when I try to replace the old pic, it gets inserted in a new row on the database instead of replacing the old row, and the new pic gets displayed on the web page alongside the old. Code: [Select] $query = "SELECT username FROM images WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main'"; $result = @mysql_query($query); $num = @mysql_num_rows($result); if ($num> 0) { //Update the image $update = mysql_query("UPDATE images SET image = '" . $image['name'] . "' WHERE member_id = '".$_SESSION['id']."' AND image_cartegory = 'main'"); $_SESSION['error'] = "File updated successfully."; //really should be session success message. header("Location: member.php"); exit; } else { // NOTE: This is where a lot of people make mistakes. // We are *not* putting the image into the database; we are putting a reference to the file's location on the server $sql = "insert into images (member_id, image_cartegory, image_date, image) values ('{$_SESSION['id']}', 'main', NOW(), '" . $image['name'] . "')"; $result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error()); $_SESSION['error'] = "File uploaded succussfully."; //really should be session success message. header("Location: member.php"); } So can anyone tell me what the problem is? Could the fact that my insert script actually uploads the image onto a folder on my server and only stores the path name in the database have anything to contribute to the mixup? Appreciate your responses in advance. This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=353002.0 I have researched all over forums from past day. Not getting correct solution. I have 2 textboxes and a button. First box is to enter value and i will click button, i need to get the value.
Here is the code, that works without input box 1
In this code i want to modify my web address, at the end after ky= i want add my first textbox value, then click event and output in second textbox, let me know where i messed.
<script type="text/javascript"> function Assign() { <?php $html = file_get_contents("http://geoportaal.maaamet.ee/url/xgis-ky.php?ky=79401:006:0812" ); preg_match_all('(<li.*?>.*?</li>)', $html, $matches); $one=$matches[0][0]; ?> document.getElementById("OutputField").value = "<?=$one?>"; } </script> <input id="InputField" type="text" style="width:200px"/> <input type="submit" value="Assign Value" onclick="Assign()"/> <input id="OutputField" type="text" style="width:200px"/> Hello
I'm using a form to collect user data (is a Cognito form) it is displayed on vacation rental properties pages, so it collects user´s email, in date, out date, etc. in the cognito form I have one text hidden field called REF, I wonder how to do in order to use the same form and be able to identify where the info comes from. I thougt that if I could put the page url into the REF variable, the problem would be solved, but I'm not sure if that is possible and how to do it. It is? If not: is it possible to handle that ? Hi guys, I need your help, I have got a problem with the if statement. When I don't insert the pass function in the url like this: Code: [Select] http://www.mysite.com/myscript.php?image=myimagelocation&strings=mystrings&user=test I will get this on my php page: Code: [Select] PASSWORD are missing Code: [Select] <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbusername'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbname'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $image = clean($_GET['image']); $strings = clean($_GET['strings']); $username = clean($_GET['user']); $pass = clean($_GET['pass']); if($username == '' && $pass) { $errmsg_arr[] = 'username are missing'; $errflag = true; }elseif($username && $pass =='') { $errmsg_arr[] = 'PASSWORD are missing'; $errflag = true; } if($username == '' && $pass == '') { $errmsg_arr[] = 'username or password missing'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $insert = array(); if(isset($_GET['image'])) { $insert[] = 'image = \'' . clean($_GET['image']) . '\''; } if(isset($_GET['strings'])) { $insert[] = 'strings = \'' . clean($_GET['strings']) . '\''; } if(isset($_GET['user'])) { $insert[] = 'user = \'' . clean($_GET['user']) .'\''; } if(isset($_GET['pass'])) { $insert[] = 'pass = \'' . clean($_GET['pass']) . '\''; } if (count($insert)>0) { $names = implode(',',$insert); $required_fields = array('image', 'strings', 'user'); if($image && $strings && $username) { echo "working 1"; } elseif($username && $pass) { echo "working 2"; } } } ?> Do anyone know how to fix this? so I need some help passing these variable from this page to final.php. how do I pass these arrays? I know if it were singled....not arrayed, I could use hidden fields in a form and echo them out....but these are multiples....not singled. The form way is prefered.....but it doesn't have to be. I just need these passed to the page where I am going to process them. I am not good at working with arrays. Thanks in advance Code: [Select] <?php include_once("connect.php"); session_start(); foreach($_POST["product"] AS $key => $val) { $product = $val; $month = $_POST['month'][$key]; $day = $_POST['day'][$key]; $year = $_POST['year'][$key]; $date = $_POST['date'][$key]; $price = $_POST['price'][$key]; $qty = $_POST['qty'][$key]; $id = $_POST['id'][$key]; $total = $_POST['total'][$key]; $academy = $_POST['academy'][$key]; $priceunit = $price * $qty; } ?> i have to pass a variable to so many pages i mean variable post from index.html to page1.php and it has to pass to other pages like page2.php, page3.php and so on, i tried session_start() but i am getting the below error so please tell me how it is possible to Warning: session_start() [function.session-start]: Cannot send session cookie - headers already sent by and Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent When i execute the following code, it works and retrieves the record info i am looking for for the appropriate record, 2155015898: $adgrouppass = 2155015898; $selector->adGroupIds = array($adgrouppass); However, when i try to get the adgroup variable from my url, which is where it is located, it doesn't work: $adgroup = trim($_GET['adgroup']); $adgrouppass = $adgroup; $selector->adGroupIds = array($adgrouppass); yet, in trying to fix it, if i echo the above variable, it says it is correct, which is 2155015898. Not sure if the $adgrouppass needs to be a string or variable to get it to pass it correctly. Can anyone help? i knw dis mst a simple question but how to pass variable to other page for eg i hav created a page where user submits username n telephone then via sms some random number goes to user mobile ...i want this random number variable in other page dis is my code for startreg.php <form action="startregprocess.php" method="post"> username:<input type="text" name="username"> telephone<input type="text" name="telephone" <input type="submit" name="submit"> </form> <?php srand ((double) microtime( )*1000000); $random_number = rand( ); echo $random_number; ?> i want tht $random_number in other page called startregprocess.php Here's my search form: <form action="search" method="get" enctype="multipart/form-data"> <input type="text" name="string" maxlength="100" /><br /><br /> <input type="submit" value="Search" /> </form> When I submit it, the URL is search.php?string=Bleh What I would like is for the URL to be search/string/Bleh I'm already doing this with some other variables, like search/tag/Bleh and search/author/Bleh, instead of search.php?tag=Bleh and search.php?author=Bleh. However, those ones are passed via a link, not a form. Any ideas? Can someone explain to me how this is done? Hi, I have this script, it all works fine apart from when I pass the variable on, it only passes the first word and not the whole variable The form that actions to the samples page passes the $siteName as My Test Site, if I type echo $siteName in the sameples page, it will print My Test Site, however when I use this Code: [Select] createSample&siteName=$siteName to pass the variable to the next page, it just echos 'My' instead of My Test Site. Even when I scroll over the link above it just shows My in the url. Code: [Select] case "samples": // Get submitted data and assign to variables $siteName = $_POST['siteName']; $adminEmail = $_POST['adminEmail']; $sendmailLoc = $_POST['sendmailLoc']; $imgdir = $_POST['imgdir']; $imgdirbase = $_POST['imgdirbase']; // Write variable data to text file $FileName = "data/server.txt"; $FileHandle = fopen($FileName, 'w') or die("can't open file"); $data = "<?php\n\$siteName = '$siteName';\n?>"; fwrite($FileHandle, $data); $data = "<?php\n\$adminEmail = '$adminEmail';\n?>"; fwrite($FileHandle, $data); $data = "<?php\n\$sendmailLoc = '$sendmailLoc';\n?>"; fwrite($FileHandle, $data); $data = "<?php\n\$imgdir = '$imgdir';\n?>"; fwrite($FileHandle, $data); $data = "<?php\n\$imgdirbase = '$imgdirbase';\n?>"; fwrite($FileHandle, $data); fclose($FileHandle); // Displays the create sample pages page ECHO <<<SAMPLES <table border=0 align=center bgcolor=#00CCFF> <tr> <td><span class=style1><b><center>Create Sample Pages</center></b></span></td> </tr> <tr> <td>Would you like Member Site Maker to create sample pages for: <ul> <li>index.html</li> <li>login.html</li> <li>search.html</li> <li>register.html</li> </ul> </td> </tr> <tr> <td><a href=installation.php?cmd=createSample&siteName=$siteName>Yes</a> </td> <td><a href=installation.php?cmd=mysql>No</a> </td> </tr> </table> SAMPLES; break; Can anybody tell me where I am going wrong please? Thanks Hello! I've got an issue! I'm trying to pass a very very long variable from JS to PHP through $_GET, but it seems too long for get. Is there any encryption function that works on both JS and PHP, so I can encrypt it in JS, send it though GET and decrypt it in PHP? Thanks ! I am passing the variable $searchno from form1.html using <form action="display1.php" method="post" Display1.php reads my database and displays the data. Now I want to link from the displayed page to display2.php and pass and capture the same variable $searchno. How do I do that? Hi guys, I have a trouble with my php snippet, when I insert the var function in the url bar something is like: http://www.mysite.com/delete.php?favorites&id=0 or http://www.mysite.com/delete.php?whateveritis&id=0 It doesn't get pass the favorites function to delete the id. It is the same things that it goes for each different function. Here's the current code: <?php Code: [Select] session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbuser'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbtablename'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $favorites = clean($_GET['favorites']); $id = clean($_GET['id']); if($favorites && $id == ''){ // both are empty $errmsg_arr[] = 'favorites id are missing.'; $errflag = true; } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $insert = array(); if(isset($_GET['id'])) { $insert[] = 'id = \'' . clean($_GET['id']) .'\''; } if(isset($_GET['favorites'])) { $insert[] = 'favorites = \'' . clean($_GET['favorites']) . '\''; } if($favorites && $id) { mysql_query("DELETE FROM favorites WHERE id='$id'"); $deleted = mysql_affected_rows(); if($deleted > 0) { echo "favorites channels is deleted"; } else { echo("favorites is already deleted"); } } } ?> If you do know how to get pass the favorites function, then please say so as i need your help. Any advice would be much appreicated. |