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PHP - Query Losing Its Data :-(
Alright, I've spent over a week trying to fix this now - And Im getting frustrated! I asked at other forums, I asked co-workers and I asked friends-of-friends, and nobody can explain what happens. Let's take a look at this first:
$name = mysql_real_escape_string($_POST['name']); mysql_query(sprintf("UPDATE em_users SET name='%s' WHERE id='" . $in_user['id'] . "'", $name)); This will insert NO data on the Name field in the database. Obviously, I thought the $_POST variable wasn't passed correctly, but echo'ing it just before the query WILL show data. And as I said, I tried everything possible for the last week. Switching variables, adding static text on the $name variable instead of using the $_POST content (this does work). I used very very simple test data on the form, such as my name "Mark" or "test" and "hey". The query is correctly executed everytime. The truely WEIRD thing is, if I ensure there is content in $name before executing the query it will work as expected everytime. Like this: $name = mysql_real_escape_string($_POST['name']); $name && mysql_query(sprintf("UPDATE em_users SET name='%s' WHERE id='" . $in_user['id'] . "'", $name)); Of course I could do this, but I want to know why my code does or doesn't work + it's a lot of work to do for something that worked fine a week ago. It has spread to a lot of forms on my website that $_POST variables aren't processed correctly - and it happened out of nowhere. Even on codes that havnt changed in months. I really need help on fixing this! This project has been in development for nearly two years, and without a fix it's pretty much lost Similar TutorialsHey there, Pretty frustrated with this. It seems so simple, but I've been staring at it and toying with it for ages, so I figured I'd post here. I have a variable called "$isTaken" that seems to be losing its value in an essential step in my function. Take a look. function subdomainTaken($inputtedSubdomain) { //Initialize $isTaken to 'true' $isTaken = true; echo "isTaken STARTS AS: <i>" . $isTaken . "</i><br />"; //FOR TESTING PURPOSES ONLY. VARIABLE HAS VALUE HERE //Check subdomain in database $subdomainDBQuery = mysql_query("SELECT * FROM `companies` WHERE company_subdomain = '$inputtedSubdomain'"); $foundResult = mysql_num_rows($subdomainDBQuery); if ($foundResult > 0) { $isTaken = true; } else { $isTaken = false; } return $isTaken; } By the time I get to the if statement, $isTaken has no value. I've echoed it to be sure. Any ideas? Thanks, Frank I recently had a customer say that one of my forms isn't working. For some reason the ID number is getting lost when she submits the form. The form code looks like: ... print "<form method='post' name='form' action='update.php'>"; ... print "<input type='hidden' name='id' value=\"$id\" />"; print "<input type='submit' name='submit' value=\"Save Session\" />"; print "</form>"; ... And the PHP code that gets executed looks like: ... //GET SESSION ID if(isset($_GET['id'])) { //ID from HTML link, before updates have been made $id = $_GET['id']; } elseif(isset($_POST['id'])) { //ID from form, after updates have been made $id = $_POST['id']; } else { $id = ''; } //IF SESSION ID IS VALID if(preg_match("/^\d+$/", $id)) { ... //ELSE, INVALID ID } else { $msg = "<span class='errorText'>Session ID not found or invalid.</span>"; } ... For some reason she usually gets the "Session ID not found..." error when submitting the form. Do you see any problems with the above code? Note that she has been able to sucessfully use the form before (in the same day); we have nearly 1,800 records submitted using these forms; and I am unable to duplicate the issue. So I'm at a loss on what to do next. Also, she talked with her IT person about the form. The IT person was able to submit data on his computer. So he reset her Internet options which seemed to fix the problem on her end temporarily. Note that she is using IE 7. She also said that she doesn't have access to any other browsers. I'm tempted to chalk this up as a personal computer issue, but wanted to get your input first. Hey all, I am having a situation where I have a singleton class with a static array in it. I insert an element into that array, and then I redirect the user to another page. At that page, I then retrieve that array (it's stored in the session vars), but the array is then empty when I retrieve it. I can't figure out why. Here are some pertinent code snippets. First, the class with the static array: class Logger { private static $instance = NULL; private static $messages = array(); //Private constructor to prevent this class from being instantiated multiple times private function __construct() { } //The clone function is private to prevent cloning of this class private function __clone() { } //Returns the singleton instance of this class public static function GetInstance() { if (!self::$instance) { self::$instance = new Logger; } return self::$instance; } public static function GetMessageCount() { return count(self::$messages); } public static function LogMessage($new_message) { self::$messages[] = $new_message; } public static function GetMessages() { return self::$messages; } public static function ClearMessages() { self::$messages = array(); } } Here is the code where I insert something into the aformentioned array (the process is that a user tries to log in, but the login fails, and so we insert a message into the array saying that the login credentials failed). //Retrieve the message logging instance from the session data to be able to pass messages to the user. $message_log = $_SESSION['user_messages']; $user = $_SESSION['user_account']; //Create a new instance of the database management class and //try to connect to the database $dbm = new DBM(); if ( !$dbm->isConnected() ) { $message_log::LogMessage("We are sorry, but we are unable to access the database at this time. Please try again later."); } else { //Retrieve the user login information $useremail_dirty = $_POST['useremail']; $password_dirty = $_POST['userpassword']; //Check to see if the email we were given exists in the database, and if the password matches $doesPasswordMatch = $dbm->doesPasswordMatch ( $useremail_dirty, $password_dirty ); if ( $doesPasswordMatch ) { //Correct login information was received. Login the user and redirect appropriately. $user->Login(); header ( "Location: ".BASE_URL."/userpage.php" ); exit; } else { //If an incorrect email or password was given, report that information to the user. $message_log::LogMessage("Incorrect login information."); } } //The user has failed to login. Redirect to the appropriate page. header ( "Location: ".BASE_URL."/loginfailed.php" ); And finally, given an unsuccessful login attempt, this is where I pick back up the array of messages to display the messages to the user: $message_log = $_SESSION['user_messages']; $all_messages = $message_log::GetMessages(); print $message_log::GetMessageCount()."<br>"; print count($all_messages)."<br>"; foreach ($all_messages as $message_string) { $prepared_string = prepare_text_for_html_output($message_string); echo $prepared_string."<br>"; } $message_log::ClearMessages(); Unfortunately, the "all_messages" variable has nothing in it...it's like the messages that I logged in the previous PHP script never even existed. I am calling session_start, so don't worry about that (even though it's not seen in these code snippets). What could be going wrong? This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=323252.0 I was just wondering if it's possible to run a query on data that has been returned from a previous query? For example, if I do Code: [Select] $sql = 'My query'; $rs = mysql_query($sql, $mysql_conn); Is it then possible to run a second query on this data such as Code: [Select] $sql = 'My query'; $secondrs = mysql_query($sql, $rs, $mysql_conn); Thanks for any help Hi, I want to pull data from db, where sometimes all rows and sometimes rows matching given "username". Here is my code:
//Grab Username of who's Browsing History needs to be searched.
if (isset($_GET['followee_username']) && !empty($_GET['followee_username']))
{
$followee_username = $_GET['followee_username'];
if($followee_username != "followee_all" OR "Followee_All")
{
$query = "SELECT * FROM browsing_histories WHERE username = \"$followee_username\"";
$query_type = "followee_username";
$followed_word = "$followee_username";
$follower_username = "$user";
echo "$followee_username";
}
else
{
$query = "SELECT * FROM browsing_histories";
$query_type = "followee_all";
$followed_word = "followee_all";
$follower_username = "$user";
echo "all";
}
}
When I specify a "username" in the query via the url: browsing_histories_v1.php?followee_username=requinix&page_number=1 I see result as I should. So far so good.
Now, when I specify "all" as username then I see no results. Why ? All records from the tbl should be pulled! browsing_histories_v1.php?followee_username=all&page_number=1 This query shouldv'e worked:
$query = "SELECT * FROM browsing_histories";
Hello, I hope all of you are safe with your families. Currently I am starting with PHP Coding and I am trying to do a simple query to MySQL DB using PHP but even when is able to bring the number of rows is not displaying the values in the DB. This is my code:
<?php
$servername = "localhost";
$database = "mydbtest";
$username = "root";
$password = "root";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $database);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
//echo "Connected successfully";
$myquery = "SELECT * FROM Country";
$result = $conn->query($myquery);
$numf = $result->num_rows;
echo "Number of rows " . $numf . "<br>";
if($numf >0){
while($row = $result->fetch_object()){
echo "Code" . $row->countrycode . "<br>";
echo "Country" . $row->countryname . "<br>";
}
}else{
echo '0 results';
}
mysql_free_result($myout);
mysqli_close($conn);
?>
What is failing?
Thanks in advance for the assistance. I've only been studying PHP for a week now and have come across this problem which I'm sure there is a simple answer to, but I just can't figure it out and would appreciate some help. I've spent far too long on this minor issue already! I have a table which contains a list of products, in this case books, which stores the date when each new item is added. I have a query that then searches through this table and extracts the 6 most recent additions. Here is the code I have so far: Code: [Select] $sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC LIMIT 6"); $productCount = mysql_num_rows($sql); // count the output amount if ($productCount > 0) { while($row = mysql_fetch_array($sql)){ $id = $row["id"]; $title = $row["title"]; $author = $row["author"]; $price = $row["price"]; $date_added = strftime("%b %d, %Y", strtotime($row["date_added"])); $dynamiclist .= //My table showing the products } } else { $dynamicList = "There are currently no Books listed in this store"; } This works well when I need to display the 5 most recent products in a normal table fashion, one below another, on the page. However, I want to display the products in a more personalised order. For example display the newest item in one section of the page and 3rd newest in another. What is the best way to select an individual row from a query? To extract the details of the 2nd newest item to display in the header, for example. Thanks for your help. I know that I can fairly easily pull my data from the database and view it in a browser. I can also 'polish' it with some HTML or put it into a table. Can I get an item from a given VARIABLE to appear inside of an INPUT box, so that it looks the same as when it was initially submitted? Can it be done with a multiple choice SELECT dropdown, so that the item chosen is viewable again? I am at a loss why my query is not inserting values into db. Even if I do echo $query or var_dump($query) there is nothing printed at all. All values are being passed successfully just not being inserted. I am getting 'Could not connect' but I do not know why. All connections are established and as a test I took this code and ran it on it's own with dummy data and it inserted the data fine. I can only think it has something to do with the $response_array. Where am I going wrong. and would appreciate any help. Thanks Code: [Select] <?php require_once('Connections/sample.php'); ?> <?php session_start(); $new = 1; $activity = 'General Contact Enquiry'; $mobile = 'Submitted from mobile'; $name = mysql_real_escape_string($_POST['GC_name']); $department = mysql_real_escape_string($_POST['GC_department']); $message = mysql_real_escape_string($_POST['GC_message']); $email = mysql_real_escape_string($_POST['GC_email']); $company = mysql_real_escape_string($_POST['GC_company']); $position = mysql_real_escape_string($_POST['GC_position']); //response array with status code and message $response_array = array(); //validate the post form //check the name field if(empty($name)){ //set the response $response_array['status'] = 'error'; $response_array['message'] = 'Name cannot be blank'; //check the name field } elseif(empty($company)) { //set the response $response_array['status'] = 'error'; $response_array['message'] = 'You must enter a company name'; //check the position field }elseif(empty($position)) { //set the response $response_array['status'] = 'error'; $response_array['message'] = 'You must enter a position'; //check the email field } elseif(empty($email)) { //set the response $response_array['status'] = 'error'; $response_array['message'] = 'You must enter a valid email address'; //check the dept field }elseif($department=="Choose Department") { //set the response $response_array['status'] = 'error'; $response_array['message'] = 'You must select a department'; //check the message field }elseif(empty($message)) { //set the response $response_array['status'] = 'error'; $response_array['message'] = 'You must enter a message'; //check the dept field } else { //set the response $response_array['status'] = 'success'; $response_array['message'] = 'Your enquiry has been sent succesfully'; $flag=1; } //send the response back echo json_encode($response_array); if($flag == 1) { mysql_select_db($database_sample, $sample); $query = 'INSERT INTO feedback (company, department, name, email, position, feedback, date, new) VALUES (\''.$company.'\', \''.$department.'\', \''.$name.'\', \''.$email.'\', \''.$position.'\', \''.$message.'\', NOW() , \''.$new.'\')'; mysql_query($query) or die("Could not connect"); } ?> I'm having trouble trying to separate the output into variables that I can use/echo on my page. when I do a print I see the two rows of data all grouped together how can I separate each result base on field and row? maybe something like $line['m_id'][0], $line['m_name'][0], $line['m_id'][1], $line['m_name'][1], etc... $bio = mysql_query("SELECT * FROM soc_meminfo WHERE m_id = '".mysql_real_escape_string($en['mm_id'])."'"); if (mysql_num_rows($bio) == 0) call404(); while ($line = mysql_fetch_assoc($bio)) { foreach ($line as $key => $value) { $en['b'.$key] = str_replace("\n",'<br/>',stripslashes($value)); } echo '<pre>'; print_r($line); echo '</pre>'; } hi guys Ive written this php to take in two variables from the http POST, the idea is that I can multiple devices submit temperature readings to the php script, the script then append the unix time stamp and then the the 3 variables - device id, temp and unix time are then stored in a mysql DB. I can get the variables to present on a php page for debugging but I cant get the variables to be stored in the mysql DB. See the code: Code: [Select] <?php $unixtime = time(); // get device variables $device_id=$_GET['device']; $device_temp=$_GET['temp']; /* //for testing purposes echo "unixtime: " . $unixtime . "<br />"; echo "device id: " . $device_id . "<br />"; echo "device_temp: " . $device_temp . "<br />"; */ // Make a MySQL Connection mysql_connect("localhost", "username", "password") or die(mysql_error()); mysql_select_db("test") or die(mysql_error()); //mysql query $query = "INSERT INTO temperature VALUES ('',$device_id,$temp,$unixtime)"; // Insert a row of information into the relevant device table mysql_query($query); or die(mysql_error()); mysql_close(); echo "Data Inserted!"; ?> I cant see where Im going wrong to correct this, but as nothing is displayed on the page i believe I am not forming the query correctly? - any ideas would be much appreciated. Thank you Mathew Hello, I seem to have some problem with my script that has a goal of outputting data about the file size when a filename is queried.
The sql table name is file
The table columns are as followed: id | name | mime | size
The file name is stored in name. The script that i have that gets the file name is:
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="content-type" content="text/html;charset=utf-8" />
<meta name="generator" content="Adobe GoLive" />
<title>File Select</title>
<!--The following script tag downloads a font from the Adobe Edge Web Fonts server for use within the web page. We recommend that you do not modify it.--><script>var __adobewebfontsappname__="dreamweaver"</script><script src="http://use.edgefonts.net/aguafina-script:n4:default.js" type="text/javascript"></script>
</head>
<body>
<div id="title">
<h3 align="center">File Upload</h3>
</div>
<form action="result.php" method="post" name="fileID" target="_self" class="inp" AUTOCOMPLETE="ON">
<h1>
<!--Input file name-->
<label for="fileID">File Name: </label>
<input type="text" name='file1' id='sampleID' list="samp"> </input><br>
<datalist id="samp">
<?php
$connect = mysql_connect('localhost', 'root', '');
mysql_select_db("test_db");
$query = mysql_query("SELECT * FROM `file` ORDER BY `file`.`name` ASC LIMIT 0 , 30");
WHILE ($rows = mysql_fetch_array($query)):
$File_name = $rows['name'];
echo "<option value=$File_name>$File_name/option> <br>";
endwhile;
?>
</datalist>
<input type="submit" class="button" >
</form>
</body>
</html>
Hi, I want to develop array like following Code: [Select] $BCD=array('type' =>'TYPE1', array( 0=>array('column1'=>'value1','column2'=>'value1','column3'=>'value1','column4'=>'value1','column5'=>'value1','column6'=>'value1','column7'=>0), 1=>array('column1'=>'value1','column2'=>'value1','column3'=>'value1','column4'=>'value1','column5'=>'value1','column6'=>'value1','column7'=>0), 2=>array('column1'=>'value1','column2'=>'value1','column3'=>'value1','column4'=>'value1','column5'=>'value1','column6'=>'value1','column7'=>0), 3=>array('column1'=>'value1','column2'=>'value1','column3'=>'value1','column4'=>'value1','column5'=>'value1','column6'=>'value1','column7'=>0), ) )); I have written following code to achieve the same but not getting result. Code: [Select] $sql = "select * from tablename "; $result = mysql_query($sql); $k=0; while ($row = $db->mysql_fetch_array($result)) { // array_push($BCD['type'],$row['type']); $BCD1=array('type' =>$row['type'], array( $k=>array('column1'=>$row['column1'],'column2'=>$row['column2'],'column3'=>$row['column3'],'column4'=>$row['column4'], 'column5'=>$row['column5'],'column6'=>$row['column6'],'column7'=>$row['column7']) )); $k++; } Hi. I am trying to get a PHP Query to refresh every 10 seconds. I have scoured the internet for days and could not find anything of much use. Plenty of Ajax going on (whatever that is) but the scripts were immense. I only want to refresh 7 lines of PHP Query Code. Can you please tell me if this is possible? Thanks in Advance. Hi guys, im trying to connect to a database and get the value for the user in the row called 'user_credit', if it equals 1 or more then i want to show the ''You have £ ....'' bit in the script. Problem is nothing shows at all, even without the if statement. I have changed the value for me in the database so in user_credit the value is 100, which is more than 1 so it should appear. I have probably done something wrong. Any ideas? Code: [Select] <? include '../admin/database/membership_dbc.php'; $r = mysql_query("SELECT * FROM users WHERE user_name='".safe($_SESSION['user_name'])."'") or die ("Cannot find table"); while( $cred = mysql_fetch_array($r) ) { if ($cred >= '1' ) { ?> <p>You have £<? echo $cred['user_credit']; ?> available on you account, would you like to use it on this order?<br> <label for="credit"></label> <select name="credit" id="credit"> <option value="Y" selected>Yes, use credit</option> <option value="N">No, save credit</option> </select> </p> <? } } ?> i want the name of a picture stored in my db after i upload it the data is not stored in the db after i run this script, but i dont get errors either i print the two vars before sending them, and they get printed fine any help on this would be greatly appreciated thanks ! <?php error_reporting(E_ALL); ini_set("display_errors", 1); // INCLUDE THE CLASS FILE include('ImageLib.Class.php'); include("./includes/egl_inc.php"); $displayMessage = ''; if($_POST){ if(isset($_FILES['image_file'])){ // SEE THE MAGIC HAPPEN $destination_path = 'uploads/'; $post_file_name = 'image_file'; $width = 600; $height = 400; $scale = false; $trim = true; $uniqueName = true; $img = ImageLib::getInstance()->upload($post_file_name, $destination_path, $uniqueName)->resize($width, $height, $scale, $trim)->save(); $imgstr = mysql_real_escape_string ($img); $fileName = $_FILES['image_file']['name']; $displayMessage = '<div class="image"><img src="'.$destination_path.$fileName.'" /><br />Uploaded And Resized...With new file name : "'.$img.'"</div><br /><br />'; $playerid=$_SESSION['tid']; $matchdetails = mysql_fetch_array(mysql_query("SELECT id FROM ffa_matches WHERE status=2 and admin=$playerid")); $id = $matchdetails[id]; print $img;print $imgstr; print $id; mysql_query(" INSERT INTO ffa_screens (imgname,match) VALUES( '" . mysql_real_escape_string($imgstr) . "', '" . mysql_real_escape_string($id) . "' )"); }} ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html> <head> <title>ImageLib Samples By Rahul Kate</title> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <style> body{font-family: arial; font-size:12px; color:#444444; padding:20px;} li{margin-top:10px;} .image{color:green;} .image img{margin-bottom:5px;} </style> </head> <body> <h1>ImageLib | Upload Image, move it to Uploads folder and Resize it and Trim</h1> <?php echo $displayMessage; ?> <form method="post" enctype="multipart/form-data"> Select Image<br /> <input type="file" name="image_file" id="image_file" /> <br /> <br /> <input type="submit" name="submit" value="Submit" /> <br /> <br /> <a href="index.html">Back TO Home</a> </form> </body> </html> I have a text box that I use to post comments and save them to my database. I can go to the database and the data I entered looks perfect but when I pull it out it all runs together.
This is how I entered it and the way it is in the database:
I have a text box that I use to post comments and save them to my database. This is how it came out: I have a text box that I use to post comments and save them to my database. I can go to the database, the data I entered looks perfect but when I pull it out and display it it all runs together. No new lines just all one paragraph. Can someone help? I got it indenting the paragraphs, but I get all the data back I entered all in 1 paragraph, is there something I am missing to be able to recognize the new lines?? I have tried several attempts and can get data to select and using echo display it, however, I need to take this data and insert it into the database in a separate table. I have the following which does hafl the job, can I get some pointers on the rest. I have looked everywhere and not found a solution, at all. // Selects the data I need <?php mysql_connect("PRIVATE INFO","PRIVATE INFO","PRIVATE INFO") or die("Could not connect: " . mysql_error()); mysql_select_db("wpdb"); $result = mysql_query("SELECT ID FROM wp_posts WHERE post_title LIKE '%future%' AND post_status = 'publish' OR post_title LIKE '%option%' AND post_content LIKE '%fundamental%' AND post_status = 'publish' ORDER BY post_date DESC"); while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo $row['ID']; echo "<br />";}; ?> Thought maybe something like the following would work, but am at a loss: INSERT INTO wp_term_relationships (object_id, term_taxonomy_id, term_order) SELECT ID FROM wp_posts WHERE post_title LIKE '%future%' AND post_status = 'publish' OR post_title LIKE '%option%' AND post_content LIKE '%fundamental%' AND post_status = 'publish' ORDER BY post_date DESC while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) VALUES ($row['ID'], '25', '0') 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>'; ?> |
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