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PHP - Php/mysql Insert Incrimental Number In Pk
I need to create a custom PK similar to 000111-GH.
whe 0001 is the counter 11 is the year GH i need randomly generated with either numbers/letters once 2012 occurs, i need the counter to be reset to 000112-DF How would i increment 0001 until 2012, then reset the first few digits back to 0001? similarly, i will have a linked table but will have a different autonumber format for client 1 the linked data would look look like this 000111-GH-001 for record 1 000111-GH-002 for record 2 but then for client 2 000211-DS-001 00211-DS-002 Can anyone provide any assistance on how to accomplish this? thank you. Similar TutorialsCan anyone tell me why this is not INSERTing? My array data is coming out just fine.. I've tried everything I can think of and cannot get anything to insert.. Ahhhh! <?php $query = "SELECT RegionID, City FROM geo_cities WHERE RegionID='135'"; $results = mysqli_query($cxn, $query); $row_cnt = mysqli_num_rows($results); echo $row_cnt . " Total Records in Query.<br /><br />"; if (mysqli_num_rows($results)) { while ($row = mysqli_fetch_array($results)) { $insert_city_query = "INSERT INTO all_illinois SET state_id=$row[RegionID], city_name=$row[City] WHERE id = null" or mysqli_error(); $insert = mysqli_query($cxn, $insert_city_query); if (!$insert) { echo "INSERT is NOT working!"; exit(); } echo $row['City'] . "<br />"; echo "<pre>"; echo print_r($row); echo "</pre>"; } //while ($rows = mysqli_fetch_array($results)) } //if (mysqli_num_rows($results)) else { echo "No results to get!"; } ?> Here is my all_illinois INSERT table structu CREATE TABLE IF NOT EXISTS `all_illinois` ( `state_id` varchar(255) NOT NULL, `city_name` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Here is my source table geo_cities structu CREATE TABLE IF NOT EXISTS `1` ( `CityId` varchar(255) NOT NULL, `CountryID` varchar(255) NOT NULL, `RegionID` varchar(255) NOT NULL, `City` varchar(255) NOT NULL, `Latitude` varchar(255) NOT NULL, `Longitude` varchar(255) NOT NULL, `TimeZone` varchar(255) NOT NULL, `DmaId` varchar(255) NOT NULL, `Code` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; What I have so far is a page that produces a list of usernames with checkboxes next to them. From the usernames that I check, I want to update a field for each of those users. My problem is that the number of checkboxes I check is not going to be static so I am pretty sure I have to run a loop based on the number of checked checkboxes ("People chosen"). Here is what I have so far: Code: [Select] <?php if (isset($_POST['submit1'])) { $pick = $_POST["pick"]; $how_many = count($pick); echo 'People chosen: '.$how_many.'<br><br>'; echo "<br><br>"; $count = 0; while ($count < $how_many) { //mysql code probably should go here but I am stuck $count++; } } else { $gender = mysql_real_escape_string($_POST['Gender']); $limit = mysql_real_escape_string($_POST['limit']); echo $gender; echo "<br />"; $count = 0; if ($gender == 'Male') { $result = mysql_query("SELECT * FROM users WHERE gender =\"1\" LIMIT 0, $limit", $connection); if (!$result) { die("Database query failed: " . mysql_error()); } ?> <form method="post"> <?php while ($row = mysql_fetch_array($result)) { ?> <input name="pick[]" type="checkbox" value="<?php echo $row['username'] ?>"><?php echo $row['username'];?> <br /> <?php } ?> <input type="submit" name="submit1" value="Make Offer" /> </form> <?php } ?> I hope this makes sense. Any help would be greatly appreciated. Hello, I'm having a bit of a problem here, all help to this issues would be much appreciated I am trying to use text boxes to insert numbers into the database based on what is inputed. If I have a string, like this for example: $variable = 09385493; And I want to insert it into the database like this: mysql_query("INSERT INTO integers(number) VALUES ('$variable')"); When checking the integers table in my database, looking at the number field, the $variable that was inserted is outputted as 9385493 Notice the number zero was taken out of the front of the number. If the number is double 0's (009385493), both of those zero's would disappear, too. Thanks I'm getting the dreaded " Invalid parameter number: number of bound variables does not match number of tokens" error and I've looked at this for days. Here is what my table looks like:
| id | int(4) | NO | PRI | NULL | auto_increment | | user_id | int(4) | NO | | NULL | | | recipient | varchar(30) | NO | | NULL | | | subject | varchar(25) | YES | | NULL | | | cc_email | varchar(30) | YES | | NULL | | | reply | varchar(20) | YES | | NULL | | | location | varchar(50) | YES | | NULL | | | stationery | varchar(40) | YES | | NULL | | | ink_color | varchar(12) | YES | | NULL | | | fontchosen | varchar(30) | YES | | NULL | | | message | varchar(500) | NO | | NULL | | | attachment | varchar(40) | YES | | NULL | | | messageDate | datetime | YES | | NULL |Here are my params:
$params = array(
':user_id' => $userid,
':recipient' => $this->message_vars['recipient'],
':subject' => $this->message_vars['subject'],
':cc_email' => $this->message_vars['cc_email'],
':reply' => $this->message_vars['reply'],
':location' => $this->message_vars['location'],
':stationery' => $this->message_vars['stationery'],
':ink_color' => $this->message_vars['ink_color'],
':fontchosen' => $this->message_vars['fontchosen'],
':message' => $messageInput,
':attachment' => $this->message_vars['attachment'],
':messageDate' => $date );
Here is my sql:
$sql = "INSERT INTO messages (user_id,recipient, subject, cc_email, reply, location,stationery, ink_color, fontchosen, message,attachment) VALUES( $userid, :recipient, :subject, :cc_email, :reply, :location, :stationery, :ink_color, :fontchosen, $messageInput, :attachment, $date);";
And lastly, here is how I am calling it:
$dbh = parent::$dbh;
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_WARNING);
if (empty($dbh)) return false;
$stmt = $dbh->prepare($sql);
$stmt->execute($params) or die(print_r($stmt->errorInfo(), true));
if (!$stmt) { print_r($dbh->errorInfo()); }
I know my userid is valid and and the date is set above (I've echo'd these out to make sure). Since the id is auto_increment, I do not put that in my sql (though I've tried that too), nor in my params (tried that too). What am I missing? I feel certain it is something small, but I have spent days checking commas, semi-colons and spelling. Can anyone see what I'm doing wrong?
I was just wondering when you run a mysql query what the easiest way to find out the number of rows returned. Just so I can do a number of results found on a search. Thanks. Hello, i made a text site, so you can text from a website. i was making a database of the texted numbers and i found that the mysql line is not inserting the correct number, its going to the max php one. Here the code: if (array_key_exists($carrier, $carriers)) { $correctCarrier = $carriers[$carrier]; $i = 0; while($i < $_POST['amount']){ $i++; $formatted_number = $to.$correctCarrier; $result = ("$i of Your Messages Has been sent to the number ". $_POST['to'] . ".<br>" . mail("$formatted_number", "$subject", "$message") . ""); } mysql_query("INSERT INTO `msgssent` (`number`, `numberofmsg`, `subject`, `message`) VALUES (". $_POST['to'] .", '". $i ."', '". $subject ."', '". $message ."')") Or die(mysql_error()); Echo $result; } I have this code: <?php $con = mysql_connect("localhost","hhh","hhh"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("hhh", $con); // -------------------- // Avatar insert check // -------------------- session_start(); $name = $_POST[name]; $group = $_POST[group]; $age = $_POST[age]; $usernameid = $_SESSION[id]; $result = mysql_query("SELECT * FROM avatars WHERE name='$_POST[name]'"); $num = mysql_numrows($result); if ($num == 0) { mysql_query("INSERT INTO avatars (id, usernameid, name, group, age, xp) VALUES ('', '$usernameid', '$name', '$group', '$age', '0')"); header( 'Location: me/' ) ; } else echo 'Sorry, please pick a new name'; ?> And it does everything but put the data into the datebase. If I add a session befor and after '$request' they both run, but the sql doesn't. No error returns, if just redirects to the other page. Any help? I don't understand where the empty value is. I've substituted the variables for text and still have the same problem. Code: Code: [Select] $sql = "INSERT INTO courses (course#, name, subject, semester, ap)VALUES('$courseNum', '$courseName', '$subject', '$semester', '$ap')"; Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 Hi guys I have a registration form working fine, my database is as below: userid username password repeatpassword I have added another column which is "name", users can update their profile once they have logged in so I have created updateprofile.php and when I login-->go to update profile and insert my name nothing adds to mysql name column this is my code below: <?php include ("global.php"); //username session $_SESSION['username']=='$username'; $username=$_SESSION['username']; //welcome messaage echo "Welcome, " .$_SESSION['username']."!<p>"; if ($_POST['register']) { //get form data $name = addslashes(strip_tags($_POST['name'])); $update = mysql_query("INSERT INTO users (name) VALUES ('$_POST[name]') WHERE username='$username'"); } ?> <form action='updateprofile.php' method='POST'> Company Name:<br /> <input type='text' name='name'><p /> <input type='submit' name='register' value='Register'> </form> can you please tell me where in this code is wrong? Im new in php so please excuse me if I have silly mistakes. thanks in advance I need help badly! What I want to do is insert into database the value from the selected radio group buttons.. All of them. There are 10 radio groups total (they can be less, but not more). Thanks! Code: [Select] <?php require_once('Connections/strana.php'); mysql_select_db($database_strana, $strana); ?> <link href="css/styles.css" rel="stylesheet" type="text/css" /> <table width="100%" height="100%" style="margin-left:auto;margin-right:auto;" border="0"> <tr> <td align="center"> <form action="" method="post" enctype="multipart/form-data" name="form1"> <table> <?php $tema = mysql_query("SELECT * from prasanja where tip=2")or die(mysql_error()); function odgovor1($string) { $string1 = explode("/", $string); echo $string1[0]; } function odgovor2($string) { $string1 = explode("/", $string); echo $string1[1]; } while ($row=mysql_fetch_array($tema)) { $id=$row['prasanje_id']; $prasanje=$row['prasanje_tekst']; $tekst=$row['odgovor']; ?> <tr> <td> </td> </tr> <tr> <td class="formaP"> <?php echo $prasanje?> </td> </tr> <tr> <td class="formaO"> <p> <label> <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor1($tekst) ?>" /> <?php odgovor1($tekst) ?></label> <br /> <label> <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor2($tekst) ?>" /> <?php odgovor2($tekst) ?></label> <br /> </p></td> </tr> <tr> <td> <br /> </td> </tr> <?php } ?> </table> <input align="left"type="submit" name="submit" value="Внеси" > </form> </td> </tr> </table> prasanje = question tekst/odgovor = answer The answer table: id - primary question_id - the questions ID whose answer is selected in the radio group user_id - cookie takes care of this answer - the value from radio group date - automatic well this is truely embarrising...i have a insert statement which works within phpmyadmin but when using mysqli_query it returns a error.
INSERT INTO users (username, timestamp) VALUES ('test', UTC_TIMESTAMP())
Unknown column 'timestamp' in 'field list'
i've been playing about with this for a few hours now ...tried changing the column name (timestamp), adding ` around column names as well as table name.
the column exists which is the strangest part, and ive even checked there is no space after the column name in the db.
whats going on please?
How do you do this? Thanks in advance. I have a code to sum up a field called quantityHand in MySQL, in my example below, the value always return 40 as a positive number instead of 40.00- 0.00 0.00 40.00- I have used intval() or floatval() function to convert the type of variable from string to int/float but still not working. Any help please? Here is the code Code: [Select] $sql = mysql_query("SELECT DISTINCT itemNumber, itemDesc, quantityHand, SUM(quantityHand) AS quantityHand FROM inventory where itemNumber like '%$term%' GROUP BY `itemNumber` ORDER BY `itemNumber`"); while ($row = mysql_fetch_array($sql)){ echo "<b>"; echo "</td><td style=\"text-align: center;\"><b>"; echo $row['itemNumber']; echo "</td><td style=\"text-align: center;\"><b>"; echo $row['itemDesc']; //echo "</td><td style=\"text-align: right;\">"; echo "</td><td style=\"text-align: center;\"><b>"; echo $row ['quantityHand']; echo "</td></tr>"; echo "</b>"; I don't want to use AI, or a primary key to sort the data, as it could create some problems if the items are visited out of order, but I want to sort it always based upon a single element for speed.
I'm creating a table that will be populated (built), when it's corresponding entry doesn't exist, how do I get it sorted out if someone visits 2, then 7 before they visit 5 to read what's there? It's not a 2d number line, but a 3d number grid.
Edited by Q695, 09 September 2014 - 11:30 PM. Hello, I am trying to create article directory for learning PHP MySql and i want to write a script for recording number of views for an article and store it in Database. Can anyone please direct me to a nice simple tutorial or help me with the steps or algorithm? Suppose the link for an article is http://www.domain.com/article.php?id=123 Do i need to put extra variables and create a tracking PHP file for it? I'm a newbie on php. I'm really a system administrator and I was just task to do this simple task. For me its hard but I'm sure for a programmer this is very simple. My agenda is to pull out data on one of my column in mysql, select it and dump it on mysql. Here is the php for retrieving mysql data Code: [Select] <?php function database_connect($users) { $resource_link = mysql_connect("localhost", "root", "root"); if (mysql_select_db($users, $resource_link)) { return $resource_link; } else { echo "Cannot connect to DB"; return false; } } function print_dropdown($query, $link){ $queried = mysql_query($query, $link); $menu = '<select username="username">'; while ($result = mysql_fetch_array($queried)) { $menu .= ' <option value="' . $result['id'] . '">' . $result['username'] . '</option>'; } $menu .= '</select>'; return $menu; } //Some other form elements, or just start a form. echo '<form method="post" action="create2.php">'; //The important bit echo print_dropdown("SELECT username FROM mailbox;", database_connect("users")); //Some other form elements, or just end the form. echo '<input type="submit" name="submit" value="submit"/></form>'; Here is the content of my create2.php. This is the php page who do the insert on my mysql. Code: [Select] <?php // open the connection $conn = mysql_connect("localhost", "root", "root"); // pick the database to use mysql_select_db("users",$conn); // create the SQL statement $sql2 = "INSERT INTO mailbox values ('','locked','','$_POST[username]','',NOW(),'','locked','')"; // for troubleshooting $result = mysql_query($sql2, $conn) or die(mysql_error()); // execute the SQL statement //if (mysql_query($sql2, $conn)) { // echo "Success"; //} else { // echo "Fail"; //} } ?> When I click the submit button, I don't see any record being inserted on my table. I'm using the create2.php on my other page though it is only an insert/fill up form not like this one that I need to pull up the date, select and insert to mysql. Hello every body,, i want to create a new database (auto generate duty assigned using form) in PHP and mysql..
i have four input fields:-
Name, Subject, Class & weekly lectures
now i want to insert name,subject & class into database, when i insert number of weekly lecture in field four..
insert automaticlly in database multiple time which i write in field four (weekly lecture)
database: table structure i have already is:::
id-name-subject-class-period-monday-tuesday-wednesday-thursday-friday-saturday
anybody please help me to create this database or just insert query ....
This is probably some obvious error I have made, but I cannot figure it out. I have made a few pages and now I am debugging them. My first page is called insert_purchase_order.php; on this page a person will enter some data in fields and hit the insert button. Then, the data is passed to another page, but when I try to insert into mysql it does not give me any errors, but I have no new rows either. The code for my 2nd page: Code: [Select] <?php session_start(); $action=$_GET[action]; if ($action==insert){ $randid=$_POST['randid']; $vendor=$_POST["vendor"]; $purchase_order_date=$_POST["purchase_order_date"]; $ship=$_POST["ship"]; $fob=$_POST["fob"]; $terms=$_POST["terms"]; $buyer=$_POST["buyer"]; $freight=$_POST["freight"]; $req_date=$_POST["req_date"]; $confirming_to=$_POST["confirming_to"]; $remarks=$_POST["remarks"]; $tax=$_POST["tax"]; $con = mysql_connect("localhost","root","pass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("main", $con); mysql_query("INSERT INTO purchase_order (randid, vendor, purchase_order, ship, fob, terms, buyer, freight, req_date, confirming_to, remarks, tax) VALUES ($randid, $vendor,$purchase_order_date,$ship, $fob, $terms, $buyer, $freight, $req_date, $confirming_to, $remarks, $tax)"); mysql_close($con); echo 'Data Accepted...'; echo '<br/>'; echo 'P.O. Inserted Successfully'; }else{ echo 'Error... Please Contact Bruce.'; echo 'Bruce, no data was passed from the insert_purchase_order.php page.'; } ?> <a href="http://localhost/insert_purchase_order_items.php?po= <?php echo $randid; ?>">Insert Purchase Order Items</a> I have permissions and everything. Thanks I have a mysql insert statement generated with php that is not populating the table. I've echoed the statement and if I copy and paste into phpmyadmin it works fine. The result of the mysql_query function is true. I've emptied the table so there are no primary key conflicts. I've put the statement in a try catch and it does not display a exception. What else can I try? Here's the statement INSERT INTO `wp_term_relationships` (object_id, term_taxonomy_id, term_order) VALUES (1597,83,0) Works absolute fine if I copy and paste into phpmyadmin. Does not populated the table if run through mysql_query |
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