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PHP - Records Don't Insert And Errors Don't Appear
When submitting the form, the records are sometimes inserted and sometimes not. I receive no errors stating what the problem is.
Code: [Select] <?php if (!function_exists("GetSQLValueString")) { function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") { if (PHP_VERSION < 6) { $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue; } $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue); switch ($theType) { case "text": $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL"; break; case "long": case "int": $theValue = ($theValue != "") ? intval($theValue) : "NULL"; break; case "double": $theValue = ($theValue != "") ? doubleval($theValue) : "NULL"; break; case "date": $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL"; break; case "defined": $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue; break; } return $theValue; } } $editFormAction = $_SERVER['PHP_SELF']; if (isset($_SERVER['QUERY_STRING'])) { $editFormAction .= "?" . htmlentities($_SERVER['QUERY_STRING']); } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) { $insertSQL = sprintf("INSERT INTO customer_tbl (customer_first_name, customer_last_name, customer_company, customer_ac, customer_phone, customer_fax_ac, customer_fax_phone, customer_email) VALUES (%s, %s, %s, %s, %s, %s, %s, %s)", GetSQLValueString($_POST['customer_first_name'], "text"), GetSQLValueString($_POST['customer_last_name'], "text"), GetSQLValueString($_POST['customer_company'], "text"), GetSQLValueString($_POST['customer_ac'], "text"), GetSQLValueString($_POST['customer_phone'], "text"), GetSQLValueString($_POST['customer_fax_ac'], "text"), GetSQLValueString($_POST['customer_fax_phone'], "text"), GetSQLValueString($_POST['customer_email'], "text")); $customer_id = false; if(mysql_query($insertSQL, $connCid)) $customer_id = mysql_insert_id(); else echo "There was an error."; } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) { $insertSQL = sprintf("INSERT INTO quote_request_tbl (customer_id, quote_trans_date, quote_from_company, quote_from_address, quote_from_city, quote_from_state, quote_from_zip, quote_to_company, quote_to_address, quote_to_city, quote_to_state, quote_to_zip, quote_comments) VALUES ('$customer_id', %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s)", GetSQLValueString($_POST['quote_trans_date'], "text"), GetSQLValueString($_POST['quote_from_company'], "text"), GetSQLValueString($_POST['quote_from_address'], "text"), GetSQLValueString($_POST['quote_from_city'], "text"), GetSQLValueString($_POST['quote_from_state'], "text"), GetSQLValueString($_POST['quote_from_zip'], "text"), GetSQLValueString($_POST['quote_to_company'], "text"), GetSQLValueString($_POST['quote_to_address'], "text"), GetSQLValueString($_POST['quote_to_city'], "text"), GetSQLValueString($_POST['quote_to_state'], "text"), GetSQLValueString($_POST['quote_to_zip'], "text"), GetSQLValueString($_POST['quote_comments'], "text")); $quote_id = false; if(mysql_query($insertSQL, $connCid)) $quote_id = mysql_insert_id(); else echo "There was an error."; } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1") && (isset($quote_id))) { $insertSQL = sprintf("INSERT INTO building_tbl (quote_id, building_quantity, building_width, building_length, building_height, building_overhang, slope_id, building_type_id, manufacture_id, foundation_id) VALUES ('$quote_id', %s, %s, %s, %s, %s, %s, %s, %s, %s)", GetSQLValueString($_POST['building_quantity1'], "int"), GetSQLValueString($_POST['building_width1'], "int"), GetSQLValueString($_POST['building_length1'], "int"), GetSQLValueString($_POST['building_height1'], "int"), GetSQLValueString($_POST['building_overhang1'], "int"), GetSQLValueString($_POST['slope_id1'], "int"), GetSQLValueString($_POST['building_type_id1'], "int"), GetSQLValueString($_POST['manufacture_id1'], "int"), GetSQLValueString($_POST['foundation_id1'], "int"); mysql_select_db($database_connCid, $connCid); $Result1 = mysql_query($insertSQL, $connCid) or die(mysql_error()); header('Location: index.php?view=article&id=7'); } mysql_select_db($database_connCid, $connCid); $query_rsSlope = "SELECT * FROM slope_tbl ORDER BY slope_id ASC"; $rsSlope = mysql_query($query_rsSlope, $connCid) or die(mysql_error()); $row_rsSlope = mysql_fetch_assoc($rsSlope); $totalRows_rsSlope = mysql_num_rows($rsSlope); $query_rsSlope = "SELECT * FROM slope_tbl ORDER BY slope_id ASC"; $rsSlope = mysql_query($query_rsSlope, $connCid) or die(mysql_error()); $row_rsSlope = mysql_fetch_assoc($rsSlope); $totalRows_rsSlope = mysql_num_rows($rsSlope); mysql_select_db($database_connCid, $connCid); $query_rsBuildingType = "SELECT * FROM building_type_tbl ORDER BY building_type_id ASC"; $rsBuildingType = mysql_query($query_rsBuildingType, $connCid) or die(mysql_error()); $row_rsBuildingType = mysql_fetch_assoc($rsBuildingType); $totalRows_rsBuildingType = mysql_num_rows($rsBuildingType); mysql_select_db($database_connCid, $connCid); $query_rsManufacture = "SELECT * FROM manufacture_tbl ORDER BY manufacture_name ASC"; $rsManufacture = mysql_query($query_rsManufacture, $connCid) or die(mysql_error()); $row_rsManufacture = mysql_fetch_assoc($rsManufacture); $totalRows_rsManufacture = mysql_num_rows($rsManufacture); mysql_select_db($database_connCid, $connCid); $query_rsFoundation = "SELECT * FROM foundation_tbl ORDER BY foundation_name ASC"; $rsFoundation = mysql_query($query_rsFoundation, $connCid) or die(mysql_error()); $row_rsFoundation = mysql_fetch_assoc($rsFoundation); $totalRows_rsFoundation = mysql_num_rows($rsFoundation); ?> I belve the problem is in this part of the code: if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1") && (isset($quote_id))) Many thanks in advance. Similar TutorialsHey, Im needing some help with an update form that I have created using PHP/MySQL. Basically the form is working great (updating records, retrieving records and showing errors) but there is a problem with the errors. If there are no errors and I edit the existing text input values, the database updates fine. However if I edit the existing text input values and there is an error, the edited input values dont save, they revert back to the values of the database. I understand why this is happening, its because the text inputs are set to show the values of the database so if i edit the existing values and an error occurs or the page is refreshed, the values revert back to the database values. However id like it so that if any text in the text field have been edited and an error occurs, the text stays to how it is until the user by passes any errors and then the database is updated. Hopefully this makes sense. I have attached the php page just incase anybody wants to look into it but any help would be much appreciated! Also, im not looking for someone to just to fix this for me but if someone knows my problem and can guide me in the right direction, that would be great. I created this database Code: [Select] <?php $mysqli = mysqli_connect('localhost', 'admin', 'jce123', 'php_class'); if(mysqli_connect_errno()) { printf("connection failed: %s\n", mysqli_connect_error()); exit(); }else{ $q = mysqli_query($mysqli, "DROP TABLE IF EXISTS airline_survey"); if($q){echo "deleted the table airline_survey....<br>";} else{echo "damm... ".mysqli_error($mysqli);} $sql = "CREATE TABLE airline_survey ( id INT NOT NULL PRIMARY KEY AUTO_INCREMENT, staff CHAR(10) NOT NULL, luggage CHAR(10) NOT NULL, seating CHAR(10) NOT NULL, clean CHAR(10) NOT NULL, noise CHAR(10) NOT NULL )"; $res = mysqli_query($mysqli, $sql); if($res === TRUE) { echo "table created"; } else { printf("Could not create table: %s\n", mysqli_error($mysqli)); } mysqli_close($mysqli); } ?> When I look at it it looks fine. I have a form that sends data to this script: Code: [Select] <?php $con = mysql_connect('localhost', 'admin', 'abc123'); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("php_class", $con); foreach ($_POST as $key => $value) { $staff = ""; $luggage = ""; $seating = ""; $clean = ""; $noise = ""; switch($key){ case "staff": $staff = $value; break; case "luggage": $luggage = $value; break; case "seating": $seating = $value; break; case "clean": $clean = $value; break; case "noise": $noise = $value; break; default: echo "we must be in the twilight zone"; } echo $staff."<br>"; [color=red] mysql_query("INSERT INTO airline_survey (staff, luggage, seating, clean, noise) VALUES ($staff, $luggage, $seating, $clean, $noise)");[/color] } ?> as you can see right before the insert query I test one of the variables to see if it has the string I'm expecting and it does. The problem is the script runs without giving me an error message but the data never gets inserted into the table. Hi,I'm hoping someone can help me:). When I add foreign keys to the database in MySQL I CAN change the primary key field and the corresponding indexed field will automatically change giving me the same value which is exactly what i want. The issue is when I try to insert the values using my php code i get the 'cannot update child error'. I have not selected a value for the unique_id field as I want it to inherit the value from the primary key. Here's my code as simplified as I could do it. <?php //this calls the class productupload and instantiates the function insert_form to insert values into the db. $product = new productUpload(); $product->insert_form();?> //This calls the insert_form function and inserts info into tables <?php class productUpload extends DatabaseObject{ protected static $table_name="itm_details"; protected static $db_fields =array('id','unique_id','itm_cat','itm_make','itm_model','itm_desc','itm_cond','itm_date_from','itm_date_to','itm_add_date'); public $id; public $unique_id; public $itm_cat; public $itm_make; public $itm_model; public $itm_desc; public $itm_cond; public $itm_date_from; public $itm_date_to; public $itm_add_date; Heres the function insert_form which grabs the form submitted data. Note I have not defined a value for $unique_id public function insert_form(){ global $database; //set the object attributes to the form the parameters if(isset($_POST['submit'])) { $result = array( $this->itm_cat =(!empty($_POST['itm_cat'])) ? trim($_POST['itm_cat']) : NULL, $this->itm_make =(!empty($_POST['itm_make'])) ? trim($_POST['itm_make']) : NULL, $this->itm_model = (!empty($_POST['itm_model'])) ? trim($_POST['itm_model']) : NULL, $this->itm_desc =(!empty($_POST['itm_desc'])) ? trim($_POST['itm_desc']) : NULL, $this->itm_cond =(!empty($_POST['itm_cond'])) ? trim($_POST['itm_cond']) : NULL, $this->itm_date_from =(!empty($_POST['itm_date_from'])) ? trim($_POST['itm_date_from']) : NULL, $this->itm_date_to =(!empty($_POST['itm_date_to'])) ? trim($_POST['itm_date_to']) : NULL, $this->itm_add_date = date("Y-m-d H:m:s")); //check date is numeric //if(is_numeric($_POST['itm_date_from'])) { //$this->itm_date_from = $_POST['itm_date_from']; //} //check date is numeric //if(is_numeric($_POST['itm_date_to'])) { //$this->itm_date_to = $_POST['itm_date_to']; //} if($result){ $result = $this->create(); } } } //here's the create function referred to above which sanitises the posted data and inserts it into the db. public function create() { global $database; $attributes = $this->sanitised_attributes(); $sql = "INSERT INTO ".self::$table_name."("; $sql .= join(", ", array_keys($attributes)); $sql .= ") VALUES ('"; $sql .= join("', '", array_values($attributes)); $sql .= "')"; if($database->query($sql)) { $this->id = $database->insert_id(); return true; } else { return false; } } Code: [Select] <?php require "db/config.php"; $fname = $_POST['fname']; $lname = $_POST['lname']; $country = $_POST['country']; $state = $_POST['state']; $city = $_POST['city']; $zcode = $_POST['zcode']; $address = $_POST['address']; $ppemail = $_POST['ppemail']; $pnumber = $_POST['pnumber']; $cemail = $_POST['cemail']; $url = $_POST['url']; $price = "$5.00"; $query = "INSERT INTO custpackage1000( id, FirstName, LastName, Country, State, City, ZipCode, Address, PayPalEmail, PhoneNumber, PrimaryEmail, WebsiteURL) VALUES ( '1', '$fname', '$lname', '$country', '$state', '$city', '$zcode', '$ppemail', '$pnumber', '$cemail', '$url')"; mysql_connect($host, $user, $pass) or die("<br /><br /><h1>Fatal error. Please contact support if this persists.</h1>"); mysql_select_db($dbname); mysql_query($query) or die ("could not open db".mysql_error()); sleep(2); ?> Why won't the code insert into my database upon submission of data? What am I doing wrong? Hi. Im searching for a way to enter multiple records at once to mysql databased on a date range. Lets say i want to insert from date 2010-11-23 to date 2010-11-25 a textbox with some info and the outcome could look in database like below. ---------------------------------------------------------- | ID | date | name | amount | ----------------------------------------------------------- | 1 | 2010-11-23 | Jhon | 1 | | 2 | 2010-11-24 | Jhon | 1 | | 2 | 2010-11-25 | Jhon | 1 | ----------------------------------------------------------- The reason i need the insertion to be like this is because if quering by month and by user to see vacation days then vacations that start in one month and end in another month can be summed by one month. If its in one record then it will pop up in both months. Help is really welcome. I am very new to PHP and haven't wriiten an update program except in the class I took about 4 years ago using MYSQL.
The company I work for is using PDO so I am having a little bit of a learning curve.
I have a table with a checkbox at the end and I want to insert the records from the table into a new file I created called
OPPSHEDT
SHEDORD Order Number 14 Char
SHEDBORD BackOrder Number 2 Char
SHEDPKD Picked Cases 9,0 Decimal
SHEDSHP Shipped Cases 9,0 Decimal
I'm just not sure how to write the update program to read the array of records and run through the loop and insert the items checked. I saw some other examples on here and all over the internet but they were using mysql. I have a copy of the code to create my table below.
I'd appreciate the help if you can.
Thanks.
ShipEstimate.php 6.41KB
4 downloads
I have a problem where I am getting these errors after a form has been submitted from join.html Here are the errors: Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'scswccla'@'localhost' (using password: NO) in /home/scswccla/public_html/insert.php on line 9 Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in /home/scswccla/public_html/insert.php on line 9 Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: Access denied for user 'scswccla'@'localhost' (using password: NO) in /home/scswccla/public_html/insert.php on line 9 Warning: mysql_real_escape_string() [function.mysql-real-escape-string]: A link to the server could not be established in /home/scswccla/public_html/insert.php on line 9 Here is the insert.php file http://pastebin.com/EVtgidQC And the join.html file - http://pastebin.com/hqDtGLe5 How would I fix those errors - I got the code from http://forum.hey0.net/showthread.php?tid=2509 The form is at scswc.com/join.html if you want to look at it. I really want to get this working so if you need any more info to help fix it just ask thanks! Hi, I"m trying to create insert images files name into the database and save images in folder, however, after i created databse connection, and excuted it... I'm getting error message said Parse error: syntax error, unexpected $end in C:\xampp\htdocs\ourdeaf\uploadimages\upload.php on line 76 That line 76 is at end of code after </html>, what went wrong! Code: [Select] <form action="<? echo $_SERVER['PHP_SELF']; ?>" method="post" ENCTYPE="multipart/form-data"> Upload:<br><br> <input type="file" name="image"><br><br> <input type="hidden" name="uploaded" value="1"> <input type="submit" value="Upload"> </form> <? }else{ //if the form hasn't been submitted then: //from here onwards, we are copying the file to the directory you made earlier, so it can then be moved //into the database. The image is named after the persons IP address until it gets moved into the database //get users IP $ip=$REMOTE_ADDR; //don't continue if an image hasn't been uploaded if (!empty($image)){ //copy the image to directory copy($image, "./temporary/".$ip.""); //open the copied image, ready to encode into text to go into the database $filename1 = "./temporary/".$REMOTE_ADDR; $fp1 = fopen($filename1, "r"); //record the image contents into a variable $contents1 = fread($fp1, filesize($filename1)); //close the file fclose($fp1); //encode the image into text $encoded = chunk_split(base64_encode($contents1)); //insert information into the database mysql_query("INSERT INTO images (img,data)"."VALUES ('NULL', '$encoded')"); //delete the temporary file we made unlink($filename1); } //end } ?> I have a form with a series of text boxes where the User enters specific data, be it text, numeric, or date/time. It is not mandatory to fill in every box, but when I submit the form to be put into a mysql table, I get errors such as "Incorrect time value: '' for column 'starttime' at row 1" "Incorrect decimal value: '' for column 'workkj' at row 1" How can I setup the mysql table to accept blank answers? Or is there something I need to do in php to fix this? Sorry if this is simple... I'm still learning. Thanks for the help! I cannot get this INSERT to work. Not records are being added to the sys_city_dev table. No query errors are being thrown. I am simply trying to add ALL records from all_illinois to sys_city_dev and Mid should have be add as 11 in all inserts. city_name of course will be field city_name from all_illinois. Here is my code: $query = "SELECT * FROM all_illinois"; if ($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 ($rows = mysqli_fetch_array($results)) { echo $rows['city_name'] . "<br />"; $mid_id = '11'; $insert_city_query = "INSERT INTO sys_city_dev (ID, Mid, cityName, forder, disdplay, cid) VALUES (' ','" . $mid_id . "','" . $rows['city_name'] . "', '', '','')"; if (!$insert_city_query) exit(mysql_error()); } } } Here is my insert table structure and 5 records dump: -- -- Table structure for table `sys_city_dev` -- CREATE TABLE IF NOT EXISTS `sys_city_dev` ( `ID` int(11) NOT NULL AUTO_INCREMENT, `Mid` int(11) NOT NULL DEFAULT '0', `cityName` varchar(30) NOT NULL DEFAULT '', `forder` int(4) NOT NULL DEFAULT '0', `disdplay` int(4) NOT NULL DEFAULT '0', `cid` int(11) NOT NULL DEFAULT '0', PRIMARY KEY (`ID`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=113970 ; -- -- Dumping data for table `sys_city_dev` -- INSERT INTO `sys_city_dev` (`ID`, `Mid`, `cityName`, `forder`, `disdplay`, `cid`) VALUES (84010, 1, 'Dothan', 0, 0, 0), (84011, 1, 'Alabaster', 0, 0, 0), (84012, 1, 'Birmingham', 0, 0, 0), (84013, 2, 'Flagstaff', 0, 0, 0), (84014, 1, 'Auburn', 0, 0, 0); And the all_illinois dump w/ 5 records: -- -- Table structure for table `all_illinois` -- CREATE TABLE IF NOT EXISTS `all_illinois` ( `state_id` varchar(255) NOT NULL, `city_name` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; -- -- Dumping data for table `all_illinois` -- INSERT INTO `all_illinois` (`state_id`, `city_name`) VALUES ('135', 'Abingdon'), ('135', 'Adair'), ('135', 'Addieville'), ('135', 'Addison'), ('135', 'Adrian'); I am having trouble showing reports for a given date range. Currently if I specify something like 11/03/2010 to 11/05/2010 I get results for all years within that month and day such as I may get results for 11/03/2008 11/03/2009 11/03/2010 11/04/2008 11/04/2009 11/04/2010 11/05/2008 11/05/2009 11/05/2010. I am using the following code $result = mysql_query("SELECT * FROM report WHERE date>='$date_begin' and date<='$date_end' ORDER BY 'date'",$db); I use the following format in my date feild mm/dd/yyyy My script is finally working as intended, but I want to add some additional data results. I am trying to resolve how I can display a count of ONLY the records updated by my query: // START :: Query to replace matches mysql_query("UPDATE orig_codes_1a AS a JOIN old_and_new_codes_1a AS b ON concat(a.orig_code_1, a.orig_code_2) = concat(b.old_code_1, b.old_code_2) SET a.orig_code_1 = b.new_code_1, a.orig_code_2 = b.new_code_2") or die(mysql_error()); // END :: Query to replace matches In this query I count ALL records selection: // START :: Create query to be displayed as final results of original codes table. $result = mysql_query("SELECT * FROM orig_codes_1a") or die(mysql_error()); I want to display a count on this part of the query: ....concat(a.orig_code_1, a.orig_code_2) = concat(b.old_code_1, b.old_code_2).... How do I display the actual errors in IIS 7.5? If I miss a semicolon, I get: "HTTP Error 500.0 - Internal Server Error" absolutely useless. In prior versions, I could see the line and get to the PHP error. How do I display PHP errors? I've added: set_ini('display_errors', '1'); but it doesn't help. Can 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($results)) else { echo "No results to get!"; } ?> Here is my all_illinois INSERT table structu CREATE TABLE IF NOT EXISTS `all_illinois` ( `state_id` varchar(255) NOT NULL, `city_name` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; Here is my source table geo_cities structu CREATE TABLE IF NOT EXISTS `1` ( `CityId` varchar(255) NOT NULL, `CountryID` varchar(255) NOT NULL, `RegionID` varchar(255) NOT NULL, `City` varchar(255) NOT NULL, `Latitude` varchar(255) NOT NULL, `Longitude` varchar(255) NOT NULL, `TimeZone` varchar(255) NOT NULL, `DmaId` varchar(255) NOT NULL, `Code` varchar(255) NOT NULL ) ENGINE=MyISAM DEFAULT CHARSET=latin1; I'm missing something here. I have a form, and when the submit is pressed, the relevant post data inserts into table one, then I want the last insert id to insert along with other form data into a second table. The first table's still inserting fine, but I can't get that second one to do anything. It leapfrogs over the query and doesn't give an error. EDIT: I forgot to add an error: I get: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'usage, why VALUES ('14', '', '123', '','1234', '', '')' at line 1 query:INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('test 14', 'asdfa', 'asdf', 'adf','asdf', '', '', '', '123', '', '') Code: [Select] if (empty($errors)) { require_once ('dbconnectionfile.php'); $query = "INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('$description12', '$sn', '$description4', '$cne','$description5', '$description6', '$description7', '$description8', '$description9', '$description10', '$description11')"; $result = @mysql_query ($query); if ($result) { $who_donated=mysql_insert_id(); $query2 = "INSERT INTO tbl_donation (donor_id, donor_expyear, donor_cvv, donor_cardtype, donor_authorization, amount, usage, why) VALUES ('$who_donated', '$donate2', '$donate3', '$donate4','$donate5', '$donate6', '$donate7')"; $result2 = @mysql_query ($query2); if ($result2) {echo "Info was added to both tables! yay!";} echo "table one filled. Table two was not."; echo $who_donated; //header ("Location: http://www.twigzy.com/add_plant.php?var1=$plant_id"); exit(); } else { echo 'system error. No donation added'; I am have a query with a limit of 3 i want it to echo the first two in one li class and the 3 in another li class. This is what i have, but i know the logic is not right, but i cant figure it out Code: [Select] </div> <?php echo "<ul class='services-list'>"; //echo "<li class='bubble'>"; $i = 1; foreach ($rows_class as $record_class ){ ?> <li class="bubble"><a href="viewclass.php?class_id=<?php echo base64_encode($record_class[class_id]) ?>" /><h6 class="title"><img src="./images/services/picture-services.png" class="picture" alt="logo" /><? echo $record_class[class_title]; ?></h6></a><p><?php $string = $record_class['description']; echo substr(strip_tags($string), 0, 200); ?></p> <?php if( $i % 2 === 0 && $record_class !== end($rows_class) ) { echo "</li>\n<li class='bubble last'>\n";?> <?php } elseif ( $record_class === end($rows_class) ) { echo "</li>\n "; } $i++; } ?> </ul> Hi there im trying to add up the number of different values i have in a table and echo the results: Ships ------ ShipID INT: auto count ShipName VARCHAR: Something Class INT: (either 1 to 4) The class row is the one i want to count, so if i have say 5 ships; first class 1, second class 2, third class 3, fourth class 4 and fith another class 1. It would total how many of each classes i have: two class 1, one class 2, one class 3 and one class 4. Ultimately i am going to do a similar count on another table and compare the results, but thats for later as im really not sure where to start. Ive created a query selecting all the appropriate ships to be tallied up: Code: [Select] } mysql_select_db($database_swb, $swb); $query_Ships = sprintf("SELECT * FROM ships WHERE PlayerName = %s", GetSQLValueString($colname_Ships, "text")); $Ships = mysql_query($query_Ships, $swb) or die(mysql_error()); $row_Ships = mysql_fetch_assoc($Ships); $totalRows_Ships = mysql_num_rows($Ships); If you could please help me that would be ace! Thank You Hi all Complete noob here..... What I have to do is create a new database every year, but I can't just import last years data completely, only records when needed. I can do it myself in PHPMyAdmin, but I don't want my employees touching mysql directly. What I am trying to do is write a php script that will serach for a reocrd in last years database and if it exists insert it into the new database if it doesn't exists bring up a form so it can be entered. Any help would be greatly appreciated. Thanx ZZ Hi there i have 4 fields in a record and 3 records in a table that i need to add up. Heres what they look like when outputted: 0024 0103 1126 Each of the fields that need to be added are called: Class1, Class2, Class3, Class4 So i want to add each number in those fields with the result of the records so above would look like: 1, 2, 4, 13 Would i need to put them in an array of somekind?? Code: [Select] $colname_resultp = "-1"; if (isset($_SESSION['MM_Username'])) { $colname_resultp = (get_magic_quotes_gpc()) ? $_SESSION['MM_Username'] : addslashes($_SESSION['MM_Username']); } mysql_select_db($database_swb, $swb); $query_resultp = sprintf("SELECT Class1, Class2, Class3, Class4 FROM planet WHERE PlayerName = %s", GetSQLValueString($colname_resultp, "text")); $resultp = mysql_query($query_resultp, $swb) or die(mysql_error()); $row_resultp = mysql_fetch_assoc($resultp); $totalRows_resultp = mysql_num_rows($resultp); <?php do { ?> <?php echo $row_resultp['Class1']; echo $row_resultp['Class2']; echo $row_resultp['Class3']; echo $row_resultp['Class4']; ?> <?php } while ($row_resultp = mysql_fetch_assoc($resultp)); mysql_free_result($resultp); ?> If you could please, please help that would be great as im finding this a real headache at the moment :( Thank you Tom |
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