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PHP - Php Insert Records Mysql Indexed Db
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; } } Similar TutorialsCode: [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($query) or die ("could not open db".mysql_error()); sleep(2); ?> Why won't the code insert into my database upon submission of data? What am I doing wrong? When submitting the form, the records are sometimes inserted and sometimes not. I receive no errors stating what the problem is. Code: [Select] <?php if (!function_exists("GetSQLValueString")) { function GetSQLValueString($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. 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. 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
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; Hello all, I am new to php coding and have a couple of problems with editing records in my database! I have two files below one test.php and edit.php. In the test.php the code outputs the records into a table. The problem is with the edit link as when it is selected I wish to be able to edit a record by a form, which is on edit.php. I am trying bring up the movie's information on the form to be edited. Currently on the form i get; Quote Movie: movie Gen Genre Year: year Any ideas how I can edit the record and then return to the test.php page? Code: [Select] test.php <html> <body style="background-color:#669999;"> <table width="490"border=0><tr> <td colspan="2" style="background-color:#FFA500;"> <div id="header" <h3 style="color:black">This is my first web-page! Below is a database of some of my favourite movies! </h3> </td> </tr> <?php //connecting to server $con = mysql_connect("localhost","root","NYOXAkly"); if (!$con) { die('could not connect: ' . mysql_error()); } //selecting movie database mysql_select_db("my_mov",$con); //Check if add button is active, start this if(isset($_REQUEST['add'])) { echo "<meta http-equiv=\"refresh\"content=\"0;URL=form.php\">"; } $result = mysql_query("SELECT * FROM Films ORDER BY filmID"); ?> <!-------------------------------creating table------------------------------------------------------------------------------------> <table width="490" border="0" cellspacing="1" cellpadding="0"> <tr> <td> <form name="test1" method="post" action="test.php"> <table width="490" border="10" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC"><tr> <td bgcolor="#FFFFFF"></td> <td align="center" colspan="6" bgcolor="#FFFFFF">Movie Database</td></tr> <td align="center" bgcolor="#FFFFFF">filmID</td> <td align="center" bgcolor="#FFFFFF">Movie</td> <td align="center" bgcolor="#FFFFFF">Genre</td> <td align="center" bgcolor="#FFFFFF">Year</td> <td align="center" bgcolor="#FFFFFF">Edit</td> <td align="center" bgcolor="#FFFFFF">Delete</td> </tr> <?php while($rows=mysql_fetch_array($result)) { ?> <tr> </td> <td bgcolor="#FFFFFF"><? echo $rows['filmID']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['movie']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['genre']; ?></td> <td bgcolor="#FFFFFF"><? echo $rows['year']; ?></td> <td bgcolor="#FFFFFF"> <a href="edit.php?filmID=<?php echo 'filmID';?>">Edit</a> <td bgcolor="#FFFFFF"> <a href="delete.php?filmID=<?php echo 'filmID';?>">Delete</a> </td> </tr> <?php } echo print_r(error_get_last()); mysql_close(); ?> <!--add button--> <tr> <td colspan="15" align="left" bgcolor="#FFFFFF"> <input name='add' type="submit" filmID="add" value="Add A New Record" action="form.php?"> </td> </tr> </table> </form> <br/> By C.M.D.W <br/> <?php echo date("Y/m/d") . "<br />"; ?> </body> </html> Code: [Select] edit.php <html> <body style="background-color:#669999;"> <!------------------Creates a form ----------------------> <br /> <form action="" method="post"> <fieldset> <legend>Enter your movies into database here!</legend> Movie: <input type ="text" name="movie" value="<?php echo 'movie';?>"> <br /> Gen <input type ="text" name="genre" value="<?php echo 'genre';?>"/> <br /> Year: <input type ="text" name="year" value="<?php echo 'year';?>"/> <br /> <input type="submit" name="name" value="Submit" /> </fieldset> </form> <?php //connecting to server $con = mysql_connect("localhost","root","NYOXAkly"); if (!$con) { die('could not connect: ' . mysql_error()); } //selecting movie database mysql_select_db("my_mov",$con); if (isset($_POST['submit'])) { // confirm that the 'id' value is a valid integer if (is_numeric($_POST['filmID'])) // get form data $filmID = $_POST['filmID']; $movie = mysql_real_escape_string(htmlspecialchars($_POST['movie'])); $genre = mysql_real_escape_string(htmlspecialchars($_POST['genre'])); $year = mysql_real_escape_string(htmlspecialchars($_POST['year'])); // check that fields are filled in if ($movie == '' || $genre == '' || $year == '') { // generate error message $error = 'ERROR: Please fill in all required fields!'; } else { // save the data to the database } mysql_query("UPDATE players SET movie='$movie', genre='$genre', year='$year' WHERE filmID='$filmID'") or die(mysql_error()); // once saved, redirect back to the view page header("Location: test.php"); } } if (isset($_GET['filmID']) && is_numeric($_GET['filmID']) && $_GET['filmID'] > 0) { // query db $id = $_GET['filmID']; $result = mysql_query("SELECT * FROM Films WHERE filmID=$FilmID") or die(mysql_error()); $row = mysql_fetch_array($result); // check that the 'id' matches up with a row in the databse if($row) { // get data from db $movie = $row['movie']; $genre = $row['genre']; $year = $row['year']; }} ?> <br/> <br/> <a href="test.php">Return To Home Page</a> <br/> <br/> By C.M.D.W <br/> <?php echo date("Y/m/d") . "<br />"; ?> </body> </html> Thanks Chris Hi guys, I am trying to UPDATE some records on a mySQL database but can't seem to find out why it is not working. This is my code. Code: [Select] <?php $latitude = $_POST['lat_location']; $longitude = $_POST['long_location']; $unique_ID = $_POST['unique_ID']; include('connect2.php'); $query = mysql_query("SELECT * FROM user_location WHERE unit = '$unique_ID'"); $numrows = mysql_num_rows($query); if ($numrows == 1) { $query2="UPDATE user_location SET lat = '$latitude', long = '$longitude' WHERE unit = '$unique_ID'"; mysql_query($query2); $test = "matches"; } else { mysql_query("INSERT INTO user_location VALUES ('','$unique_ID','$latitude','$longitude')"); $test = "not match"; } echo $test . "<br />"; echo $numrows; ?> The script receives the data via the POST method and assigns it to variables. Then I query the database for one of those variables and check how many results are found. If 0 results are found then a new record is created on the database, but if there is 1 record found then the record that is found has to be UPDATED. When the result is 0 the scripts creates the new record fine, but if the result is 1 it doesn't update. I just can't figure out why. Any help will be greatly appreciated. Thanks in advanced, Hi guys, I need your help. I am checking on a database as I want to see if the records have been updated or not. Code: [Select] <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbuser'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbtable'); $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)); } $username = clean($_GET['user']); $password = clean($_GET['pass']); $test = clean($_GET['test']); $public = clean($_GET['public']); if (isset($_GET['user']) && (isset($_GET['pass']))) { if($username == '' || $password == '') { $errmsg_arr[] = 'username or password are missing'; $errflag = true; } } elseif (isset($_GET['user']) || (isset($_GET['test'])) || (isset($_GET['public']))) { if($username == '' || $test == '' || $public == '') { $errmsg_arr[] = 'user or others are missing'; $errflag = true; } } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $qry="SELECT * FROM members WHERE username='$username' AND passwd='$password'"; $result=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if ($username && $password) { if(mysql_num_rows($result) > 0) { $qrytable1="SELECT images, id, test, links, Public FROM user_channel_list WHERE username='$username'"; $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { echo "<p id='test'>"; echo $row['test'] . "</p>"; echo '<p id="images"> <a href="images.php?test=test&id='.$row['id'].'">Images</a></td> | <a href="http://' . $row["links"] . '">Link</a> </td> | <a href="delete.php?test=test&id='.$row['id'].'">Delete</a> </td> | <span id="test">'.$row['Public'].'</td>'; } } else { echo "user not found"; } } elseif($username && $test && $public) { $qry="SELECT * FROM members WHERE username='$username'"; $result1=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if(mysql_num_rows($result1) > 0) { $qrytable1="SELECT Public FROM user_channel_list WHERE username='$username' && test='$test'"; $result2=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if(mysql_num_rows($result2) > 0) { $row = mysql_fetch_row($result2); mysql_query("UPDATE user_list SET Public=('$_GET[public]') WHERE username='$username' AND test='$test'"); echo "update!"; } else { echo "already updated!"; } } else { echo "user not found"; } } } ?> When I run debug the code on my php, if i input the data in a url bar while the records are the same as the data that I enter in the url, i should get the print out on my php "already updated", but I keep getting "update!". Do you know how i can check on mysql database to see if the records have been updated or not?? I have a problem . I 've been trying for a long time to make an update for php mysql to change the data. but every time I do not manage to make it work with a form. but it works if I only if I put this ($ sql = "UPDATE users SET username = 'value' WHERE id = 10 " ; ) so it only works when I put the value of the id. but I want in an html form to indicate what I want to change and what id goes. but I have tried so long that I do not feel like I so want someone help me. make the same database and same as my records and make the code and test it if it works show me please my database name : web test my table called : users my records are called : id int ( 11) AUTO_INNCREMENT username , varchar ( 255 ) password , varchar ( 255 ) first_name , varchar ( 255 ) last_name , varchar ( 255 ) email, varchar ( 255 ) Age, int ( 11) Look, my update.php is like this now <?php $servername = "localhost"; $username = "root"; $password = "....."; $dbname = "webtest"; $conn = new mysqli($servername, $username, $password, $dbname); if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } $sql = "UPDATE users SET password='cotton candy' WHERE id=10"; if ($conn->query($sql) === TRUE) { echo "Record updated successfully"; } else { echo "Error updating record: " . $conn->error; } $conn->close(); ?> but now i have still have to go into the php file to change the valeu or the id but i looked on site and youtube how to put it in a simple html form but it still does not work. i want it in a html from. I want that when I enter the ID that the data of the user appears and that I can change any valeu separately. I'm sorry to be back so soon, but I'm up against another mystery. I'm using the code below to enter a bunch of css data from a spreadsheet into a mysql table. I think the data file is OK. The array created by the script checks out with print_r. (There are many more records than shown. I truncated it to save space.) The problem is that I get this error regarding my sql statement, not the data or anything else: Fatal error: Uncaught PDOException: SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'check, name, phone, email, entry_fee, print_fee, image_name, description, med...' at line 1 in /Users/studio/Sites/BannerProject/b-as/_test_site/csv_to_array.php:242 Stack trace: #0 /Users/studio/Sites/BannerProject/b-as/_test_site/csv_to_array.php(242): PDO->prepare('INSERT INTO tbl...') #1 {main} thrown in /Users/studio/Sites/BannerProject/b-as/_test_site/csv_to_array.php on line 242 I've typed it in a dozen times to make sure there are no errors and keep getting the same error. I tried running a test file and gradually increasing the number of placeholders and at some point I always end up getting the same error, I can delete the most recent addition and it works again. Then I can add another placeholder exactly as before and it works the second time. It feels like a ghost in the machine. Any idea what I am doing wrong? An I typing something I don't see?
<?php
require '__classes/Db.php';
$csvData = '1,FALSE,Carol Lettko,,,TRUE,FALSE,Carol_Lettko-DSC_3022.jpg,Baby Herons/Brickyard,photo,,,
,,,925-285-0320,cjl164@aol.com,,,Carol_Lettko-DSC_0164.JPG,Heron/Brickyard,photo,,,
,,,,,,,Carol_Lettko-IMG_5723.jpg,Kayaker/Brickyard,photo,,,
,,,,,,,,,,,,
2,FALSE,Louise Williams,,,TRUE,FALSE,Louise_Williams-BirdsOfAFeatherAOPR.jpg,Alligator with Words,Book Excerpt,,,
,,,510-232-9547,lkw@louisekwilliams.com,,,Louise_Williams-Hope-TheFairyChickenAOPR.jpg,Hope The Fairy Chicken,,,,
,,,The d exatrfrfvct/.*tygrvurr,,,,,,,,,
,,,,,,,,,,,,
3,TRUE,Dorothy Leeland,,lelanddorothy@gmail.com,TRUE,FALSE,DJ_Lee-bridge at dusk 700px width.jpg,Bridge,photo,,,
,,,,,,,DJ_Lee-friends 700px width.jpg,Friends,photo,,,
,,,,,,,DJ_Lee-hybiscus 700 px wide.jpg,Hibiscus,photo,,,
,,,,,,,,,,,,
4,FALSE,Rita Gardner,,,TRUE,FALSE,Rita_Gardner-Explosion - Gardner photo.JPG,Explosion,photo,,,
,,,,tropicrita@msn.com,,,Rita_Gardner-Ferry Point tables and chair - Gardner.JPG,Ferry Point Tables,photo,, ,
,,,,,,,Rita_Gardner-Forks - Gardner photo.JPG,Forks,photo,,,
,,,,,,,,,,,,
';
$lines = explode(PHP_EOL, $csvData);
$array1 = array();
foreach ($lines as $line) {
$array1[] = str_getcsv($line);
}
$stmt = $pdo->prepare("INSERT INTO tbl_person_data (number, check, name, phone, email, entry_fee, print_fee, image_name, description, medium, select, orient, site) VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?)");
foreach ($array1 as $row)
{
$stmt->execute('$row');
}
echo '<pre>';
print_r($array1);
echo '</pre>';
?>
Hello, This is driving me crazy, as I cannot get it working. I have a database that has a table called 'categories'. I need to retrieve all categories with category_id != '1,2,3'. For this, I can this query: Code: [Select] $q = "SELECT * FROM categories WHERE category_id NOT IN (1,2,3)"; The query executes fine. I need this categories retrieved, to be passed to a Smarty Template Engine powered page. I have: Code: [Select] $q = "SELECT * FROM categories WHERE category_id NOT IN (1,2,3)"; $res = mysql_query($q); while($row=mysql_fetch_array($res)) { Assign('categories',$row); } But when retrieving the array in the Smarty page, with a loop, I only see the last item of the array (the last category). If I go back to the while loop, and add a Code: [Select] print_r($row);, I see two arrays (I have two categories). But the question is: How can I work with those two categories, as I need to use them individually outside the while loop? Any help is appreciated. Best Regards, Richi How do you do this? Thanks in advance. Hi, I'm new to the forum so this could go in the MySQL section but I'm not sure. I am trying to make a page that will list all records from a column in HTML table and have a delete button to remove a specific record. I have got to the part where I have listed the records in a table. Note: Only records from a specific column ('links') are printed. Here is the code: Code: [Select] $con = mysql_connect("localhost","***","***"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("***", $con); $result = mysql_query("SELECT * FROM main"); echo "<table border='1'> <tr> <th>Current links</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['links'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); How would I go about adding a delete button next to each record to delete that specific record? Thanks for any help. I wonder whether someone can help me please. I've found http://www.plus2net.com/php_tutorial/ajax-listbox.php tutorial to create a drop down menu using mySQL table data, which, in turn returns a list of results on the page. Following this tutorial I've put together the tables in my database and the required scripts as shown in the tutorial with the one exception, the "z_db.php" file, which I've assumed to be: Code: [Select] <?php mysql_connect("host", "user", "password")or die(mysql_error()); mysql_select_db("database"); ?> The problem I have, is that when I try and run this, I receive the following error: Quote Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /homepages/2/d333603417/htdocs/development/catsearch.php on line 91 which is this line in the search form: echo "</head><body onload="ajaxFunction()";>";. I must admit I've guessed as to the structure of the 'z_db.php' file should look like because this is not shown so perhaps this is the problem. I just wondered wether someone could perhaps take a look at this please and let me know where I've gone wrong. Many thanks and kind regards I want to pull some records plus unique key from one mysql table and then use them as questions in a form/questionnaire. The post form will have six radio buttons to the right of the fields pulled with the key posted but hidden and will wirte to another table. I guess this is a templating question- as I've successfully retrieved the table 1 records via a query - I just can't seem to use them in the form. Below is the code I'm using to view table "PCM1" <?php // query.php require_once 'login.php'; $db_server = mysql_connect($db_hostname, $db_username, $db_password); if (!$db_server) die("Unable to connect to MySQL: " . mysql_error()); mysql_select_db($db_database) or die("Unable to select database: " . mysql_error()); $query = "SELECT * FROM pcm1"; $result = mysql_query($query); if (!$result) die ("Database access failed: " . mysql_error()); $rows = mysql_num_rows($result); for ($j = 0 ; $j < $rows ; ++$j) { echo 'RID: ' . mysql_result($result,$j,'mcde') ; echo 'Statement: ' . mysql_result($result,$j,'oppclub') . '<br />'; ; } Hi, I need to compare records in a table based on the 'datetime' field, like if the difference between time is less than 15min. need to combine the records(rows) as a single resultant row until the time difference between them is less than 15min and when a record's time difference is >15min that should return as another row and so on. Please find following sample table and the required output format Sample Table: Code: [Select] ID NAME DATETIME 1 aaa 2011-10-10 06:30:00 2 bbb 2011-10-10 06:33:00 3 ccc 2011-10-10 06:38:00 4 ddd 2011-10-10 06:40:00 5 eee 2011-10-10 07:10:00 6 ffff 2011-10-10 07:14:00 7 sss 2011-10-10 08:16:00 8 jjj 2011-10-10 08:26:00 9 kkk 2011-10-10 08:28:00 10 mm 2011-10-10 09:46:00 11 ppp 2011-10-10 09:49:00 12 qqq 2011-10-10 09:52:00 Output Needed : Code: [Select] IDs START DATETIME END DATETIME 1,2,3,4 2011-10-10 06:30:00 2011-10-10 06:40:00 5,6 2011-10-10 07:10:00 2011-10-10 07:14:00 7,8,9 2011-10-10 08:16:00 2011-10-10 08:28:00 10,11,12 2011-10-10 09:46:00 2011-10-10 09:52:00 You can see 1st row in the output table having IDs 1,2,3,4 coz the time difference between them is less than 15minutes (it doesn't means time difference between start and end; it is actually the time difference between current record and previous one ; ie; ID-1 and ID-2 have difference less than 15min and ID-2 & ID-3 have difference less than 15min and ID-3 & ID-4 have difference less than 15min, so those 4 records combine together as a single row also shows their start time and end time. Then you can see diff between ID-4 and ID-5 is greater than 15 min so it should display as new row, and so on) How can I acheive above Output with mysql query ?? Please help.. Not sure if this is possible, but.. As most of you likely know, if a person searches Google, they can see how many webpages are indexed by using: site:techcrunch.com Is there any way that I could do this via a php script for 10 different sites and then have the results emailed to me on a daily basis? I'm not asking for a coded solution.. just an idea on how this might be achieved, if it could? Thanks. |
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