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PHP - Add Records In Xml File And Use As Database Table
Hi,
I have contact form in my website project. Users fill First name ,Last Name ,Email etc information. I do not want to create database and table to store this information. I want to store this information in XML file. When user will add information then information will be added in XML file. Any solution for this ? - Thanks Similar TutorialsYes, I am going to be one of those guys today. I have never done XML anything, besides past 2-3 hours or so, and I am getting frustrated. I have learned how to create simple XML files and save them. I am however having hard time figuring out how to create and XML file when trying to access an table from databse. I am looking to export from "registered_users" table all "registrations" and include "first_name", "last_name" and "age"
Sample:
<registered_users>
<registration>
<first_name>Bob</first_name>
<last_name>Smith</last_name>
<age>36</age>
</registration>
<registration>
<first_name>Jack</first_name>
<last_name>Miller</last_name>
<age>45</age>
</registration>
</registered_users>
It would be nice if the file would be saved as spearate "all_users.xml"
Any help would be appreciated
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
Hi friends i need a small help in importing csv files into mySQL database table i tried all possible options but i am no where... here is the example for what i am trying to do Mysql DB name = raw Table name is = dump +--------------------+--------------------+ + Account + Bal + +--------------------+--------------------+ + + + +--------------------+--------------------+ CSV Format 50************13, 11095 When they upload the file it should automatically get inserted into appropriate fields any help would be great i tried all possible scripts found in the net could nothing is happening 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 Code: [Select] <?php ob_start(); session_start(); $pagerank=2; if ($rank < $pagerank){ header('Location:main.php?id=lowrank.php'); } else{ $name1 = $_GET["name"]; $type2 = $_GET['type']; Echo "Your seach for ".$name1." gave the following results: "; $records_per_page = 15; $total = mysql_result(mysql_query("SELECT COUNT(*) FROM systems WHERE '$type2' LIKE '%$name1%'"), 0) or die(mysql_error()); $page_count = ceil($total / $records_per_page); Echo $total." Records Found"; echo "<table border=0>"; $page = 1; if (isset($_GET['page']) && $_GET['page'] >= 1 && $_GET['page'] <= $page_count) { $page = (int)$_GET['page']; } $skip = ($page - 1) * $records_per_page; $result = mysql_query("SELECT * FROM systems WHERE $type2 LIKE '%$name1%' ORDER BY Security DESC LIMIT $skip, $records_per_page") or die(mysql_error()); echo "<table border=1>"; echo "<td>System Name</td><td>Security</td><td>Class</td>"; while ($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td valign =top>".$row['System Name']."</td>"; echo "<td valign =top>".round($row['Security'],1)."</td>"; echo "<td valign =top>".$row['System Type']."</td>"; echo "</tr>"; } echo "</table>"; echo "<br>"; for ($i = 1; $i <= $page_count; ++$i) { echo '<a href="main.php?id=' . $_GET['id'] .'&name='.$name1.'&type='.$type1. '&page=' . $i . '">' . $i . '</a> '; } } ?> The above code takes data entered in a form and is suppose to show the matching results, but no records are showing, it works when I replace the $name1 and $type2 with the actual values, any ideas? Google Chrome Developer Tool causes an error when I run the following and there are not any records. I keep getting errors when I try to change it.
<?php
$host = 'localhost';
$user = 'root';
$pass = '';
$database = 'ecommerce';
$options = array(
PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
PDO::ATTR_EMULATE_PREPARES => false
);
$keyword = $_GET['keyword'];
$dbo = new PDO("mysql:host=$host;dbname=$database", $user, $pass, $options);
$q1 = "SELECT * FROM products INNER JOIN keywords on keywords.keywordID = products.KeywordID and keywords.KeyWord1 = \"$keyword\" ";
$counter = 10;
$counter1 = 0;
foreach ($dbo->query($q1) as $row) {
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. When a record is added its done via a form and processed via inserts.php. I need to use the same form but to recall the data (by id) so it can be editted. Cant work it out though. Help would be well appreciated! edit.php: <CENTER><B>Update a Vehicle</B></CENTER> <BR> <?php $query="SELECT * FROM cars"; $result=mysql_query($query); $i=0; while ($i < $num) { $carname=mysql_result($result,$i,"CarName"); $cartitle=mysql_result($result,$i,"CarTitle"); $carprice=mysql_result($result,$i,"CarPrice"); $carmiles=mysql_result($result,$i,"CarMiles"); $cardesc=mysql_result($result,$i,"CarDescription"); ?> <form action="showroomedit.php" method="post"> <CENTER>Vehicle Name:</CENTER> <CENTER><input type="text" name="CarName" value="<?php echo $carname; ?>"></CENTER> <br> <CENTER>Vehicle Type:</CENTER> <CENTER><input type="text" name="CarTitle" value="<?php echo $cartitle; ?>"></CENTER> <br> <CENTER>Vehicle Price:</CENTER> <CENTER><input type="text" name="CarPrice" value="<?php echo $carprice; ?>"></CENTER> <br> <CENTER>Vehicle Mileage:</CENTER> <CENTER><input type="text" name="CarMiles" value="<?php echo $carmiles; ?>"></CENTER> <br> <CENTER>Vehicle Description:</CENTER> <CENTER><textarea name="CarDescription" rows="10" cols="30" value="<?php echo $cardesc; ?>"></textarea></CENTER> <br> <CENTER><input type="Submit"></CENTER> </form> </TD> cant work out why it isnt working... hello every body...
two
I'm creating an IPN in paypal for my membership site but the problem I'm facing is that on successfull verification of the purchase, four rows are getting inserted in the database...
The code is
<?php
require '../db.php';
$paypalmode = '.sandbox';
$req = 'cmd=' . urlencode('_notify-validate');
foreach ($_POST as $key => $value) {
$value = urlencode(stripslashes($value));
$req .= "&$key=$value";
}
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, 'https://www'.$paypalmode.'.paypal.com/cgi-bin/webscr');
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_RETURNTRANSFER,1);
curl_setopt($ch, CURLOPT_POSTFIELDS, $req);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, 1);
curl_setopt($ch, CURLOPT_SSL_VERIFYHOST, 2);
curl_setopt($ch, CURLOPT_HTTPHEADER, array('Host: www'.$paypalmode.'.paypal.com'));
$res = curl_exec($ch);
curl_close($ch);
if (strcmp ($res, "VERIFIED") == 0)
{
$transaction_id = $_POST['txn_id'];
$payerid = $_POST['payer_id'];
$firstname = $_POST['first_name'];
$lastname = $_POST['last_name'];
$payeremail = $_POST['payer_email'];
$paymentdate = $_POST['payment_date'];
$paymentstatus = $_POST['payment_status'];
$mdate= date('Y-m-d h:i:s',strtotime($paymentdate));
$otherstuff = json_encode($_POST);
$date = date("y-m-d");
$q = $pdo->connect()->query("INSERT INTO payment (mid,username,amount,paypal_id,txn_id,received_date) VALUES('{$_SESSION['user_id']}','{$_SESSION['uname']}','{$_POST['mc_gross']}','{$_POST['payer_email']}','{$_POST['txn_id']}','$date')");
$q->execute();
$q1 = $pdo->connect()->query("UPDATE members SET amount_loaded = amount_loaded + {$_SESSION['amount']} WHERE mid = '{$_SESSION['user_id']}'");
$q1->execute();
//header("Location: funds.php");
echo "verified";
}
?>
two for the payment and in the members table, the amount is getting doubled. (i.e if anybody purchases For $2, it shows $4 in the database....)
Any help will be really appreciated...
Hello all. Please I need help to update customer table. I get the texboxes populated with select query no problem. I have html page with a search box and when the user enter their first name or last name and click search it will call up the search.php file here is the file: This code below populate the textboxes to be updated: Code: [Select] <?php if(isset($_POST['submit'])){ if(isset($_GET['go'])){ if(preg_match("^/[A-Za-z]+/", $_POST['name'])){ $name=$_POST['name']; } } else{ echo "<p>Please enter a search query</p>"; } } //connect to the database $con = mysql_connect("localhost","dbusrn","dbpwd"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("mydb", $con); $name = $_POST['name']; $result = mysql_query ("select * from Customer_Registration where Firstname like '%$name%' or lastname like '%$name%' "); $row = mysql_fetch_row($result); $cf_uid = $row[0]; $Firstname = $row[6]; $lastname = $row[7]; $Address = $row[8]; $Postcode = $row[9]; $Phone = $row[10]; $Email = $row[11]; $Customer_Type = $row[12]; $Computer_type = $row[13]; $Computer_maker = $row[14]; $Model = $row[15]; $OS = $row[16]; $Appointment_date = $row[17]; $Problem = $row[18]; $Solution = $row[19]; $Comment = $row[20]; ?> <form action="updatecus1.php" method="post"> <table width="100%" border="2" cellspacing="0" cellpadding="8"> <tr><td width="45%" class="FormText">Customer ID:</td> <td width="55%"><?php echo $cf_uid;?></td></tr> <tr><td width="45%" class="FormText">First name:</td> <td width="55%"><input name="Firstname" type="text" value="<?php echo $Firstname;?>"?> </td></tr> <tr><td width="45%" class="FormText">Last name:</td> <td width="55%"><input name="lastname" type="text" value="<?php echo $lastname;?>"?> </td></tr> <tr><td width="45%" class="FormText">Address:</td> <td width="55%"><input name="Address" type="text" value="<?php echo $Address; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Postcode:</td> <td width="55%"><input name="Postcode" type="text" value="<?php echo $Postcode; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Phone:</td> <td width="55%"><input name="Phone" type="text" value="<?php echo $Phone; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Email:</td> <td width="55%"><input name="Email" type="text" value="<?php echo $Email; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Customer:</td> <td width="55%"><input name="Customer_type" type="text" value="<?php echo $Customer_Type; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Computer :</td> <td width="55%"><input name="Computer_type" type="text" value="<?php echo $Computer_type; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Manufactural:</td> <td width="55%"><input name="Computer_maker" type="text" value="<?php echo $Computer_maker; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Model:</td> <td width="55%"><input name="Model" type="text" value="<?php echo $Model; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Operating System:</td> <td width="55%"><input name="First name" type="text" value="<?php echo $OS; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Appointment:</td> <td width="55%"><input name="Operating System" type="text" value="<?php echo $Appointment_date; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Problem:</td> <td width="55%"><textarea name="Problem" rows="10" cols="20" ><?php echo $Problem; ?></textarea></td></tr> <tr><td width="45%" class="FormText">Solution:</td> <td width="55%"><textarea name="Solution" rows="10" cols="20" ><?php echo $Solution; ?></textarea></td></tr> <tr><td width="45%" class="FormText">Comment:</td> <td width="55%"><textarea name="Comment" rows="10" cols="20" ><?php echo $Comment; ?></textarea></td></tr> <td width="55%"><input name="submit" value="submit" type="submit" <br> <input type="Submit" value="Cancel"</br></td> </form> <?php -------------------------------------------------------------------------------------- The update.php file have the following code: Code: [Select] <?php //connect to the database $con = mysql_connect("localhost","dbusrn","dbpwd"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("mydb", $con); if (isset($_POST['submit'])) { $cf_uid = $_POST['cf_uid']; $Firstname = $_POST['Firstname']; $lastname = $_POST['lastname']; $Address = $_POST['Address']; $Postcode = $_POST['Postcode']; $Phone = $_POST['Phone']; $Email = $_POST['Email']; $Customer_Type = $_POST['Customer_Type']; $Computer_type = $_POST['Computer_type']; $Computer_maker = $_POST['Computer_maker']; $Model = $_POST['Model']; $OS = $_POST['OS']; $Appointment_date = $_POST['Appointment_date']; $Problem = $_POST['Problem']; $Solution = $_POST['Solution']; $Comment = $_POST['Comment']; $query = "UPDATE Customer SET Firstname='$Firstname',lastname='$lastname',Address='$Address',Postcode='$Postcode',Phone='$Phone',Email='$Email',Customer_Type='$Customer_Type',Computer_type='$Computer_type',Computer_maker='$Computer_maker',Model='$Model',OS='$OS', Appointment_date='$Appointment_date',Problem='$Problem',Solution='$Solution',Comment='$Comment' WHERE cf_uid='$cf_uid '"; mysql_query($query) or die(mysql_error()); mysql_close($con); echo "<p>Congrats Record Updated</p>"; } ?> I got congrats record updated but in actual fact no record are updated. Please help MOD EDIT: code tags added. This is the form that I'm using and it populates just fine but when you make the changes I can't figure out how to get it to update each record in my database. Code: [Select] <table width="100%" cellspacing="1" cellpadding="2" border="0"> <form action="<?php echo $_SERVER["PHP_SELF"] . "?update=1"; ?>" method="post"> <tr> <td><b>Category Name</b></td> <td><b>Category Description</b></td> <td><b>Order</b></td> <td align="right"><input type="submit" value="Update"></td> </tr> <?php read_cat_list($cat); for ($i = 0; $i < $cat["count"]; $i++) { echo "<tr>\n"; echo "<td width=\"30%\" valign=\"top\"><input type=\"text\" name=\"cat_name_" . safe_string($cat[$i]["cat_id"]) . "\" size=\"30\" maxlength=\"250\" value=\"" . safe_string($cat[$i]["cat_name"]) . "\"></td>\n"; echo "<td width=\"36%\" valign=\"top\"><textarea name=\"cat_desc_" . safe_string($cat[$i]["cat_id"]) . "\" rows=\"2\" cols=\"30\">" . safe_string($cat[$i]["cat_desc"]) . "</textarea></td>\n"; echo "<td width=\"10%\" valign=\"top\"><input type=\"text\" name=\"cat_order_" . safe_string($cat[$i]["cat_id"]) . "\" size=\"3\" maxlength=\"3\" value=\"" . safe_string($cat[$i]["cat_order"]) . "\"></td>\n"; echo "<td width=\"24%\" valign=\"top\">[ <a href=\"#\" onmouseover=\"window.status = 'Delete " . safe_string($cat[$i]["cat_name"]) . "'; return true;\" onmouseout=\"window.status = ''; return true;\" onclick=\"javascript:del_cat(" . $cat[$i]["cat_id"] . ", '" . safe_string($cat[$i]["cat_name"]) . "'); return false;\">Delete</a> ]</td>\n"; echo "</tr>\n"; } ?> <tr> <td colspan="4" align="right"><input type="submit" value="Update"></td> </tr> </form> </table> the safe_string function just cleans up the output/input from the database, This next block is my form processor for this form. Code: [Select] <?php function update_cats($vars) { $err = ""; #if ($SESSION["level"] != ADMIN) { # $err = ERR_NOT_ENOUGH_ACCESS; #} else { $temp = array_keys($vars); for ($i = 0; $i < count($temp); $i++) { if (substr($temp[$i], 0, 9) == "cat_name_") { if ($vars[$temp[$i]] == "") { $err = "Category names cannot be blank."; break; } else { $name_query["cat_name"] .= substr($temp[$i], 9) . ", "; } } } for ($i = 0; $i < count($temp); $i++) { if (substr($temp[$i], 0, 9) == "cat_desc_") { $desc_query["cat_desc"] .= substr($temp[$i], 9) . ", "; } } for ($i = 0; $i < count($temp); $i++) { if (substr($temp[$i], 0, 10) == "cat_order_") { if ($vars[$temp[$i]] == "") { $err = "Category orders cannot be blank."; break; } else { $order_query["cat_order"] .= substr($temp[$i], 10) . ", "; } } } #} if (!$err) { if ($name_query["cat_name"]) { $update_name_query = "update category"; $update_name_query .= " set cat_name = '" . $name_query["cat_name"] . "'"; $update_name_query .= " where cat_id in (" . substr($name_query["cat_name"], 0, -2) . ")"; update_db($update_name_query); } if ($desc_query["cat_desc"]) { $update_desc_query = "update category"; $update_desc_query .= " set cat_desc = '" . $name_query["cat_desc"] . "'"; $update_desc_query .= " where cat_id in (" . substr($desc_query["cat_desc"], 0, -2) . ")"; update_db($update_desc_query); } if ($order_query["cat_order"]) { $update_order_query = "update category"; $update_order_query .= " set cat_order = '" . $name_query["cat_order"] . "'"; $update_order_query .= " where cat_id in (" . substr($order_query["cat_order"], 0, -2) . ")"; update_db($update_order_query); } } return $err; } ?> update_db is my database caller, if you need any of the functions that I use that are not here please post back and tell me. now when I process the form all my fields change to this: cat_name: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, cat_desc: cat_id: 0 for all the records in the database table, I'm just trying to update each record with the values it needs Here is my database table with the data. Code: [Select] CREATE TABLE `category` ( `cat_id` int(11) NOT NULL AUTO_INCREMENT, `cat_name` varchar(255) NOT NULL, `cat_desc` text NOT NULL, `cat_order` int(11) NOT NULL, PRIMARY KEY (`cat_id`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=16 ; -- -- Dumping data for table `category` -- INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(1, 'Hand Tossed Pizza', '', 1); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(2, 'Hand Tossed Specialty Pizzas', '', 2); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(3, 'Chicago Style Deep Dish', '', 3); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(4, 'Specialty Chicago Style Deep Dish', '', 4); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(5, 'Chicken Wings & Tenderloins', '', 5); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(6, 'Hot Sides', '', 6); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(7, 'Hot Sandwiches', '', 7); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(8, 'Cold Sandwiches', '', 8); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(9, 'Pastas', '', 9); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(10, 'Fresh Salads', '', 10); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(11, 'Fresh Breads', '', 11); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(12, 'Soups', '', 12); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(13, 'Kids Menu', '', 13); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(14, 'Drinks', '', 14); INSERT INTO `category` (`cat_id`, `cat_name`, `cat_desc`, `cat_order`) VALUES(15, 'Desserts', '', 15); Hope I have provided plenty of info on what I got and what I need for it todo Thanks 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. 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? the second database found on the cloud
i try to get JSON data but how to insert and update them to another online database with the same table my php script to return json data <?php include_once('db.php'); $users = array(); $users_data = $db -> prepare('SELECT id, username FROM users'); $users_data -> execute(); while($fetched = $users_data->fetch()) { $users[$fetchedt['id']] = array ( 'id' => $fetched['id'], 'username' => $fetched['name'] ); } echo json_encode($leaders);
i get
{"1":{"id":1,"username":"jeremia"},"2":{"id":2,"username":"Ernest"}} Edited March 24 by mahenda First of all excuse me if this topic is inappropriate in this forum. But I think it's rather a PHP problem. I can't figure out multiple duplicate database records on submitting a form. The database table have two columns: the first one 'Id' with AUTO_INCREMENT and the second one 'Name'. Here's the php code for database insertion and the form: ------------------------------------------------------------------ <?php if($_GET['add_name']){ $host = *******; $user = *******'; $pass = *******; $db = *******; $con = mysql_connect($host,$user,$pass) or die; mysql_select_db($db,$con); $name = $_GET['add_name']; $sql = "INSERT INTO names (Name) VALUES ('$name')"; mysql_query($sql); } ?> <form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="GET"> Your Name: <input name="add_name" type="text" /> <input type="submit" value="Submit" /> </form> ------------------------------------------------------------------ After submitting the form to itself once I have multiple Name entries with different Ids. The curious thing is that with Chrome browser I get two duplicate records, with Mozilla - three of them. Seems like mysql_query runs multiple times. It works fine when submitting the form to a separate script and not to itself. Do I miss something? It must be very basic. Hello, i am having an issue when i try to go through multiple pages from a database. The database contains apartment information. I currently have it set to limit my results to 30 per page and this is working correctly, however when i try to scroll to page two like 1..2..3 i am displayed with every apartment listed in the database, and not the certain category that i would like selected. //assigning variables $SESuser_lng = $_COOKIE['SESuser_lng']; $SESuser_admlevel = $_COOKIE['SESuser_admlevel']; $id = $_GET['id']; $pgrow = $_GET['pgrow']; $msg = $_GET['msg']; $ht_filter = $_GET['ht_filter']; //sql query to retrieve result set $htfilter selects the distinct apartment categories $sqlFilter = ($ht_filter > 0 ? " WHERE p.idhotel = $ht_filter" : ""); $sql = "SELECT p.id, p.idhotel, p.nombre, p.thmb, p.archivo, h.nombre FROM prensa p LEFT JOIN hoteles h ON p.idhotel = h.id $sqlFilter ORDER BY p.nombre, h.nombre"; //DB_pager $nav = new DB_Pager('30'); $result = $nav->execute($sql, $dbcnx); //Javascript link creation <span id="display-links"><?php echo $nav->display_links(); ?></span> any help will be greatly appreciated Hi all, I am having difficulty getting my UPDATE script to work. Basically the info is being pulled from a form and then being UPDATED, but it doesn't actually update and I am presented with no errors. It finds the record correctly and displays in the fields in the form as it should, just no update! Any ideas? Thanks in advance... Code: [Select] <?php error_reporting (E_ALL ^ E_NOTICE); $usr = "fdgdg"; $pwd = "dgdg"; $db = "dgdg"; $host = "dgdg.ddg.dg.50dgdg /***********************Get Record***********************/ $ref = (!empty($_GET['ref']))?trim($_GET['ref']):""; # connect to database $cid = mysql_connect($host,$usr,$pwd); if (!$cid){ echo("ERROR: " . mysql_error() . "\n"); } $db_selected = mysql_select_db(database, $cid); $query = sprintf("SELECT * FROM fleet where fleetref like '%s' LIMIT 1","%".mysql_real_escape_string($ref)."%"); $result = mysql_query($query) or die($query."<br>".mysql_error()); $record = mysql_fetch_assoc($result); if(mysql_num_rows($result) > 0) { echo "<strong>Updating fleet vehicle</strong>"; $fleetref=$record["ref"]; $fleetmake=$record["fleetmake"]; $fleetmodel=$record["fleetmodel"]; $fleetberth=$record["fleetberth"]; $fleetyear=$record["fleetyear"]; $fleetlength=$record["fleetlength"]; $fleetchassis=$record["fleetchassis"]; $fleetengine=$record["fleetengine"]; $fleetlayout=$record["fleetlayout"]; $fleettype=$record["fleettype"]; $fleetcomments=$record["fleetcomments"]; $pricelow=$record["pricelow"]; $pricemid=$record["pricemid"]; $pricehigh=$record["pricehigh"]; $fleetof=$record["fleetof"]; }else{ echo "<div id=\"pages\"><strong>Record not found</strong> - Edit Directory</div>"; $fleetmake=""; $fleetmodel=""; $fleetberth=""; $fleetyear=""; $fleetlength=""; $fleetchassis=""; $fleetengine=""; $fleetlayout=""; $fleettype=""; $fleetcomments=""; $pricelow=""; $pricemid=""; $pricehigh=""; $fleetof=$record["fleetof"]; } if ($fleetof == $account) { /***********************Save Record***********************/ # this is processed when the form is submitted # back on to this page (POST METHOD) if (isset($_POST['submit'])){ $fleetmake=$_POST["fleetmake"]; $fleetmodel=$_POST["fleetmodel"]; $fleetberth=$_POST["fleetberth"]; $fleetyear=$_POST["fleetyear"]; $fleetchassis=$_POST["fleetchassis"]; $fleetengine=$_POST["fleetengine"]; $fleetlayout=$_POST["fleetlayout"]; $fleetlength=$_POST["fleetlength"]; $fleettype=$_POST["fleettype"]; $fleetcomments=$_POST["fleetcomments"]; $pricelow=$_POST["pricelow"]; $pricemid=$_POST["pricemid"]; $pricehigh=$_POST["pricehigh"]; # setup SQL statement $query = sprintf("UPDATE `fleet` SET `fleetmake` = '%s',`fleetmodel` = '%s',`fleetberth` = '%s',`fleetyear` = '%s',`fleetchassis` = '%s',`fleetengine` = '%s',`fleetlayout` = '%s',`fleetlength` = '%s',`fleettype` = '%s',`fleetcomments` = '%s',`pricelow` = '%s',`pricemid` = '%s',`pricehigh` = '%s' WHERE `fleetref` ='%s' LIMIT 1", mysql_real_escape_string($fleetmake),mysql_real_escape_string($fleetmodel),mysql_real_escape_string($fleetberth),mysql_real_escape_string($fleetyear),mysql_real_escape_string($fleetchassis),mysql_real_escape_string($fleetengine),mysql_real_escape_string($fleetlayout),mysql_real_escape_string($fleetlength),mysql_real_escape_string($fleettype),mysql_real_escape_string($fleetcomments),mysql_real_escape_string($pricelow),mysql_real_escape_string($pricemid),mysql_real_escape_string($pricehigh),mysql_real_escape_string($fleetref)); #execute SQL statement $result = mysql_db_query($db,$query,$cid) or die($query."<br>".mysql_error()); # check for error if (!$result){ echo("ERROR: " . mysql_error() . "\n$SQL\n"); } echo "<P>Fleet vehicle updated</P>\n"; } } else {echo "Sorry, you are not authorised to edit this vehicle, please contact the system administrator for further help.";} ?> Hi all. I built an online finance tracking thingy for myself and I got it to work using the following database table and PHP code. I didn't write the PHP code, rather, I edited it to suit my needs. It works great for my needs. I have a way to enter records and view the last 180 records in a table. The thing is, while viewing the table, I would lke to be able to edit and delete records. I don't know enough to be able to do this. Can it be done? I realize my PHP code would probably, for the most part, be totally different. Thanks for any help with this. Here is my DB table structu Code: [Select] DROP TABLE IF EXISTS `money`; CREATE TABLE IF NOT EXISTS `money` ( `id` int(5) NOT NULL AUTO_INCREMENT, `date` date NOT NULL, `type` varchar(18) NOT NULL, `checking` decimal(5,2) NOT NULL DEFAULT '0.00', `cash` decimal(5,2) NOT NULL DEFAULT '0.00', `description` varchar(25) NOT NULL, `who` varchar(25) NOT NULL, `note` varchar(50) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ; And here is the PHP: Code: [Select] <?php $result = mysql_query ('SELECT date, type, checking, cash, who, description, note' . ' FROM money' . ' ORDER BY date DESC LIMIT 0,180') or die(mysql_error()); echo "<table border='1' cellpadding='5'>"; echo "<tr><th>Date</th><th>Type</th><th>Checking</th><th>Cash</th><th>Who?</th><th>Description</th><th>Note</th></tr>"; while($row = mysql_fetch_array( $result )) { // Print out the contents of each row into a table echo "<tr><td>"; echo $row['date']; echo "</td><td>"; echo $row['type']; echo "</td><td>"; echo $row['checking']; echo "</td><td>"; echo $row['cash']; echo "</td><td>"; echo $row['who']; echo "</td><td>"; echo $row['description']; echo "</td><td>"; echo $row['note']; echo "</td></tr>"; } echo "</table>"; ?> Hi I'm using this code to insert multiple records. The code executes but nothing is entered into my database. This usually happens when there's a mismatch in data types.
How do I ensure that description goes in as text which in sql is wrapped in single quotes, but also make sure the other variables go in as numeric.
// an array items to insert
$array = array( 'theid' => $theid,
'descr' => $descr,
'costperunit' => $costperunit,
'quantity' => $quantity,
'costperlot' => $costperlot
);
// begin the sql statement
$sql1 = "INSERT INTO descriptions (jobid, description, costperunit, quantity, costperlot) VALUES ";
$it = new ArrayIterator( $array );
// a new caching iterator gives us access to hasNext()
$cit = new CachingIterator( $it );
// loop over the array
foreach ( $cit as $value )
{
// add to query
$sql1 .= "('".$cit->key()."','" .$cit->current()."')";
if( $cit->hasNext() )
{
$sql1 .= ",";
}
}
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>';
?>
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