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PHP - Mysql + Php Help Needed For Beginner.
Hei everyone.
I am new on php development and i do not have budget on offering a course so i am trying to learn my self. i need some help on the following : i want to create infopanel.php where the script will get info from. For example if user logs in the script will have to call : <?php include '/incl/infopanel.php' ; ?>The info panel to load all details from database: <?php $name='$myname'; $sur_name='$sname'; $age='$myage'; $country='$mycountry'; ?>So the info with my has to be from database . so can some one help me how to call all information from database to a specific file that i will not have to call always information on every page I hope i have been so clean with my request Similar TutorialsI am getting this error : You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 I have absolutely not a clue what that means or where I went wrong, but ANY help would be appreciated. I've checked my DB username, pass and db name and they are all correct. I am getting the error when I type in $subject_set = mysql_query("SELECT * FROM pages WHERE subject_id = ${subject["id"]}", $connection); if (!$subject_set) { die("database query failed: " . mysql_error()); } this is my code: <?php require_once("includes/connection.php"); ?> <?php require_once("includes/functions.php"); ?> <?php include("includes/header.php"); ?> <div id="sidebar"> <ul> <?php $subject_set = mysql_query("SELECT * FROM subjects", $connection); if (!$subject_set) { die("database query failed: " . mysql_error()); } while ($subject = mysql_fetch_array($subject_set)) { echo "<li>{$subject["menu_name"]}</li>"; } $subject_set = mysql_query("SELECT * FROM pages WHERE subject_id = ${subject["id"]}", $connection); if (!$subject_set) { die("database query failed: " . mysql_error()); } while ($subject = mysql_fetch_array($subject_set)) { echo "<li>{$subject["menu_name"]}</li>"; } ?> </ul> </div> <div id="content"> <h1 class="sidebar">Content Area</h1> </div> </body> </html> <?php mysql_close($connection); ?> Hello, all! I am trying to learn PHP and MySQL on my own, and need some debugging help. What exactly is going wrong here? I am following a tutorial and trying to write the code as it says, but am still having trouble with syntax. Running a WAMP, PHP5.3, and MySQL5.5. This is my code: Code: [Select] <html> <body> <form name = "newVenue" method = "post"> Establishment name: <input type = "text" name = "name"> <br> Street Address: <input type = "text" name = "streetAddress"> <br> City: <input type = "text" name = "city"> <br> State: <select name="state"> <option value="AL">AL</option> <option value="AK">AK</option> <option value="AZ">AZ</option> <option value="AR">AR</option> <option value="CA">CA</option> <option value="CO">CO</option> <option value="CT">CT</option> <option value="DE">DE</option> <option value="DC">DC</option> <option value="FL">FL</option> <option value="GA">GA</option> <option value="HI">HI</option> <option value="ID">ID</option> <option value="IL">IL</option> <option value="IN">IN</option> <option value="IA">IA</option> <option value="KS">KS</option> <option value="KY">KY</option> <option value="LA">LA</option> <option value="ME">ME</option> <option value="MD">MD</option> <option value="MA">MA</option> <option value="MI">MI</option> <option value="MN">MN</option> <option value="MS">MS</option> <option value="MO">MO</option> <option value="MT">MT</option> <option value="NE">NE</option> <option value="NV">NV</option> <option value="NH">NH</option> <option value="NJ">NJ</option> <option value="NM">NM</option> <option value="NY">NY</option> <option value="NC">NC</option> <option value="ND">ND</option> <option value="OH">OH</option> <option value="OK">OK</option> <option value="OR">OR</option> <option value="PA">PA</option> <option value="RI">RI</option> <option value="SC">SC</option> <option value="SD">SD</option> <option value="TN">TN</option> <option value="TX">TX</option> <option value="UT">UT</option> <option value="VT">VT</option> <option value="VA">VA</option> <option value="WA">WA</option> <option value="WV">WV</option> <option value="WI">WI</option> <option value="WY">WY</option> </select> <br> Zip: <input type = "text" name = "zip"> <br> email: <input type = "text" name = "email"> <br> password: <input type = "text" name = "password"> <br> <input type="submit" name="Submit" value="Submit"> </form> <?php //If the form isn't empty, assign the value to a variable if (!empty($_POST['name'])) { $name = $_POST['name']; $address = $_POST['streetAddress']; $city = $_POST['city']; $state = $_POST['state']; $zip = $_POST['zip']; $email = $_POST['email']; $password = $_POST['password']; //Connect to the 'Users' database and store the new bar into the 'Venue' table... mysql_connect ("localhost", "newbar", "Jpr5HJ2K5fWvPLXq") or die ('Oh, fuck: '.mysql_error()); mysql_select_db ("users"); $query = "INSTERT INTO venues VALUES ('NULL', 'testPic.jpg', '".$name."', '".$address."', '".$city."', '".$state."', '".$zip."', '".$email."', '".$password."', 0)"; mysql_query($query) or die ('Oh, fuck: '.mysql_error()); echo "Damn, Nathan. This shit actually worked..."; } ?> </body> </html>This is the error I receive: Code: [Select] Oh, fuck: 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 'INSTERT INTO venues VALUES ('NULL', 'testPic.jpg', 'Nathan's house', '666', 'DAY' at line 1This is my table: Code: [Select] id INT PRIMARY KEY pic_location VARCHAR name VARCHAR address VARCHAR city VARCHAR state VARCHAR zip VARCHAR email VARCHAR password VARCHAR event_name INT Any thoughts as to what is causing this error? Thanks in advance... Im trying to get a code to find an id by name and then find a value by id this kinda works can you help? <?php $con = mysql_connect("localhost","root","password"; if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("ryanteck", $con); $result = mysql_query("SELECT * FROM forum_members WHERE member_name='ryan'"); while($row = mysql_fetch_array($result)) { echo $row['member_name'] . "s Id is " . $row['id_member']; echo "<br />"; $result2 = mysql_query("SELECT * FROM forum_themes WHERE id_member= $id && variable= 'cust_uberis'"); while($row = mysql_fetch_array($result2)) or die (mysql_error()); { echo $id . "Uber Island Is " . $row['value']; echo "<br />"; } } ?> My hosts version of php just seems to automatically add the backslashes. Adding the escape string just seems to add like 3 of them rather than 1 Is this command out-of-date? Code: [Select] <?php $verificate = $_GET["ver"]; $username = $_GET["user"]; $password = $_GET["pass"]; $dbh = mysql_connect("localhost","XXXXXX_dtbusre","my database password here") or die(mysql_error()); mysql_select_db("databasename_zxq_dtb") or die(mysql_error()); $sql = "SELECT loggedin FROM entityTable WHERE username=$username"; $result = mysql_query($sql) or die(mysql_error()); while ($line = mysql_fetch_array($result)){ echo $line[0]."\t".$line[1]."\n"; } mysql_close($dbh); if($verificate === "144356455343"){ echo "SURE"; } else{ echo "NOPE"; } ?> I get this: Code: [Select] You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 Hi, Ive stumbled apon a Mongo database connected to a broadcast script. I would like to change it to a Mysql database can anybody please show me an example how it will look in Mysql format
<?php
/* Require the PHP wrapper for the Mxit API */
require_once ('MxitAPI.php');
/* Function to count the number of users in MongoDB */
function count_users() {
$mongo = new Mongo('127.0.0.1');
$collection = $mongo->sampleapp->users;
$collection->ensureIndex(array('uid' => 1));
return $collection->count();
}
/* Function to get batches of users from MongoDB */
function get_users($skip=0, $limit=50) {
$mongo = new Mongo('127.0.0.1');
$collection = $mongo->sampleapp->users;
$collection->ensureIndex(array('mxitid' => 1,
'created_at' => 1));
$users = $collection->find()->sort(array('created_at' => 1))->skip($skip)->limit($limit);
return iterator_to_array($users);
}
/* Instantiate the Mxit API */
$api = new MxitAPI($key, $secret);
/* Set up the message */
$message = "(\*) Congratulations to our winners (\*)\n\n";
$message .= "1st place - Michael with 100 points\n";
$message .= "2nd place - Sipho with 50 points\n";
$message .= "3nd place - Carla with 25 points\n\n";
$message .= 'Good Luck! :) $Click here$';
/* Mxit Markup is included in the message, so set ContainsMarkup to true */
$contains_markup = 'true';
/* Count the number of users in the database */
$count = count_users();
/* Initialise the variable that counts how many messages have been sent */
$sent = 0;
/* Keep looping through the user list, until the number of messages sent equals the number of users */
while ($sent < $count) {
/* Get users in batches of 50 */
$users = get_users($sent, 50);
/* The list where the user MxitIDs will be stored */
$list = array();
foreach ($users as $user) {
$list[] = $user['mxitid'];
$sent++;
}
/* If there is a problem getting an access token, retry */
$access_token = NULL;
while (is_null($access_token)) {
/* We are sending a message so request access to the message/send scope */
$api->get_app_token('message/send');
$token = $api->get_token();
$access_token = $token['access_token'];
// Only attempt to send a message if we have a valid auth token
if (!is_null($access_token)) {
$users = implode(',', $list);
echo "\n$sent: $users\n";
$api->send_message($app, $users, $message, $contains_markup);
}
}
}
echo "\n\nBroadcast to $sent users\n\n";
Here is my HTML code: Code: [Select] <html> <head> <title>Simple Search Form</title> </head> <body> <form name="searchform" method="get" action="/search.php"> Select Gender: <select name="gender"> <option value="Male">Male</option> <option value="Female">Female</option> </select> Select City: <select name="gender"> <option value="all">All Cities</option> <option value="newyork">New York</option> <option value="toronto">Toronto</option> <option value="london">London</option> <option value="paris">Paris</option> </select> <form> </body> </html> Here is my PHP code: <?php // get the data from the search form # get the gender (male or female) $gender = $_GET['gender']; # get the city $city = $_GET['city']; // connect to mysql and select db mysql_connect('localhost', 'root', 'pass') or die(mysql_error()); mysql_select_db($test_db); // send query $query = mysql_query("SELECT * FROM `visitors_location` WHERE gender='$gender' AND city='$city'"); $count = mysql_num_rows($query); // display data while ( $show = mysql_fetch_assoc($query) ) { echo $gender . " " . $city; } ?> My script basically shows # of males or females in a specific city. How can I show all males in all cities? In other words, let's say I want to show # of Females from all those 4 cities combined. I don't know how to do that. Can someone please help me? I have a webpage where the candidates can attach their resumes and send to the admin.These attachments are saved in the mysql db as blob datatype.In another webpage the admin needs to download all this resumes and see the content. How will i code for that. Hi! I hope somebody can help me what im do wrong. i have checked that the data from the file is in $source_file but nothing imports to the database Code: [Select] <?php include('config.php'); include('opendb.php'); if(isset($_POST['upload'])) { $source_file = @$_POST['userfile']; //$source_file = fopen('http://localhost/test/upload/test.csv', 'r'); $target_table = 'foretag'; function csv_file_to_mysql_table($source_file, $target_table, $max_line_length=10000) { if (($handle = fopen("$source_file", "r")) !== FALSE) { $columns = fgetcsv($handle, $max_line_length, ","); foreach ($columns as $column) { $column = str_replace(".","",$column); } $insert_query_prefix = "INSERT INTO $target_table (".join(",",$columns).")\nVALUES"; while (($data = fgetcsv($handle, $max_line_length, ";")) !== FALSE) { while (count($data)<count($columns)) array_push($data, NULL); $query = "$insert_query_prefix (".join(",",quote_all_array($data)).");"; mysql_query($query); } fclose($handle); } } function quote_all_array($values) { foreach ($values as $key=>$value) if (is_array($value)) $values[$key] = quote_all_array($value); else $values[$key] = quote_all($value); return $values; } function quote_all($value) { if (is_null($value)) return "NULL"; $value = "'" . mysql_real_escape_string($value) . "'"; return $value; } } include('closedb.php'); echo "<br>done<br>"; ?> Hello. I'm a newbie so sorry if this isn't the best forum to post my problem.
I am using a MySQL and PHP to create a web app. I have authentication, and I can register users. I also have a form that users provide information and it is successfully inserting data into a table in my database.
I will use fictional fields for my database table called meal_info:
username
dateStartedDiet
numberMealsPerDay
costPerMeal
Problem: Select user-specific data from the MySQL database, using Session username to select only the current user's data, then display it and do some calculations.
Here is thecode at the top, and I am fairly sure it's working:
session_start(); //execute commone code
require("common.php"); //includes code to connect to database, etc.
if(empty($_SESSION['user'])) I'm attempting to implement a simple social networking system but at the moment am confused about how to create a multiple query which will display a certain user's friends list. The database contains four tables, the two tables that I'm using at the moment at 'usersTable' and 'friendshipsTable' are detailed below. usersTable | Table that stores all the user data UserID | Default primary key Forename | Surname | Username | Password | Email Address | friendshipTable | Table that stores information about friendships between users FriendshipID | Default primary key userID_1 | UserID userID_2 | UserID Status | Either Pending or Confirmed. The user's id is parsed into the url, and then saved into a variable. blah.com/userprofile.php?id=6 $id = $_GET['id']; I am familiar with creating simple queries, but can't quite work out how to set up multiple table queries. What the query needs to do is to check the userID that is parsed with the url, and then check the friendshipsTable by checking if either the userID_1 or userID_2 field matches the userID to grab the records from the table where there is a match. The next step is to check to see if the friendship is 'Confirmed' or 'Pending' and if it is 'Pending' to ignore it. Once the records have then been chosen I need the query to then check the value in either userID_1 or userID_2 that doesn't match userID and then pull the user's username and name from the usersTable so it can be displayed on a webpage. I've no idea hoe much I may or may not be overcomplicating this, an example of the code that I've got so far for this query can be found below, but that's as far as I've got at the moment. $displayFriends = mysql_query("SELECT * FROM friendshipTable, usersTable WHERE friendshipTable.userID_1='$id' OR friendshipTable.userID_2='$id' "); Cheers for any help. I am creating a site to display some products. For ease of updating I want it to run off a MySQL database. I have created the database and php scripts to output and input data etc. I know want to show that data in my web pages. My question is.... Is it best to insert HTML into the php output script to display the information and make the site look how I want....OR ....... should I create a template of the site in HTML and then somehow call the php output script (and the particular row of the database...) Basically... should I put the html code into the php - OR - put the php into the HTML?? I hope this make sense...... thanks This topic has been moved to Miscellaneous. http://www.phpfreaks.com/forums/index.php?topic=333523.0 First of all, sorry for my bad english. I have this: PHP Code: [Select] $link = file_get_contents("http://www.imdbapi.com/?i=tt1285016"); $json = json_decode($link,true); echo $json["Title"]; and I want to replace tt1285016 with $info["id"]; (this is something from mysql, first time when i use ). If I put echo $info["id"]; it return exactly what i need: tt1421545. How can I do that? thank u very much and sorry again for bad enlighs, not native language. EDIT: Sorry for bad section, first time when i come here. Hello, For all purposes, I am a complete beginner. I just know the basics of passing form data through to an email, and displaying the text of a variable on the next page via ECHO, POST, etc.. My situation is that I do not know what sort of code to use to accomplish the job of what I want done. I have created a form to be used for displaying insurance information. This form allows users to select the following: To Get Started: State You Live In: Car Information: Model Year: Original Listing Price: Your Car Currently Is: Your Information: Do You Rent Or Own? Are You Married? Do You Have Children Under 18 Years Of Age? Gross Annual Household Salary Do You Have Health Insurance? How Much Are You Worth In Total Assets (Savings, Equity, Stocks, 401K, Car, Home, etc...) ------------------------ Next to each question, I have a box from which they can select their options (standard html form code). The form has an action and the method is post. Some code is displayed below.. Now so far I am able to display the text of whatever item they selected, by using Echo $_Post etc and the answer page... Here is my problem... I want to display to the user, "recommended insurance limits" based on the data they select from the drop down boxes. So... If a user lives in Alabama, and has an income of $150K+, I want to display a different answer than someone who only earns $30K and lives in Alabama, and the list goes on, (range of options for each question). I have no idea what sort of code to use to display the appropriate answer. I thought I could use the "if, elseif, else" statement, and simply do hundreds of variations on it for each state, but surely there is a more refined and less bloated code for doing such a thing? I would need to display a different answer for the user, for each separate option they choose. Example) Alabama, Model Year of 1995, Salary of $50K would be DIFFERENT then Alabama, Model Year of 2000, Salary of $50K, etc... Thank you for your time. Here is my php code. Code: [Select] <?php $state = $_POST["state"]; $modelyear = $_POST["modelyear"]; $carprice = $_POST["carprice"]; $carownership = $_POST["carownership"]; $homeownership = $_POST["homeownership"]; $marriage = $_POST["marriage"]; $childrenunder18 = $_POST["childrenunder18"]; $salary = $_POST["salary"]; $healthinsurance = $_POST["healthinsurance"]; $assets = $_POST["assets"]; ?> <?php if ($state=="Alabama") { echo "Alabama requires the following minimum insurance limits:<br /> $25,000 - Liability Per Person<br /> $50,000 - Liability For Total Persons<br /> $25,000 - Property Damage Total"; } elseif ($state=="Alaska") { echo "Alaska requires etc..."; } else { echo "You did not select a state."; } ?> Hi, I'm trying to set up a mysql database on my laptop, XP, and connect to a test database using php. I've created the database called 'testdb' with user account 'test' and password 'password. The code I'm using to try and connect to the test datase is the following; $dbconnect=mssql_connect("localhost", "test", "password"); But when I open the page in a web browser, I am getting the following error; Fatal error: Call to undefined function mssql_connect() in <file_location> I'm hoping I'm doing something simple here, but, I've looked up forums and this seems to be the code to use to connect. Anyone who could help would be greatly appreciated. Thanks. hey guys, I am very new to PHP and wanted to create a simple form script. Somehow it doesnt work... can you help me? Help is highly appreciated!! here is the code: Code: [Select] <?php $admin= 'name@email.com'; // 1. Message to the admin $subject1= 'You have one new subscription'; $message1.= 'Email: '.$email."\n\n"; $message1.= 'Name: '.$name."\n\n"; // Sending mail mail($admin, $subject1, $message1, "From: $email"); header('Location: http://www.youtube.com'); ?> Hi guys, sorry for the beginner issues here. after following a tutorial i come across a little problem! Everything works just fine, no errors but my page source only shows the xml tags and not the actual products <?php $link = mysql_connect("localhost","Joao","password"); mysql_select_db("brimelow_store"); $query = 'SELECT * FROM products'; $results = mysql_query($query); echo "<?xml version=\"1.0\"?>\n"; echo "<products>\n"; while ($line = mysql_fetch_assoc($results)); { echo "<item>" . $line["product"] . "</item>\n"; } echo "</products>\n"; mysql_close($link); ?> source = <?xml version="1.0"?> <products> <item></item> </products> I have checked the database names and they all match... im confused. can i get some help pls? hi there i am a beginner in PHP and i really would like some help with this.....
i need to make use of the date() function to retrieve the current date. Use the split() function to retrieve the day month and the year from the current date. and the calculations to display the age.
if anyone could help me with this it would be amazing.
thank you!!
Attached Files
newagecalc.php 1.56KB
0 downloads Hi there, I'm new to PHP about 5month (previously i don't have any programming background), i study the basic PHP through online. i found a tutorial from : http://www.phpwebcommerce.com/ , and there is some error in this tutorial + i need customize this tutorial for my site. I'm here to ask , is the tutorial suitable beginner like me ? is the tutorial consider for advanced used ? i able to solve some error in this tutorial but it take too long. Can give a solution ? should i give up this tutorial on my site ? or just continue find solution ? but i'm already spent almost 2month in this tutorial. so far, i left shipping cost cant find solution...(but i have try do it for 2weeks) Thank you. |
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