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PHP - Need Help About Code To Retrieve Image And Other Data From Mysql
I m trying to fetch a image from mysql (blob) with header..here is my coding..."<?php
include("db.php"); $query=mysql_query("select * from table where id='3' "); $row=mysql_fetch_array($query); $r=$row['image']; header("content-type:image"); echo $r; ?>" i want to fetch another fields from the database....but when i try to echo another fields...the page shows error or it does not echo other fields of database....please help me...how can i resolve it...i want to fetch other fields from database,like'username'password'firstname'lastname and image...thanks in advance... Similar Tutorialshi guys, im new to this forum I'm new also to php, I need help from you guys: I want to display personal information from a certain person (the data is on the mysql database) using his name as a link: example: (index.php) names 1. Bill Gates 2. Mr. nice Guy i want to click Bill Gates (output.php) Name: Bill Gates Country:xxxx Age: xx etc. How can i make this or how to learn this? I want to use session to do a query and will I be able to do this? I have a session that was gathered from login and now i was to use this session to do a query If Yes, How? I insert multiple id from my checkbox to mysql database using php post form. in e.x i insert id (checkbox value table test) to mysql. no i need to any function for retrieve data from mysql and print to my page with my e.x output.(print horizontal list name of table test where data = userid) my checkbox value ( name table is test ) : Code: [Select] 01 ---id----- name ---- 02 ---1 ----- test1 ---- 03 ---2 ----- test2 ---- 04 ---3 ----- test3 ---- 05 ---4 ----- test4 ---- 06 ---5 ----- test5 ---- 07 ---6 ----- test6 ---- 08 ---7 ----- test7 ---- 09 ---8 ----- test8 ---- 10 ---9 ----- test9 ---- mysql data Insert ( name of table usertest ): Code: [Select] 1 ---id----- data ---- userid ----- 2 ---1 ----- 1:4:6:9 ---- 2 ----- 3 ---2 ----- 1:2:3:4 ---- 5 ----- 4 ---3 ----- 1:2 ---- 7 ----- example outout : ( print horizontal list name of table test where data = userid ) print? Code: [Select] 1 user id 2 choise : test1 - test4 - test6 - test9 Thanks Hi folks, Complete No0b here when it comes to PHP and MySQL. I am in the middle of creating a PHP website. What I want to do is have the contents of a page in a MySQL table and have PHP gather the page content and display it on the page. Here is what I have in my home.tpl file: Code: [Select] <?php $query = "SELECT home, FROM $database_name"; $result = mysql_query($query); ?> And, in my index.php I simply call that home.tpl to display the data by using: Code: [Select] <?php include 'templates/default/home.tpl'; ?> Now, all I get is a blank page on index.php. Is there something else I should be doing? Remember, I am a complete No0b! Thanks folks. I am not sure if the title is correct; I tried my best.
I'm a PHP/MySQL beginner and I really need some help.
I have a small script that I am using for sending SMS. I recently added a phonebook. The problem with the phonebook right now is that it's available to all users, i.e. they can all update and delete all rows. What I would like to do is make it so that each user can update and delete only their own contacts.
I have a table call contacts. Inside that table there is first name, last name, company and phonenumber.
How can I accomplish this with PHP & MySQL?
CREATE TABLE IF NOT EXISTS `contacts` ( `contact_id` int(10) NOT NULL AUTO_INCREMENT, `firstname` varchar(255) NOT NULL, `lastname` varchar(255) NOT NULL, `company` varchar(255) NOT NULL, `cell_no` text NOT NULL, PRIMARY KEY (`contact_id`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ; CREATE TABLE IF NOT EXISTS `users` ( `user_id` int(11) NOT NULL AUTO_INCREMENT, `users_name` varchar(30) NOT NULL, `uname` varchar(30) NOT NULL, `u_pass` varchar(60) NOT NULL, `utype` varchar(30) NOT NULL, `timezone` varchar(30) NOT NULL, `uapi_user` varchar(30) NOT NULL, `uapi_pass` varchar(60) NOT NULL, PRIMARY KEY (`user_id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=20 ;
<?php
if (isset($_POST['submit'])){
//form has been submitted1
$firstname = trim($_POST['firstname']);
$lastname = trim($_POST['lastname']);
$company = trim($_POST['company']);
$cellno = trim($_POST['cell_no']);
if($firstname == ''){
echo '<div class="alert alert-danger">First Name is not Valid!</div>';
exit;
}elseif($lastname == ''){
echo '<div class="alert alert-danger">Last Name is not Valid!</div>';
exit;
}elseif($company == ''){
echo '<div class="alert alert-danger">Company is not Valid!</div>';
exit;
}elseif($cellno == ''){
echo '<div class="alert alert-danger">Cellphone Number is not Valid!</div>';
exit;
}else{
$query = "Select cell_no from contacts where cell_no = '".$cellno."' ";
$result = mysql_query($query);
if (!mysql_num_rows($result)) {
$sql = "INSERT INTO contacts(firstname, lastname, company, cell_no)
values('{$firstname}','{$lastname}', '{$company}', '{$cellno}')";
$result = mysql_query($sql);
confirm_query($result);
//echo '<div class="alert alert-success">Successfully added.</div>';
//exit;
?>
<script type="text/javascript">
window.location = "contact_list.php";
</script>
<?php
}
else{
echo '<div class="alert alert-danger">Username. already exist!.</div>';
echo '<p><a href="new_contact.php" class="btn btn-success"> Back </a></p>';
exit;
}}
}else{
$firstname = "";
$lastname = "";
$company = "";
$cellno = "";
}
?>
Hello guys, I am a new programmer and i am building a new website. I would like to give me your advice for the following problem: I want to build a webpage, in which there will be a <div id="book-content"> ...............</div> part. Inside this div i would like to dynamically display pages from a book. Each page will have text, scripting code blocks, blocks with the output of each scripting code, and images. There will be a bar on the left of the webpage in which the user can select which page of the book he wants to load. For example...My web page will look something like this... Code: [Select] <HTML> <HEAD> </HEAD> <BODY> <DIV ID="PAGE-HEADER"> //LOGO OF THE WEB PAGE </DIV> <DIV ID="PAGE-MENU"> //PAGE MENU </DIV> <DIV ID="BOOK-INDEX" WITH FLOAT:LEFT > //HERE A WILL SHOW THE CHAPTERS OF THE BOOK AND THE PAGES OF EACH CHAPTER </DIV> <DIV ID="BOOK-PAGE"> //THE PAGE SELECTED FROM THE PREVIOUS 'BOOK-INDEX' MENU WILL BE DISPLAYED HERE WITH A MYSQL QUERY ****** </DIV> </BODY> </HTML> Then in a mysql database, i would like to have records with a text field, that will contain for example the followng: <h1> Chapter 1: bla bla </h1> <p> in this chapter we will speak about bla bla bla.... </p> <div id="code"> int main() { int x,y; x=2; y=3; x=x+y; } </div> <p> this will outpout the following:</p> <div id="code output"> x=5! </div> I would like to get this html code from the database, and then show it in the <div id=BOOK-PAGE> div. But i dont want to use php eval(). Also, if i store the code to a file and then include it, i will have too may files(equal to the book's number of pages etc 100). Any ideas? i have 8 division (div), i want to display 4 rows in 4 division and the remain 4 rows in the next 4 division here is my code structure for carousel
<div class="nyie-outer"> second row third row
fourth row fifth row sixth row seven throw eighth row
</div><!--/.second four rows here-->
sql code
CREATE TABLE product( php code
<?php how can i echo that result in those rows
Hey everyone. I am creating a website so my family can select from the list of present my daugther has asked for. I have them logging in, and that works. I have the table data set and the search and table display works. But I would like to display the list (as in the code below) but with a checkbox in the first column (add a column). From that the authenticated user can click on the item they have purchased and that selection will be moved to another table (called purchased). I have that code. The only thing I cannot figure out is to present the data in list form with a checkbox (like you would see on an order form). Here is the display code (there is no error checking at this point): <head><LINK REL="SHORTCUT ICON" HREF="cmwschl.ico"></head> <body bgcolor="#C0C0C0"> <font face="Arial" color="#000080">Books from the Selected Series</font> </h1> <hr font color="Navy" font size="3"> <center> <table border="1" cellpadding="5" cellspacing="0" bordercolor="#000000"> <tr> <td width="170" bgcolor="#FFCC66"><center><b><font color="navy" size="2" face="Trebuchet MS"><b><center>Book Number</center></font></b></center></td> <td width="140" bgcolor="#FFCC66"><center><b><font color="navy" size="2" face="Trebuchet MS"><b><center>Book Group</center></font></b></center></td> <td width="100" bgcolor="#FFCC66"><center><b><font color="navy" size="2" face="Trebuchet MS"><b><center>Book Title</center></font></b></center></td> </tr> <?php $con = mysql_connect("localhost","twilson","R00tb33r!") or die('Connection: ' . mysql_error());; mysql_select_db("CMWWeb", $con) or die('Database: ' . mysql_error()); ?> <? $group = $_POST['bookgrp']; //if ($Prod <> 0) { $sql = "SELECT * FROM `OpenBooks` WHERE `bookgrp`= '$group'"; $results = mysql_query($sql); if ($results) { //this will check if the query failed or not if (mysql_num_rows($results) > 0) { //this will check if results were returned while($copy = mysql_fetch_array($results)) { $variable1=$copy['booknum']; $variable2=$copy['bookgrp']; $variable3=$copy['booktitle']; //table layout for results print ("<tr>"); print ("<td><center>$variable1</center></td>"); print ("<td><center>$variable2</center></td>"); print ("<td><left>$variable3</left></td>"); print ("</tr>"); } } else { echo "No results returned"; } } else { echo "Query error: ".mysql_error(); } mysql_close($con); ?> </table> </center> Hello everyone, I begin in everything web related but I have been programming for years. I tried to code something simple : small Mysql DB (works fine) and to begin a search bar to browse data. I adapted a code that I understood provided here : https://www.cloudways.com/blog/live-search-php-mysql-ajax/. Base principle is simple : as you type in your query, it will pass the text to script.js that will forward this request to ajax.php file. In the ajax.php, a javascript function named “fill()” will pass the fetched results. This function will also display the result(s) into “display” div in the “search.php” file. The problem is that when I type anything it displays, below the search bar, at the moment I type a character: Quote
'; //Fetching result from database. while ($Result = MySQLi_fetch_array($ExecQuery)) { ?> ")'>
instead of the actual answer from my database (no error in the browser console). I tested the SQL query + the user I provide and everything seems fine. Any clue what could be the root cause ? I strongly suspect a mistake in the code as I already corrected one (script.js instead of scripts.js) but I really cannot figure out where. Thanks in advance,
problematic code (ajax.php):
<?php
//Including Database configuration file.
include "db.php";
//Getting value of "search" variable from "script.js".
if (isset($_POST['search'])) {
//Search box value assigning to $Name variable.
$Name = $_POST['search'];
//Search query.
$Query = "SELECT Name FROM search WHERE Name LIKE '%$Name%' LIMIT 5";
//Query execution
$ExecQuery = MySQLi_query($con, $Query);
//Creating unordered list to display result.
echo '
<ul>
';
//Fetching result from database.
while ($Result = MySQLi_fetch_array($ExecQuery)) {
?>
<!-- Creating unordered list items.
Calling javascript function named as "fill" found in "script.js" file.
By passing fetched result as parameter. -->
<li onclick='fill("<?php echo $Result['Name']; ?>")'>
<a>
<!-- Assigning searched result in "Search box" in "search.php" file. -->
<?php echo $Result['Name']; ?>
</li></a>
<!-- Below php code is just for closing parenthesis. Don't be confused. -->
<?php
}}
?>
</ul>
I have some php code on view invoice and it's getting all the data I need to from three database tables but for some reason it's not getting the data for one specific column and unsure why I have done a echo sql and it is echoing the sql query ok and have ran the sql query in phpmyadmin and is displaying the column in phpmyadmin that is not displaying on the php page so am bit confused, below is the coding I have. I took out a lot of it and tried to keep to the relevant parts <?php
$sql = "SELECT t1.id, t1.invoice_number, DATE_FORMAT(t1.invoice_date,'%d/%m/%Y') AS invoice_date, t1.user_id, t2.invoice_number, t2.service_name, t2.service_price, t3.user_id, t3.customer_name, t3.customer_phone, t3.customer_email, t2.service_name, t2.service_price, t1.invoice_total, t1.balance, t1.notes, DATE_FORMAT(t1.payment_date,'%d/%m/%Y') AS payment_date, t1.payment_method
FROM invoices as t1
LEFT JOIN invoice_products as t2
ON t1.id = t2.invoice_id
LEFT JOIN users as t3
ON t1.user_id = t3.user_id
WHERE t1.user_id = '".$_SESSION['user_id']."' and t1.id = '".$_GET['id']."'";
$result = mysqli_query($connect, $sql);
if(mysqli_num_rows($result) > 0)
{
if($row = mysqli_fetch_array($result))
{
?>
Notes: <?php echo $row['notes'];?>
<?php
}
}
?>
I have code for displaying result from database:
foreach ($_SESSION["products"] as $cart_itm)
{
$product_code = $cart_itm["code"];
$results = $mysqli->query("SELECT * FROM products WHERE product_code='$product_code' LIMIT 1");
$obj = $results->fetch_object();
Now, I display data with: $obj->product_name
When I get the list, how to put inline numbers (1 2 3 4 ...)?
1 products1
2 products1
3 products1
4 products1..
This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=322859.0 Hi all, I have the following code that generates a table of results from a MySQL query: $i=1; while($arr = mysql_fetch_array($result, MYSQL_NUM)) { $table .= "<tr>" ."<td width='5px';><input type='checkbox' name='transcheck".$i."'></td>" ."<td id='parent".$i."A'>".$arr[0]." ".$arr[1]."</td>" ."<td id='parent".$i."B'>".$arr[2]."</td></tr>" ."<tr style='display:none'><input type='hidden'; name='transemail".$i."'; value='".$arr[9]."'></tr>"; $i++; } $table .= "</table>"; echo $table; As you can see it generates a table containing (amongst other things): checkboxes: transcheck1,2........9...etc. hidden inputs: transemail1,2........9...etc. This table is inside a form, so that the above gets posted to another php file. What I now want to do in this 2nd php file, is to retrieve all the checkboxes and hidden inputs and then to display the values of the hidden inputs, where the corresponding checkbox has been checked. So e.g. if transcheck1, transcheck3 and transcheck12 have been checked, then I want to display transemail1, transemail3 and transemail12. I can see that this should be relatively straightforward, but I'm fairly new to this stuff, could someone pls help me out? Thanks! Hi I have coded a drop down menu with php and i am trying to retrieve the data when a user select a option from the menu and the data is retrieved from the database. So far i have tried and nothing is displaying when i tried to process the php form. Sales.php Page <form action="saleprocess.php" method="GET"> <?php echo 'Product Model:'; $query="SELECT * FROM products"; /* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */ $result = mysql_query ($query); echo "<select name=product_model value=Select>Product Model</option>"; // printing the list box select command while($rows=mysql_fetch_array($result)){//Array or records stored in $nt echo "<option name=product_model value='.$rows[product_id].'>$rows[product_model]</option>"; /* Option values are added by looping through the array */ } echo "</select><br>"; ?> <input type='submit' name='submit' value='Create'></input> <br> </form> ******************************************************************************** Salesprocess.php page <?php include("connect.php"); if(isset($_GET['product_id'])){ $product_id = $_GET['product_id']; $query = mysql_query("SELECT * FROM products WHERE product_id= $product_id"); while($rows = mysql_fetch_assoc($query)) { echo 'Product Model<br>'; echo $rows['product_id']; echo $rows['product_model']; } } ?> Muchly appreciated if someone can help me Hi, I have an HTML form with a "submit" button that sends form data to a PHP script. Firstly, the form doesn't seem to send the data to the PHP script, unless I include a line in the PHP script as follows: $var1 = $_REQUEST['var1']; Is this always the case or is there a way to automatically have the form send through data? The reason I ask is because the form includes elements that can be disabled, in which case (if element 'var1' is disabled) the above PHP line returns an error "Notice: Undefined index: var1" i.e. I want to only send through data from enabled elements and not from disabled elements. Alternatively, is there a line I can include in the PHP script something along these lines: if (defined($var1)==FALSE) {$var=" ";} so that when the PHP script tries to retrieve var1, which is undefined because the corresponding form element has been disabled, it sets it to a particular value. my HTML form is along these lines: Code: [Select] <form action=something.php method=POST> <select id="var1" name="var1"> <option> </option> <option selected="selected"> option1</option> <option> option2</option> <option> option3</option> </select> <input type="submit" value="next"> </form> Thanks! I am trying to create a code where the user enters the zip code and if found in the database, a location will be retrieved. I am getting two undefined variables as errors. I am still learning the POST and GET method and I fear that is where I am getting something mixed up. I have also attached an image of my DB. Thank you so much any input is greatly appreciated. Notice: Undefined variable: address in C:\xampp\htdocs\index.php on line 53 Notice: Undefined variable: zip in C:\xampp\htdocs\index.php on line 54
<?php
include ('header.php');
include ('function.php');
?>
<div class="wrapper">
<div>
<?php
echo'<form method="POST" action"'.getLocations($conn).'">
<input type="text" name="zip" class="search" placeholder="Zip code"><br>
<button type="submit" value="submit" id="submit">Submit</button>
</form>';
getLocations($conn);
?>
</div>
</div>
<div>
<!--foreach($zip as $zip) : -->
<?php
echo $address['address'];
echo $zip['zip'];
?>
</div>
function.php file code:
<?php
$dBServername = "localhost";
$dBUsername = "root";
$dBPassword = "";
$dBName = "addresses";
// Create connection
$conn = mysqli_connect($dBServername, $dBUsername, $dBPassword, $dBName);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
function getLocations($conn) {
if (isset($_POST['submit'])){
// Validate Zip code field
if (!empty ($_POST['zip']) && is_numeric ($_POST['zip'])) {
$zip = (int)$_POST['zip'];
$query = "SELECT * FROM locations WHERE zip = '$zip'";
//get the results
$result = mysqli_query($conn, $query);
//fetch the data
$zip = mysqli_fetch_all($result, MYSQLI_ASSOC);
var_dump($zip);
mysqli_free_result($result);
//close connection
mysqli_close($conn);
}
}
}
so what i have going on is that i need help writing a basic script to upload original and copy with resize for thumbnail. then also need to rename both image files with content in the form. My server supports GD Library and that imagemagik or whatever it is lol. upload dirs orig: ../media/photos/ thumb: ../media/photos/thumb/ mySQL DB: name: m_photos fields: id(INT) m_cat(varchar) m_sub_cat(varchar) pic(varchar) description(varchar) p_group(varchar) I have the form written up for how it should look like: Code: [Select] <form action="upload.php" method="post" enctype="multipart/form-data"> Upload an image for processing<br /> <input type="file" name="Image"><br /> <select name="sub_group"> <option value="Photo Shoot">Photo Shoot</option> <option value="Live Performances">Live Performances</option> <option value="Randoms">Randoms</option> <option value="Fan Photos">Fan Photos</option> </select><br /> Location: <input type="text" name="p_group" /><br /> Date of Pic: <input type="text" name="date" /><br /> Description: <input type="text" name="description" /><br /> <input type="submit" value="Upload"> </form> so what needs to happen is for the file upload name. it needs to grab a count from "p_group" database to give it a starting number in a "00" pattern and then the date of pic needs to go in there next then the p_group name. so when the files gets uploaded it would look something like this after upload. "03 - Oct 21, 2011 - The Mex.jpg" NOTE: all pics must be converted to ".jpg" extension. both the orig and thumb will use that same file name cuz they just go into different dirs re-sizing for the thumbnail should be done by aspect ratio and needs to either be 300px width or 400px height. so if anyone would like to help me out please do. im not the greatest at writing in php yet I'm creating a report page and can't figure out how to retrieve multiple rows from a table and display them in html. There are 5 data elements in each row and I'm thinking that using a form in the html might be the best way as I'm totally ignorant about tables in html. I'm a newbie to php and can't figure out how to accomplish this. Here's the code that I have so far:
<?php
$filename = NULL;
session_start();
// start of script every time.
// setup a path for all of your canned php scripts
$php_scripts = '../php/'; // a folder above the web accessible tree
// load the pdo connection module
require $php_scripts . 'PDO_Connection_Select.php';
//*******************************
// Begin the script here
// Connect to the database
if (!$con = PDOConnect("foxclone")):
{
echo "Failed to connect to database" ;
exit;
}
else:
{
$sql = 'SELECT COUNT(IP_ADDRESS) FROM download WHERE FILENAME IS NOT NULL';
$sql1 = 'Update download t2, ip_lookup t1 set t2.country = t1.country, t2.area = t1.area, t2.city = t1.city where ((t2.IP_ADDRESS) = (t1.start_ip) OR (t2.IP_ADDRESS) > (t1.start_ip)) AND ((t2.IP_ADDRESS) = (t1.end_ip) OR (t2.IP_ADDRESS) < (t1.end_ip)) AND (t2.FILENAME is not null and t2.country is null)';
$sql2 = 'SELECT (IP_ADDRESS, FILENAME, country, area, city) from download where FILENAME is not null';
// Update the table
$stmt = $con->prepare($sql1);
$stmt->execute();
// Get count of rows to be displayed in table
$stmt = $con->prepare($sql);
$stmt->execute() ;
$cnt = $stmt->fetch(PDO::FETCH_NUM);
// retrieve one row at a time
$i = 1;
while($i <= $cnt){
$stmt = $con->prepare($sql2);
$row->execute(array('')); // Do I need an array here?
// from here on, I'm lost
$i++;
I'd appreciate any guidance you can provide or understandable tutorials you can point me to. Larry I have been given a task, and I gotta say it is kicking my butt.
Here is what I have to do.
1. Have user fill out and submit a form.
2. Data gets sent to: http://www.ffiec.gov...de/Default.aspx
3. Data is set as values for input fields in the sites form.
4. Form executes.
5. Retrieve result data.
6. Display data back to my site.
I have no idea how to do this.
Usually when I have done something like this I use an API.
Hope my question is clear.
Thanks for the help.
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