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PHP - Db Query Wont Return Data
Hi guys, im trying to connect to a database and get the value for the user in the row called 'user_credit', if it equals 1 or more then i want to show the ''You have £ ....'' bit in the script.
Problem is nothing shows at all, even without the if statement. I have changed the value for me in the database so in user_credit the value is 100, which is more than 1 so it should appear. I have probably done something wrong. Any ideas? Code: [Select] <? include '../admin/database/membership_dbc.php'; $r = mysql_query("SELECT * FROM users WHERE user_name='".safe($_SESSION['user_name'])."'") or die ("Cannot find table"); while( $cred = mysql_fetch_array($r) ) { if ($cred >= '1' ) { ?> <p>You have £<? echo $cred['user_credit']; ?> available on you account, would you like to use it on this order?<br> <label for="credit"></label> <select name="credit" id="credit"> <option value="Y" selected>Yes, use credit</option> <option value="N">No, save credit</option> </select> </p> <? } } ?> Similar TutorialsHey, In my script I am currenlty working on I have 2 classes one which calls the second in the hope it will return a value to it but doesnt send the variable back. The example code below give you and idea of what I am looking for as my script it too long to add. Code: [Select] class oneClass{ function bar(){ $var1 = "test" $two = twoClass(); $result = $two->foo($var1); echo $result; } } class twoClass{ function foo($var1){ $result = $var1 . ' is successful!'; $this->fooTwo($result); } function fooTwo($result){ $result = $result . ' Pass me back now?'; return($result); //Want to pass the variable back to "oneClass->bar();" } } // Starts script $testme = new oneClass(); $testme->bar(); This is a very simple example but states what i am wanting to. Any ideas would be much appreciated! Thanks Here's the code that deals with the client side:
<?php
session_start();
if(!isset($_SESSION['Logged_in'])){
header("Location: /page.php?page=login");
}
?>
<!DOCTYPE Html>
<html>
<head>
<!--Connections made and head included-->
<?php require_once("../INC/head.php"); ?>
<?php require_once("../Scripts/DB/connect.php"); ?>
<!--Asynchronously Return User Names-->
<script>
$(document).ready(function(){
function search(){
var textboxvalue = $('input[name=search]').val();
$.ajax(
{
type: "GET",
url: 'search.php',
data: {Search: textboxvalue},
success: function(result)
{
$("#results").html(result);
}
});
};
</script>
</head>
<body>
<div id="header-wrapper">
<?php include_once("../INC/nav2.php"); ?>
</div>
<div id="content">
<h1 style="color: red; text-align: center;">Member Directory</h1>
<form onsubmit="search()">
<label for="search">Search for User:</label>
<input type="text" size="70px" id="search" name="search">
</form>
<a href="index.php?do=">Show All Users</a>|<a href="index.php?do=ONLINE">Show All Online Users</a>
<div id="results">
<!--Results will be returned HERE!-->
</div>
search.php
<?php //testing if data is sent ok echo "<h1>Hello</h1><br>" . $_GET['search']; ?>This is the link I get after sending foo. http://www.family-li...php?&search=foo Is that mean it was sent, but I'm not processing it correctly? I'm new to the whole AJAX thing. Hi I have a query where it returns a few fields based on the location. I created a class for the function and an index page. It does not return any values. Can someone please advise/ THE CLASS Code: [Select] <?php /**************************************** * * WIP Progress Class * * ****************************************/ class CHWIPProgress { var $conn; // Constructor, connect to the database public function __construct() { require_once "/var/www/reporting/settings.php"; define("DAY", 86400); if(!$this->conn = mysql_connect(DB_HOST, DB_USERNAME, DB_PASSWORD)) die(mysql_error()); if(!mysql_select_db(DB_DATABASE_NAME, $this->conn)) die(mysql_error()); } public function ListWIPOnLocation($location) { $sql = "SELECT `ProgressPoint.PPDescription` AS Description ,`Bundle.WorksOrder` AS WorksOrder, `Bundle.BundleNumber` AS Number, `Bundle.BundleReference` AS Reference,`TWOrder.DueDate` AS Duedate FROM `TWOrder`,`Bundle`,`ProgressPoint` WHERE `Bundle.CurrentProgressPoint`=`ProgressPoint.PPNumber` AND `TWOrder.Colour=Bundle.Colour` AND `TWOrder.Size=Bundle.Size` AND `TWOrder.WorksOrderNumber`=`Bundle.WorksOrder` AND `ProgressPoint.PPDescription` LIKE '" . $location . "%' ORDER BY TWOrder.DueDate DESC"; mysql_select_db(DB_DATABASE_NAME, $this->conn); $result = mysql_query($sql, $this->conn); echo $sql; while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { $return[] = $row; } return $return; } } ?> The index page Code: [Select] <?php // First of all initialise the user and check for permissions require_once "/var/www/users/user.php"; $user = new CHUser(7); // Initialise the template require_once "/var/www/template/template.php"; $template = new CHTemplate(); // And create a cid object require_once "/var/www/WIPProgress/DisplayWIPOnLocation.php"; $WIPProgress= new CHWIPProgress(); $content = "Check WIP Status on Location <br>"; $content = "<form action='index.php' method='get' name ='location'> <select id='location' > <option>Skin Room</option> <option>Clicking</option> <option>Kettering</option> <option>Closing</option> <option>Rushden</option> <option>Assembly</option> <option>Lasting</option> <option>Making</option> <option>Finishing</option> <option>Shoe Room</option> </select> <input type='submit' /> </form>"; $wip = $WIPProgress->ListWIPOnLocation($_GET['location']); // Now show the details $content .= "<h2>Detail</h2> <table> <tr> <th>PPDescription</th> <th>Works Order</th> <th>Bundle Number</th> <th>Bundle Reference</th> <th>Due Date</th> </tr>"; foreach($wip as $x) { $content .= "<tr> <td>" . $x['Description'] . "</td> <td>" . $x['WorksOrder'] . "</td> <td>" . $x['Number'] . "</td> <td>" . $x['Reference'] . "</td> <td>" . $x['DueDate'] . "</td> </tr>"; } $template->SetTag("content", $content); echo $template->Display(); ?> thank you MySQL connection works and it connects to my database but it doesnt insert values into the table that I created. <form action="phplogin2.php" method="post"> Username: <input type="text" name="user" style="color: white; background-color: blue;"/><br/> Password: <input type="password" name="pass" style="color: grey; background-color: black;"/><br/> <button>Login</button> </form> <?php $con = mysql_connect('localhost', 'root', 'eagles1') or die("did not connect"); $dbc = mysql_select_db('mysql') or die("did not connect to database"); $query = mysql_query("INSERT INTO login VALUES('', '$user', '$pass')") or die("query did not work"); $user = $_POST['user']; $pass = $_POST['pass']; if ($con==true){ echo "MySQL Connection Succesful"; } if ($dbc==true){ echo "MySQL Database Connection Succesful"; } if ($query==true){ echo "MySQL Query Succesful"; } ?> ok, this is clearly 1st grade code for some, but i'm not there yet - I'm querying posts in WordPress, and the post_content will always have an image in the beginning of the post followed by the content. i don't want to get the image, just the content that's after the image, which is wrapped in anchor tags, of course. Code: [Select] <a href="http://path/to/image.jpg"><img src="http://path/to/image.jpg" /></a> <p>Post content yadda, yadda, hoowie</p> obviously a character count won't work, so i need to get anything that follows the first "</a>", say...? is this the best way, or is there an easier way? thanks for anyone's help. GN Given the following query,
SELECT t1.a, t1.b, t2.c FROM t1 INNER JOIN t2 ON t2.id=t1.t2_id WHERE t1.pk=123;I get the following three records:
array(
array('a'=>1,'b'=>2,'c'=>4),
array('a'=>1,'b'=>2,'c'=>5),
array('a'=>1,'b'=>2,'c'=>8)
)
What would be the best way to get just one record such as the following?
array('a'=>1,'b'=>2, 'c'=>array(4,5,8))
My thoughts were to use MySQL's GROUP_CONCAT, and then use implode() to turn it into an array, but didn't know if there was a better way.
Thanks
Hi all
we have an ajax query which queries a MySQL database for cities around the world for our hotel search. We are running on a Linux web hosted server.
The problem we are having is when you first visit out site and type in Brisbane or Sydney it takes around 15 seconds maybe to return. After that what ever city you query is lightening fast. I have had a few friends try this and when they first visit our site they experience the same then second time around it is fast.
Could this be some sort of caching issue or setting somewhere that I am missing?
I have attached a file so you can see what I mean
Thanks in advance
Attached Files
ajaxqry.jpg 169.31KB
0 downloads hi there alll newbie here lol just needed help with the code for the next page of results ? IE i have say 30 results returned from a sql query but only want to show ten at a time i no how to limit the result with the LIMIT 10 but how do i get it to put link show the other ten and so on. the code i have is here at the bottom of page http://www.phpfreaks.com/forums/index.php/topic,307971.0.html thanks in advance for any help kaine This is the query which should likes salestrack productid with printers productid, but even though there are data no result is showing and I really dont know why. Please help Code: [Select] SELECT salestrack.orderid AS orderid, salestrack.salesman AS salesman, printers.rrp AS price, salestrack.name AS name, salestrack.phone AS phone, salestrack.email AS email, printers.name AS printername, salestrack.orderdate AS orderdate, salestrack.status AS STATUS FROM printers, salestrack WHERE salestrack.productid = printers.productid AND salestrack.status = 'OPEN' ORDER BY salestrack.orderdate I'm using a switch with multiple nested switches. I tested the switch and nested switches by setting up the structure, and using basic text for the cases. For example, instead of one of the nested switch cases needing to = $_GET['id'], I just entered two cases, each equaling 1 and 2, then the default case. Once I begin to enter my SQL coding to connect to the databases, only certain information prints. For example; on my switch that is $_GET['page'], I have a case for 'news'. Inside this case, I have a nested switch called $_GET['id'], which will display information if the page=nes&id=whatever. That works, but when I go do the page=news, nothing displays. My coding is omitting displaying the default information if there is no $_GET['id']. Attached, I am going to have my .php file. The reason for not posting it here is because my coding is rather messy. Please help! hello i have this code here to delete people form a call list but it is not deleting form it could I get a some help? the form Code: [Select] <?php include '../config.php'; $query="SELECT * FROM call_list WHERE ecs = 'Jam' order by date desc"; $result=mysql_query($query); echo mysql_error(); //////////////// Now we will display the returned records in side the rows of the table///////// while($row = mysql_fetch_array($result)) { echo " <table id='call_list'> <form name='Call_delet' action='del.php' method='get'> <tr> <td class='call_names'> $row[Fname] $row[Lname] </td> <td class='call_numbers'> $row[phone] </td> <td class='call_email'> $row[email] </td> <td class='call_email'> $row[calltime] </td> <td> <a href='del.php?del=$row[id]'>Del</a> </td> </tr> </form> </table> "; } ?> del.php page Code: [Select] <?php include '../config.php'; if (isset($_GET['id']) && is_numeric($_GET['id'])) { $id = $_GET['id']; $result = mysql_query("DELETE FROM call_list WHERE id=$id") or die(mysql_error()); header("Location: view.php"); } else { echo"dident work"; } ?> Hey all Hope ya can help. its almost done just having problems with the new data getting inserted into the current table. Code: [Select] foreach($_SESSION as $key => $value) { $key = $value; } echo $item_number . "<br>" . $po_number . "<br>" . $lot_number . "<br>" . $amount ; $query = "SELECT * FROM invintory WHERE item_number ='$item_number'"; $result = mysql_query($query) or die ("couldnt execute query." . mysql_error()); while ($row = mysql_fetch_array($result)) { $onhand = $row['onhand']; } echo $onhand ; $newamount=$onhand+$amount; echo $newamount; [color=red]mysql_query("INSERT INTO invintory WHERE item_number ='$item_number'(onhand) values ('$newamount')");[/color] mysql_query("INSERT INTO recvied(item_number,po_number,lot_number,amount) values ('$item_number','$po_number','$lot_number','$amount')"); ?>The part in red seems to be the problem Given the below 3 database tables I am trying to construct a SQL query that will give me the following result: customer_favourites.cust_id customer_favourites.prod_id OR product.id product.code product.product_name product.hidden product_ section.section_id (MUST BE ONLY THE ROW WITH THE LOWEST SECTION ID, I.E. ROW ID #44108) product_ section.catpage (MUST BE ONLY THE ROW WITH THE LOWEST SECTION ID, I.E. ROW ID #44108) product_ section.relative_order (MUST BE ONLY THE ROW WITH THE LOWEST SECTION ID, I.E. ROW ID #44108) I currently have.... SELECT customer_favourites.cust_id, customer_favourites.prod_id, product.code, product.product_name, product.hidden, product_section.section_id, product_section.relative_order, product_section.catpage FROM customer _favourites INNER JOIN product ON customer_favourites.prod_id = product.id INNER JOIN product_section ON product_section.product_code = product.code WHERE `cust_id` = '17' AND `hidden` = '0' GROUP BY `code` ORDER BY `section_id` ASC, `relative_order` ASC, `catpage` ASC LIMIT 0,30 This gives me what I want but only sometimes, at other times it randomly selects any row from the product_section table. I was hoping that by having the row I want as the last row (most recent added) in the product_section table then it would select that row by default but it is not consistent. Somehow, I need to be able to specify which row to return in the product_section table, it needs to be the row with the lowest section_id value or it should by the last row (most recent). Pulling my hair out so any help is gratefully received. customer_favourites id cust_id prod_id 70 4 469 product id code product_name hidden 469 ABC123 My Product 0 product_section id section_id catpage product_code relative_order recommended 44105 19 232 ABC123 260 1 44106 3 125 ABC123 87 1 44107 2 98 ABC123 128 1 44108 1 156 ABC123 58 0 I have a code that works but it wont add the data to my database. It says the data has been added but the actual data is not actually added. I have checked through this code to see any errors but I cant find any. Can someone help me Code: [Select] $sqll= takeanexam($_SESSION['username1'], $_SESSION['ssubject'], $_SESSION['smodule']); // mysql_data_seek(sqll, 0); while($info = mysql_fetch_array( $sqll )) { $questionId = $info['Que_ID']; $choice = array(); for ($i =1; $i < 5; $i++) { if (empty($_POST['choice'][$questionId][$i])) { $choice[$i] = 0; } else { $choice[$i] = 1; } } $username= $_SESSION['username1']; mysql_query("INSERT INTO answer (Ans_Answer1, Ans_Answer2, Ans_Answer3, Ans_Answer4, Que_ID, Use_ID) VALUES ({$choice[1]}, {$choice[2]}, {$choice[3]}, {$choice[4]}, $questionId, $username)"); } I was just wondering if it's possible to run a query on data that has been returned from a previous query? For example, if I do Code: [Select] $sql = 'My query'; $rs = mysql_query($sql, $mysql_conn); Is it then possible to run a second query on this data such as Code: [Select] $sql = 'My query'; $secondrs = mysql_query($sql, $rs, $mysql_conn); Thanks for any help I have a problem with my rest service as below: Hi, I want to pull data from db, where sometimes all rows and sometimes rows matching given "username". Here is my code:
//Grab Username of who's Browsing History needs to be searched.
if (isset($_GET['followee_username']) && !empty($_GET['followee_username']))
{
$followee_username = $_GET['followee_username'];
if($followee_username != "followee_all" OR "Followee_All")
{
$query = "SELECT * FROM browsing_histories WHERE username = \"$followee_username\"";
$query_type = "followee_username";
$followed_word = "$followee_username";
$follower_username = "$user";
echo "$followee_username";
}
else
{
$query = "SELECT * FROM browsing_histories";
$query_type = "followee_all";
$followed_word = "followee_all";
$follower_username = "$user";
echo "all";
}
}
When I specify a "username" in the query via the url: browsing_histories_v1.php?followee_username=requinix&page_number=1 I see result as I should. So far so good.
Now, when I specify "all" as username then I see no results. Why ? All records from the tbl should be pulled! browsing_histories_v1.php?followee_username=all&page_number=1 This query shouldv'e worked:
$query = "SELECT * FROM browsing_histories";
Hello, Curious to know if someone could point me in the right direction, been struggling with this for a bit now. I have a HTML page with a search field, I can enter a search term and hit the submit button and I am directed to my search.php page with the appropriate results. What I am looking to accomplish is having the search results from the search.php page displayed in a text area below my search field in my HTML page. I have included an image to better describe what I am looking to accomplish: Additionally below is the source from my HTML page and search.php page: page.html <form name="search" action='search.php' method="post"> <input type="text" class="myinputstyle" name="search" value="search" onClick="this.value=''"/><br> <input type="submit" value="submit" class="myinputstyle"> </form> search.php <?php $search = "%" . $_POST["search"] . "%"; mysql_connect ("localhost", "game_over", "Ge7Ooc9uPiedee3oos9xoh4th"); mysql_select_db ("game_over"); $query = "SELECT * FROM game_over WHERE first_name LIKE '$search'"; $result = mysql_query ($query); if ($result) { while ($row = mysql_fetch_array ($result)) { echo "Name: {$row['name']} " . "{$row['lname']} <br>" . "Email: {$row['email]} <br>" . } } ?> Any insight would be most appreciated. Thank you. ok..ive done this a million times..i have a working example here and i copied it and amended it for this new project but for some reason i cant get a form to post data to another page. this is the error message i get Notice: Undefined index: username in C:\wamp\www\uni\fyp\site\mobile\login.php on line 16 Notice: Undefined index: password in C:\wamp\www\uni\fyp\site\mobile\login.php on line 17 here is my form code: <form method="post" action="login.php"> <table align="center" cellpadding="0" cellspacing="0"> <tr> <td style="vertical-align:top;">Username: </td><td><input type="text" name="username" value="" /></td> </tr> <tr> <td style="vertical-align:top;">Password: </td><td><input type="password" name="password" value="" /><br /><input type="submit" id="submit" value="Login" /></td> </tr> </table> </form> and here is the code within the login.php where the form should post to $username = $_POST['username']; $password = $_POST['password']; // Help protect against MySQL injection $username = stripslashes($username); $password = stripslashes($password); $username = mysql_real_escape_string($username); $password = mysql_real_escape_string($password); // Selecting data from database where correct username and password are found $sql="SELECT * FROM customer WHERE username='$username' and password='$password'"; $result=mysql_query($sql) or die(mysql_error()); i cant see anything wrong..been looking for hours...please please help me Here is something i dreamt up, i know there is a way for this to work, but i'm not having any great ideas at this point. I want to be able to auto increment through the $_POST[] array data sent from a PayPal IPN. Sample return data looks like item_number1=val item_name1=val so on and so forth. The idea is: While there are $_POST item_numbers, do something with it. The loop continues until there are no more item_numbers. I don't want to use the static methods eg: explicitly defining item_number1, item_number2, ect. This is what i thought would work, but keeps failing -> the script doesn't enter the first condition: $x = 1; // set the initial item number while(isset($_POST['item_number$x'])){ $qty = $_POST['quantity$x']; while($qty > 0){ // step through each item response result, setting each item to paid $package_id = $_POST['item_number$x']; update_paid_status($transaction_id, $package_id); $qty--; } $x++; } Any Ideas? |
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