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PHP - Php Mysql Recordset . . . Setting Variables
Two part question:
1) I have a MySql table called "lang_key", which will store all of the common text for a site, which will allow easy site modifications from the "non tech" admin. Now, I am creating a recordest that will pull the information from the table, and create a basic list of variables in php. For example: I connect to the database, and do a wildcard select. I can figure out how to set a basic php variable like the example below: Code: [Select] $query1 = "SELECT * FROM lang_key"; $result = mysql_query($query1); while($row = mysql_fetch_array($result)) { $template = $row['common_text']; } There are three fields in my lang_key table (unique_id, string_id, and common_text') What I want to do is take it a step further from the example and I want $template to be the value of "common_text" Where string_id = "curr_template" I hope this makes sense? I really just want to create my recordset and then populate the website with the recordset info. Part II Do you recommend that I use a lang_key database, or would it be simpler and more efficient to just create a lang page with static variables, and use the php file write option to update the page? Thanks!!! Similar TutorialsRecord set has 2 text fields in the form which is set in a full repeat recordset browse. So, we get a long list of every record in the database. However, I want to be able to click on a single record and make another page appear. I can do this if the display is set as a table without using a text field form -- just the record variable and using a hyperlink. But, I want to use the text field. Wrapping the form only gets me the value of the last record displayed. Help would be appreciated. I have created a test account in my database with a user level of -1 and i think my code might be wrong but i am hoping someone can spot where i have gone wrong as i cannot, also a similar problem with another session variable loggedIn this is what i get when i login this is on the index page.
Notice: Undefined index: loggedIn in C:\xampp\htdocs\Login\index.php on line 11 Notice: Undefined index: loggedIn in C:\xampp\htdocs\Login\index.php on line 17 You must be logged in to view this page!Index page source code:
<?php
session_start();
error_reporting(E_ALL | E_NOTICE);
ini_set('display_errors', '1');
require 'connect.php';
if($_SESSION['loggedIn'] == 1) {
//Do Nothing
exit();
} else if($_SESSION['loggedIn'] != 1) {
echo "You must be logged in to view this page!";
exit();
}
if($_SESSION['user_level'] == -1) {
header("Location: banned.php");
} if(isset($_SESSION['username'])) {
echo "<div id='welcome'> Welcome, ". $_SESSION['username'] ." <br> </div> ";
}
?>
Also if you need my login source code:
<?php
error_reporting(E_ALL | E_NOTICE);
require 'connect.php';
session_start();
if (isset($_POST['submit'])) {
$username = trim($_POST['username']);
$password = trim($_POST['password']);
if (empty($username)) {
echo "You did not enter a username, Redirecting...";
echo "<meta http-equiv='refresh' content='2' URL='login.php'>";
exit();
}
if (empty($password)) {
echo "You did not enter a password, Redirecting...";
echo "<meta http-equiv='refresh' content='2' URL='login.php'>";
exit();
}
//Prevent hackers from using SQL Injection to hack into Database
$username = mysqli_real_escape_string($con, $_POST['username']);
$password = mysqli_real_escape_string($con, $_POST['password']);
$result = $con->query("SELECT * FROM $tbl_name WHERE username='$username' AND password='$password'");
$row = $result->fetch_array();
$user_level = $row['user_level'];
// check to make sure query did execute. If it did not then trigger error use mysqli::error to see why it failed
if($result->num_rows > 0)
{
//Set default user
$_SESSION['loggedIn'] == 1;
$_SESSION['user_level'] == 1;
$_SESSION['username'] == trim($_POST['username']);
header("Location: index.php");
exit();
} else if($row['user_level'] == 1) {
$_SESSION['user_level'] == 1;
//Location admin
header("Location: admin.php");
exit();
} else if($row['user_level'] == -1) {
$_SESSION['user_level'] == -1;
$_SESSION['username'] == trim($_POST['username']);
//Location banned
header("Location: banned.php");
exit();
} else if($_SESSION['loggedIn'] == true) {
//Location default user home page
header("index.php");
} else {
echo "Invalid Username/Password";
}
//Kill unwanted session
} if(isset($_POST['killsession'])) {
session_destroy();
echo "<br> <br> The Session Destroyed. (Basically means you have been logged out)";
exit();
}
?>
I appreciate all help
Hi all.
i am trying to use php to include a javascript onto different pages, and then "sort of " pass it a var.
The bulk of the code will be in an included footer php file.
The var will be set in the main page.
i have it working as follows: by just using echo 3 times.. the 1st with the first part of the script, the 2nd is the variable, and the 3rd is the rest of the script.
The same endcode.php file needs to also be used for pages that wont have a var set, and wont be using the script - hense the isset.
<!-- mainPage.php -->
<?php $sion_gallery_id = '450'; ?>
<?php include(endcode.php); ?>
<!-- endcode.php -->
<?php if (isset($sion_gallery_id))
{echo "start of javascript.......album/"; echo $sion_gallery_id; echo"/end of script"; }
else { echo "var not on set";} ?>
HI. I am talking about this website. http://www.elfster.com/ Please browse this website. Can any one tell me how database structure / queries will work? (i know PHP / MYSQL) Can any one guide me a little bit. I would really appreciate it, Thanks Waiting for reply. Hey guys, Currently Im using: $row = mysql_fetch_array($result) or die(mysql_error()); echo $row['user_family']. " - ". $row['user_registered']; $row['user_family'] = $fam; $_SESSION['family'] = $fam; to take data from a mysql table & set it as SESSION family. However, I cant seem to get this to set. The information IS being taken from mysql because its being echo'd earlier up in the code, but its just not passing to the session. Any ideas? Hi I'm having a problem getting a query to work. I have a simple form with user input for start and end date with format: 2009-03-19 (todays date): $Startdate = $_POST['date']; This works well when something is entered into the form, and afterwards using my query: SELECT COUNT(*) as total FROM mydb WHERE Date BETWEEN '$Startdate' AND '$EndDate' ........ Problem is if user submits the form without entering anything in the date input fields, which makes sense. I want to check if inputs has been made, and if not set af default date, but can't make it work:
if (isset($_POST['date']) && $_POST['date'] !='') {
$Startdate = $_POST['date'];}
else { $Startdate = '1980-01-01';}
How can I set $Startdate to something that can be used in the query as below doesn't work? Hi guys I am working on adding a third party php members register and login into a clients web site but every time I try the regidter page is shows me this error message. Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'username'@'localhost' (using password: YES) in /home/cpassoc2/public_html/register-exec.php on line 15 Failed to connect to server: Access denied for user 'username'@'localhost' (using password: YES) This is what i have in the config.php page. <?php define('DB_HOST', 'localhost'); define('DB_USER', 'cpassoc2-memark'); define('DB_PASSWORD', '?????????'); password replaced define('DB_DATABASE', 'cpassoc2-me'); ?> I have set up the my SQL within the control panel of the site, So do i need to do any thing else???, what am i missing???. Mark..... When I run 'select 1700-price as blah from goldclose as t2 order by dayid desc limit 1' by itself in mysql I get a numerical result: one row, one column. In my php script, the 1700 is actually a variable. so here it is $changequery = sprintf("select $goldprice-price as change from goldclose order by dayid desc limit 1"); $change = mysql_query(changequery); while ($row = mysql_fetch_array($change)) { printf("$row[0]"); } mysql_free_result($changeresult); I get the following error, Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b>/root/fuzzy/htmlmain4.php on line 99 Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in <b>/root/fuzzy/htmlmain4.php on line 103 Not sure why? All i want is to get the result of that select statement into a variable such as $change How to update table records one by one.... I have a table with five records .. I am getting values of month from another function..I would like to fill the values of month into the table.(like jan feb mar april may) respectively in the first row instead of Average. my code is below $insquery = "select * from reporttmp"; $insresult = mysql_query($insquery, $link) or die("Error in sql sytax."); $fel=mysql_num_fields($insresult); $nro=mysql_num_rows($insresult); $i=1; if ($nro > 0) { while ($i < $nro - 1) { $monname = getmonthname($fmonth); $monname = $monname. " ". $fyear; $upquery = ("UPDATE reporttmp SET vdate ='$monname' where mysql_num_row($insresult)=$i"); if(!mysql_query($upquery, $link)) die ("Mysql error ....<p>".mysql_error()); $fmonth = $fmonth + 1; $i = $i + 1; } } [attachment deleted by admin] Hi I have a recordset from a database which outputs a value of '% THOBEKA MFEKA::12110235675?' I need to seach the database based on the number between the :: and ? only. So for example '12110235675'. I have tried this with a sql statement using LIKE but its not working. How can I trim the value from the database so I only get the number out. The amount of characters before the :: can vary. Please can someone give me some advice asap. Rob I don't know if this is the correct Forum to post this under but I'm trying to fill the values of a recordset into an array. Here is what I have but it doens't work so far. function fetch($info) { return mysql_fetch_array($info); } $Proddb = new mysql; $Proddb->connect(); $result = $Proddb->query("SELECT * FROM Colors WHERE id = '1'"); $row = $Proddb->fetch($result) foreach ($row as $value) { echo $value. "<br>"; } $Proddb-> close(); The array is outputting everything twice does anyone know why? Hi,
I'm very new to php and to mysql and I was hoping someone would be kind enough to give me a hand. I'm trying to write a query that returns only one record set.
I'll paste what I have below and the error. I'm sure that I am doing something wrong that is right in front of my face but for the life of me I can't figure it out.
The relevant part of the code is:
require_once ('../mysqli_connect.php'); $movie_id = $_GET['movie_id']; $q = "SELECT * FROM movies WHERE movie_id = $movie_id"; $r = mysqli_query($dbc, $q); echo '<table cols="2" width="1100px" align="center"> <tr> <td align="left" width="240px"><img src="img/boxcovers/small/movie_' . $r['movie_id'] .'_small.jpg"></td> <td></td> </tr> The error I'm getting is: PHP Fatal error: Cannot use object of type mysqli_result as array Help! Hi again probably a very simple code but not working. I am trying to show a default image "fabric.jpg" if the recordset is empty. If not empty it shows the recordset image at a size of 100X100. This is the code I am using and have probably left something out, any ideas? <img src="<?php if ($row_Recordset3['fabricpicture']==null); echo "<img src='graphics/fabric.jpg'>"; ?>" alt="" name="fabric" width="100" height="100" border="0" align="bottom" id="fabric" title="Selected Bottom Up Blind Fabric" /> I have tried switching the img scr both "" and ' ' but still no joy? I have a table called parties with 3 fields - partyid, partyname, linkid. For any linkid there will be a number of parties somewhere between 2 and 10. I can get a vertical list but have a couple of issues which I can't fix: I want to have a page that displays the list of partynames (alphabetically) horizontally rather than vertically within a piece of text. For example, "The parties linked to you are Party1, Party2, Party3" Also, ideally, I would like the word 'and' before the last Party name so, using the above example, I would get "The parties linked to you are Party1, Party2 and Party3" I have no idea how to do this or if it can be done. Any ideas would be VERY gratefully received. Thanks I have the code below: ---------------------------------------------------------------------- ------------------------------------ <div id="content"> <table width="998" border="0" cellspacing="4" id="stuff"> <?php do { ?> <tr> <td><?php echo $row_Recordset1['name']; ?></td> </tr> <?php } while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)); ?> </table> <div id="spryTable" spry:region="ds1"> <table align="center"> <tr> <th spry:sort="name"> </th> </tr> <tr spry:repeat="ds1" spry:setrow="ds1" spry:hover="hover" spry:select="selected"> <td height="40" align="center">{name}</td> // green line <td height="40" align="center">{name}</td> //orange line </tr> </table> </div> ---------------------------------------------------------------------- -------------------------------- At the moment the green line of code will display the entire record of names, and the orange line displays the entire record again in another column of the table. The problem is, I want half the names to be printed in one column and the other half of names to be displayed in the other column. Can someone show me how to set the php code to do this? I've been trying various things for a while now without success. Thank you. Hi All! I've written up a script for my website. It\ is basically a virtual job quest. My queries are all correct it just isn't registering the variable for the session. It is $-SESSION[theid']. I want to be bale to use it in my table but I get an error. How do I write this in my SQL query for it to work. The page (when no errors), doesn't show my data. Here is my ocde: Code: [Select] <?php session_start(); include("config536.php"); ?> <html> <head> <link rel="stylesheet" type="text/css" href="style.css" /> </head> <?php if(!isset($_SESSION['username'])) { echo "<ubar><a href=login.php>Login</a> or <a href=register.php>Register</a></ubar><content><center><font size=6>Error!</font><br><br>You are not Logged In! Please <a href=login.php>Login</a> or <a href=register.php>Register</a> to Continue!</center></content><content><center><font size=6>Messages</font><br><br></center></content>"; } if(isset($_SESSION['username'])) { echo "<nav>$shownavbar</nav><ubar><img src=/images/layout/player.gif><a href=status.php>$showusername</a>.......................<img src=/images/layout/coin.gif> $scredits</ubar><content><center><font size=6>Basic Quests</font><br><br>"; $startjob = $_POST['submit']; $jobq = "SELECT * FROM jobs WHERE username='$showusername'"; $job = mysql_query($jobq); $jobnr = mysql_num_rows($job); if($jobnr == "0") { ?> <form action="<?php echo "$PHP_SELF"; ?>" method="POST"> <input type="submit" name="submit" value="Start Job"></form> <?php } if(isset($startjob)) { $initemidq = "SELECT * FROM items ORDER BY RAND() LIMIT 1"; $initemid = mysql_query($initemidq); while($ir = mysql_fetch_array($initemid)) { $ids = $ir['itemid']; } mysql_query("INSERT INTO jobs (username, item, time, completed) VALUES ('$showusername', '$ids', 'None', 'No')"); $wegq = "SELECT * FROM items WHERE itemid='$ids'"; $weg = mysql_query($wegq); while($wg = mysql_fetch_array($weg)) { $im = $wg['image']; $nm = $wg['name']; $id = $wg['itemid']; } $_SESSION['theid'] = $id; echo "<font color=green>Success! You have started this Job!</font><br><br>Please bring me this item: <b>$nm</b><br><br><img src=/images/items/$im><br><br><br>"; echo $_SESSION['theid']; } if($jobnr == "1") { $finish = $_POST['finish']; $okgq = "SELECT * FROM items WHERE itemid='$yes'"; $ok = mysql_query($okgq); while($ya = mysql_fetch_array($ok)) { $okname = $ya['name']; $okid = $ya['itemid']; $okimage = $ya['image']; } echo "Where is my <b>$okname</b>?<br><br><img src=/images/items/$okimage><br><br><br>"; echo $_SESSION['theid']; ?> <form action="<?php echo "$PHP_SELF"; ?>" method="POST"> <input type="submit" name="finish" value="I have the Item"></form> <?php } } if(isset($finish)) { $cinq = "SELECT * FROM uitems WHERE theitemid='$_SESSION[theid]'"; $cin = mysql_query($cinq); $connr = mysql_num_rows($cin); if($connr != "0") { echo "<font color=green>Success! You have the item.</font>"; } else { echo "<font color=red>Error! You do not have my item!</font>"; } } ?> . I basically just want to know how I can set this session as a variable. Also..I have a user login on every page and I want to be able to destroy JUST THE "theid" session and NOT the username session. How would I do that too? thanks for the help in advance! Hey Guys, I have to insert some data in MySQL but it wont work . Please have a look. <?php // to values are set to empty $vatsim=""; $ivao=""; // values from form in other page are set if(isset($_POST["pilotid"])) $pilotid=$_POST["pilotid"]; if(isset($_POST["network"])) $network=$_POST["network"]; if(isset($_POST["vid"])) $vid=$_POST["vid"]; if(isset($_POST["pilot"])) $pilot=$_POST["pilot"]; // if value is that copy data in this value, otherways in that value if ($network == "IVAO") { $ivao="$vid";} if ($network == "VATSIM") { $vatsim="$ivao";} // connect db include(dbconnect.inc.php); // first sql to update some data in one table $sql = "UPDATE `360283`.`jos_users` SET `IPS` = \'1\' WHERE `jos_users`.`id` = \'$pilotid\'"; $result1 = mysql_query($sql); // 2nd sql to insert some data in other table $sql2 = "INSERT INTO `360283`.`IPS_Pilots` (`ID`, `Name`, `Hours`, `Flights`, `LastFlight`, `IVAO`, `VATSIM`, `Enabled`, `Rating`) VALUES ('$pilotid', '$pilot', NULL, NULL, NULL, '$ivao', '$vatsim', '1', '0');"; $result2 = mysql_query($sql2); // sql to check if it was succesful $sql3 = "SELECT * FROM `IPS_Pilots` WHERE `ID` = '$pilotid' LIMIT 0, 30 "; $result3 = mysql_query($sql3); $num3 = mysql_numrows($result3); // echo succesfull or not if (!$num3) { echo "Sorry, but I failed to apply this pilot."; } else { echo "Pilot succesfully applied."; } ?> Thanks I have a table and the structure is Code: [Select] ID, UID, Site, Uname, PassI already have this <?php $result = mysql_query(sprintf("SELECT * FROM Logins WHERE UID = %d", $_COOKIE['UID_WatsonN'])); //check login table against cookie if(($num = mysql_num_rows($result)) > 0){ mysql_close(); ?> <table border="0" cellspacing="2" cellpadding="2"> <tr> <th>ID</th> <th>UID</th> <th>Site</th> <th>Uname</th> <th>Pass</th> </tr> <center> <?php $i=0; while ($i < $num) { $f1=mysql_result($result,$i,"ID"); $f2=mysql_result($result,$i,"UID"); $f3=mysql_result($result,$i,"Site"); $f4=mysql_result($result,$i,"Uname"); $f5=mysql_result($result,$i,"Pass"); ?> <tr> <td><?php echo $f1; ?></td> <td><?php echo $f2; ?></td> <td><?php echo $f3; ?></td> <td><?php echo $f4; ?></td> <td><?php echo $f5; ?></td> </tr> <?php $i++; } } ?> which only gets the data from that one person. What is in the table is usernames and passwords for diffrent sites and i want to take all the usernames and passwords and put them into variables to pass on to the login form. So, I'm working on a quiz system. The text and choices are stored in a MySQL database. I haven't gotten to the choices yet, I'm still having trouble with the text. Here's my code: Code: [Select] $querytext = "SELECT text FROM quiz id = '$prob'"; $result = mysql_query($querytext) or die(mysql_error()); echo $result; Yes, I'm already connected and stuff, that's just the snippet. I made sure that $prob is 1. Here's the MySQL Error I'm getting: 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've looked all over the web and through the MySQL/PHP section of a 991-page PHP book. What am I doing wrong? I bet that I really, really failed epicly this time, but I never catch my epic fails and have to ask questions about them in order to fix the problem. So please reply Hi, I'm only just getting started with php and mysql but I like to go off the deep end. something is wrong with my script. it seems to work up untill it comes to the UPDATE part. It wont actually update the table and the mysql_affected_rows bit returns -1 I've searched for an explanation and can't find one. Can someone tell me what I'm doing wrong Thanks. //connect to mysql $con = mysql_connect("xxx","yyy","zzz"); if (!$con) { die('Could not connect: ' . mysql_error()); } //use the right database mysql_select_db("foo_bar"); //select all cancelled tickets $status_query = "SELECT * FROM jos_vm_orders where order_status = 'X'"; $status_result = mysql_query($status_query) or die(mysql_error()); while ($newArray = mysql_fetch_array($status_result, MYSQL_ASSOC)) { $order_id = $newArray['order_id']; $quantity_query = "SELECT * FROM jos_vm_order_item where order_id = $order_id"; $quantity_result = mysql_query($quantity_query) or die(mysql_error()); while ($newArray2 = mysql_fetch_array($quantity_result, MYSQL_ASSOC)) { $product_quantity = $newArray2[product_quantity]; $order_item_sku = $newArray2[order_item_sku]; echo ("$order_id cancelled $product_quantity tickets sku was $order_item_sku \n"); mysql_query("UPDATE LOW_PRIORITY jos_vm_product SET product_in_stock = product_in_stock + $product_quantity WHERE product_sku = $order_item_sku AND STR_TO_DATE($order_item_sku, '%Y-%d-%m') $quantity_query = "SELECT * FROM jos_vm_order_item where order_id = $order_id"; $quantity_result = mysql_query($quantity_query) or die(mysql_error()); while ($newArray2 = mysql_fetch_array($quantity_result, MYSQL_ASSOC)) { $product_quantity = $newArray2[product_quantity]; $order_item_sku = $newArray2[order_item_sku]; echo ("$order_id cancelled $product_quantity tickets sku was $order_item_sku \n"); mysql_query("UPDATE LOW_PRIORITY jos_vm_product SET product_in_stock = product_in_stock + $product_quantity) WHERE product_sku = $order_item_sku AND STR_TO_DATE($order_item_sku, '%Y-%d-%m') >= CURDATE()"); echo ("$product_quantity tickets released into $order_item_sku \n"); echo ("Affected rows "); echo (mysql_affected_rows()); echo ("\n"); } } mysql_free_result($status_result); mysql_free_result($quantity_result); mysql_close($con); ?> |
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