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PHP - Using A Mysql Function To Select A Multidemi Array?
Here is my query:
Code: [Select] $db->query("SELECT cache.*,u.username from search_cache cache LEFT Join users as u ON u.id = 'NEED HELP HERE' where ident = '{$username}' ORDER BY DATE DESC") or error('Unable to send the message.', __FILE__, __LINE__, $db->error()); Here is my data inside my search_cache Here is my while loop: Code: [Select] while($recent = $db->fetch_assoc($rec)){ $data = unserialize($recent['search_data']); } Now I can use $data['search_type'] as a array and If I use echo $data['search_type']['2'] it echo's out the user_id 32 as seen in the screenshot at the very end. Problem is, I want to use that 'NEED HELP HERE' to join with that id 32 dynamically, how the hell is that possible with mysql or is it? Or would it be easier to just add a new row/column named user_id and just join of that instead of trying to do all this? would that be faster? Similar TutorialsHi Guys I don't know if this is possible but can someone point me in the right direction. I have a php function which takes two inputs and returns an output. for simplicity's sake let's say it's an addition function. What I want to do is use a mysql select statement to show all the rows from a database where field1 and field2 equal '3'. Here's the sort of thing I mean. function addNumbers($one,$two) { return $one + $two; } mysql_query("SELECT * FROM table WHERE 'addNumbers(field1,field2)' = '3'"); What I actually want to do is a lot more complex than this but I am trying to understand how to make the syntax work in simple terms first. Can anybody help? Many Thanks Dan I am simply trying to insert a value generated from an array in a while loop, but it seems the value is not global and I can't pass it as I like. I did some research on this, but could not find an answer that solved my issue... Here is my select box code which is working perfect: <select name="city"> <?php $sql = "SELECT id, city_name FROM cities ". "ORDER BY city_name"; $results_set = (mysqli_query($cxn, $sql)) or die("Was not able to produce the result set!"); while($row = mysqli_fetch_array($results_set)) { echo "<option value=$row[id]>$row[city_name]</option>"; } ?> </select> Are any variables defined in a while loop global to the while loop only? Here is my SQL which you can see my $row[id] being passed thru field city_id... The value is being generated as supposed to based on value like: value="1", value="2" etc.. for the select options for each city name. So the values are there... But I CANNOT get that numerical id to pass to the database when submitting my form. Any ideas for a workaround to get this value passing as normal? if (isset($_POST['addPosting'])) { $query = "INSERT INTO Postings (id, city_id, title, description) VALUES ('','$row[id]','$_POST[title]','$_POST[description]')"; Hi, I have come up with the following code, I need it to get the details of several scattered products and echo the results, the trick is I don't want it to echo the results one after the other... I want to have the products scattered between unique text on the page but don't want to run the query several times for performance reasons. E.g.- PAGE to look like this: $Product_1 unique text/images $Product_2 $Product_3 unique text/images $Product_4 Current Code: Code: [Select] <? $result = mysql_query("SELECT * FROM products where Product_ID IN (475, 465, 234, 567, 845)"); while($row = mysql_fetch_array($result)) { $x = "1"; while ($x<=3) { echo $x; $Product = "Product_"; $Product = $Product.$x; echo $Product; $Product = $row['Product_ID']; echo $Product; $x++; echo $x; } } At the moment it returns the following results: Quote 1 Product_1 465 2 2 Product_2 465 3 3 Product_3 465 4 1 Product_1 475 2 2 Product_2 475 3 3 Product_3 475 4 A few problems... In Blue... it duplicates for product 465 In Red... It repeats again for 475 Also.... it starts with 465, but I want it to go in order as how it appears - $result = mysql_query("SELECT * FROM products where Product_ID IN (475, 465, 234, 567, 845)"); so should start with 475 I want to get the following result: Quote 1 Product_1 475 2 2 Product_2 465 3 3 Product_3 234 4 4 Product_4 567 4 (and so on.....) If anyone could provide me assistance with my troubled 'while loop' statement that would be much appreciated! I hope I'm not being a vampire here sucking the life out of you guys, but I can't find any resources in google that explains how to properly use this function in a mysql array. All it does is add a url for any urls added in plaintext. Code: --- <?php function hyperlink ($string) { $string = preg_replace('#(^|\s)([a-z]+://([^\s\w/]?[\w/])*)#is', '\\1<a href="\\2" target="_blank">\\2</a>', $string); $string = preg_replace('#(^|\s)((www|ftp)\.([^\s\w/]?[\w/])*)#is', '\\1<a href="http://\\2" target="_blank">\\2</a>', $string); $string = preg_replace('#(^|\s)(([a-z0-9._%+-]+)@(([.-]?[a-z0-9])*))#is', '\\1<a href="mailto:\\2">\\2</a>', $string); return $string; } // mysql connection $result = mysql_query("SELECT * FROM top_web_articles ORDER BY ID DESC"); echo "<table border='0'>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td class='newstitle' align='left' valign='top' bgcolor='#D6D6D6'>".$row['Category']."</td></tr>"; echo "<td class='newstitle' align='left' valign='top' bgcolor='#D6D6D6'>". $row['ArticleName'] ."</td></tr>"; echo "<td class='newstitle' align='left' valign='top' bgcolor='#D6D6D6'>"hyperlink(. $row['ArticleDescription'] .);"</td></tr>"; echo "<tr><td class='newstitle' align='left' valign='top' bgcolor='#D6D6D6'><a href='". $row['URL'] . "' target='_blank'><img src='http://www.thenewsguys.ca/images/read entire article.png' alt='' width='130' height='11' border='0' /></a></td>"; echo "</tr>"; echo "<td align='left' valign='top'> </td>"; } echo "</table>"; mysql_close($con); ?> --- Error: "Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/doofyd5/public_html/trevor/playground/displayarticles.php on line 146" (Clearly an error where I'm abusing the heck out of the bolded line. Any ideas on how to properly code this? Okay, this is getting on my nerves now, my head is throbbing probably just need a break from it all now. I'm currently getting the following message; Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in C:\xampp\htdocs\duff3\test3.php on line 25 Can anyone figure out and explain why this isn't working for me, I believe I am calling the function correctly and then putting the result into a while loop. Code: [Select] <?php function getPageContent($selectedPage) { $sql = "SELECT * "; $sql .= "FROM tbl_pages "; $sql .= "WHERE tbl_pages.pageCategoryId = ".$selectedPage." "; $sql .= "AND active = 1 "; $sql .= "LIMIT 1"; $result = mysql_query($sql) or die ("Error in page sql query:". $sql); return $pageSet; } ?> Code: [Select] <?php require_once ''.$_SERVER['DOCUMENT_ROOT'].'/duff3/commonResources/dbConnection/dbQueryClass.php'; ?> <?php #FUNCTIONS IS A TEMP FILE UNTIL OO PHP COMES INTO PLAY require_once ("commonResources/includes/functions.php"); # if page isn't set then set it to default. if(isset($_GET["page"])) { $selectedPage = $_GET["page"]; } else { $selectedPage = 1; } #call function to select all page $pageSet = getPageContent($selectedPage); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title> <?php while ($row = mysql_fetch_array($pageSet)) { # Start while loop echo $row["pageTitle"]; ?> </title> <link rel="shortcut icon" href="commonResources/images/container/favicon.ico" /> <link rel="stylesheet" type="text/css" href="<?php $_SERVER["DOCUMENT_ROOT"] ?>/duff3/commonResources/css/style.css" /> <meta http-equiv="Content-Type" content="text/html; charset=windows-1252" /> <!-- PUT COMMON JAVASCRIPT FILES IN AN INCLUDE 05/05/2011 --> <script type="text/javascript" src="commonResources/javaScript/jQuery.js"></script> <script type="text/javascript" src="commonResources/javaScript/jQueryTest/menu.js"></script> <?php require_once ("".$_SERVER["DOCUMENT_ROOT"]."/duff3/commonResources/includes/tinymce.php"); ?> </head> <body> <div id="viewContainer"><!--OPEN DIV FOR VIEW CONTAINER --> <div id="headerContent"><!--OPEN DIV FOR HEADER CONTENT --> <div class="logoContent"><!--OPEN DIV FOR LOGO CONTAINER --> <a href="index.html"><img src="<?php $_SERVER["DOCUMENT_ROOT"] ?>/duff3/commonResources/images/container/logo.png" alt="Hannah Jane Duff" longdesc="Hannah Jane Duff Home Link" /></a> </div><!--CLOSE DIV FOR LOGO CONTAINER --> </div><!--CLOSE DIV FOR HEADER CONTENT --> <!-- TEMP TAKE OUT AND PUT INTO HEADER AREA --> <div id="mainContent"><!--OPEN DIV FOR MAIN CONTENT--> <div class="centreContent"><!--OPEN DIV FOR CENTRE CONTENT--> <div id="menuContent"><!--OPEN DIV FOR MENU CONTENT--> <div id="menu"> <ul class="menu"> <li><a href="index.html" class="main"><span>Home</span></a></li> <li><a href="" class="main"><span>bio</span></a> <div><ul> <li><a href=""><span>violin</span></a></li> <li><a href=""><span>piano</span></a></li> <li><a href=""><span>singing</span></a></li> <li><a href="" class="parent"><span>teaching</span></a> <div><ul> <li><a href="" class="parent"><span>aberdeen</span></a> <div><ul> <li><a href="#"><span>aberdeen 1</span></a></li> </ul></div> </li> <li><a href="" class="parent"><span>bradford</span></a> <div><ul> <li><a href=""><span>bradford 1</span></a></li> </ul></div> </li> <li><a href="" class="parent"><span>leeds</span></a> <div><ul> <li><a href=""><span>leeds 1</span></a></li> </ul></div> </li> </ul></div> </li> <li><a href="" class="parent"><span>influences</span></a> <div><ul> <li><a href=""><span>classical</span></a></li> <li><a href=""><span>folk</span></a></li> </ul></div> </li> <li><a href="" class="parent"><span>other</span></a> <div><ul> <li><a href=""><span>folk′d</span></a></li> <li><a href=""><span>string quaret</span></a></li> </ul></div> </li> </ul></div> </li> <li><a href="" class="main"><span>publicity</span></a> <div><ul> <li><a href="" class="parent"><span>news</span></a> <div><ul> <li><a href=""><span>may 2011</span></a></li> <li><a href=""><span>march 2011</span></a></li> </ul></div> </li> <li><a href=""><span>gallery</span></a></li> <li><a href=""><span>other</span></a></li> </ul></div> </li> <li><a href="" class="main"><span>recordings</span></a> <div><ul> <li><a href=""><span>all</span></a></li> <li><a href=""><span>new york dolls</span></a></li> <li><a href=""><span>classical</span></a></li> </ul></div> </li> <li><a href="" class="main"><span>contact</span></a></li> </ul></div> </div><!--CLOSE DIV FOR MENU CONTENT--> <div class="paraBlock"><!--OPEN DIV FOR PARA BLOCK --> <p><?php echo $row["pageContent"]; ?></p> </div><!--CLOSE DIV FOR PARA BLOCK--> <div class="clearArea"><!--OPEN DIV FOR CLEAR AREA--> </div><!--CLOSE DIV FOR CLEAR AREA--> </div><!-- CLOSE DIV FOR CENTRE CONTENT--> </div><!--CLOSE DIV FOR MAIN CONTENT--> <div id="footerContent"><!--OPEN DIV FOR FOOTER CONTENT--> <div class="leftFooter"><!--OPEN DIV FOR LEFT FOOTER--> © <?php echo date(Y); ?> All Rights Reserved | Design by <a href="http://www.innovationation.co.uk/">Innovationation UK</a> </div><!--CLOSE DIV FOR LEFT FOOTER--> <div class="rightFooter"><!--OPEN DIV FOR RIGHT FOOTER--> <a href="">Copyright | </a> <a href="">Disclaimer | </a> <a href="">Privacy Policy </a> <a href="">Terms of Use | </a> <a href="">Site Map | </a> <a href="loginArea/login.php">Admin</a> </div><!--CLOSE DIV FOR RIGHT FOOTER--> </div><!--CLOSE DIV FOR FOOTER CONTENT--> <?php } ?> </div><!--CLOSE DIV FOR VIEW CONTAINER--> </body> </html> I am having some problems getting a query correct. Basically I have two tables, one with listings and another with categories. In the listings table I have a column called shortdescription. I am trying to pull the shortdescription from the listings table with the query $shortdesc = array_shift(mysql_fetch_row(mysql_query("select shortdescription from links where category = '".$scat->id."' "))); The shortdecription display properly on the pages display the listings, however I am getting the following error on any pages that only display the parent categories Warning: array_shift() [function.array-shift]: The argument should be an array in /home/...path/file.php on line 1462 The listings id numbers begin at 75+ because the initial parent category id ends at 74. The query seems to be searching for listing ids below 75 and spitting out an error because it is not finding them. Any ideas on how to eliminate this error and/or stop the query from looking for non-existant data? How can i save array from inputs, witch is $igraci and it have 5 values, to mysql base, i tryed but in mysql it shows Array. This is form: <form action="dodaj_klan.php" method="post" > <p> <label>Naziv klana:</label> <input name="naziv" type="text" size="20%" /> <label>Web Sajt:</label> <input name="website" type="text" size="20%" /> <label>E-Mail:</label> <input name="email" type="text" size="20%" /> <br /><br /> <label>Igraci klana(5):</label> <input name="lider" type="radio" value="1" /> <input name="igraci[]" type="text" size="20%" /><br /> <input name="lider" type="radio" value="2" /> <input name="igraci[]" type="text" size="20%" /><br /> <input name="lider" type="radio" value="3" /> <input name="igraci[]" type="text" size="20%" /><br /> <input name="lider" type="radio" value="4" /> <input name="igraci[]" type="text" size="20%" /><br /> <input name="lider" type="radio" value="5" /> <input name="igraci[]" type="text" size="20%" /> <label><i>(obelezi lidera klana)</i></label> <br /><br /> <input class="button" type="submit" name="submit" value="Dodaj" /> </p> </form> Now i wonna save this igraci[] array in mysql, i tried like this but it doesn't work: $igraci = $_POST['igraci']; $query = "INSERT INTO klanovi (naziv, website, email, igraci) VALUES ('{$naziv}', '{$website}', '{$email}', '{$igraci}')"; $result = mysql_query($query, $connection); How can i do this? Thanks.. hy I want to make a select in a class: Code: [Select] class test { var $unu = array(); function __construct(){ $link = mysql_connect("", "", "") or die("Couldn't make connection."); $db = mysql_select_db("", $link) or die("Couldn't select database"); //$this->unu = $textin; } function show($col){ $q = mysql_query("SELECT $col FROM categorie"); $qw = mysql_fetch_array($q); $this->unu = $qw; return $this->unu; } } If I want to show how many rows are in that table: Code: [Select] $box = new test(); echo count($box->show("titlu")); it show me that are only 2, although in "categorie" table are 10 rows. Can you please tell me where is the error ? Thanks I am designing my own framework, and I have created a select function. I have set a parameter to be an array by default, with the default value of *. When I check if it's an array (with is_array) it says that it isn't. Does anyone have any idea why it would say that? Code: [Select] function select($table, $rows = array('*'), $cond = "") { $query = "SELECT {$rows} FROM `{$table}`"; if (isset($cond)) { $query .= " WHERE {$cond}"; } if (is_array($rows)) { if (isset($rows) && $rows != "*") { foreach ($rows as $v) { print_r($v); } } } //return mysql_query($query); } Any help is appreciated. Thanks. I want to define a function instead of repeating query in all my php pages. I call a function by passing an $id value and from that function i have to get all the info related to that id, like name, description and uom.
I am trying to do this, but i dont know how to get these values seperately.
here is my function
function items($item_id)
{
$details = array();
$result = mysql_query("select item_id, name, uom, description from items where item_id=".$item_id."") or die (mysql_error());
while($row = mysql_fetch_array($result))
{
$details[] = array((stripslashes($row['name'])), (stripslashes($row['uom'])), (stripslashes($row['description'])));
}
return $details;
}
and i call my function like this
$info = items($id);Can somebody guide me in this I have a DB It gets records from a sensor every 5 min This is then displayed on a Chart When I request 4 Hrs I get 48 records x 12 Sensors, Not a issue When I request 12 Hrs I get 144 records x 12 Sensors, A little Slower but Still OK But as I start to request longer periods Like 7 Days I start getting larger and Larger data sets causing time out issues with the AJAX Calls (4-30MB per sensor downloads of data) When looking at 4 weeks I really cant see the fine detail so is there a way to request records from MYSQL that drops some records in between Kinda like the Step function in a For loop If I request a period that has 2048 records but only want 204 records is there a way to say give me From -> To dates Step 10 and only return every 10th Record in the data set Thanks When i use this code it show everything except for the first record in the table. I have also tried it with the users for the login system im using and then it doesnt show the person im logged in to or also the first record in the table and i need it too show all the records How do i solve this? I thought it had something to do with: Code: [Select] <?php while ($rij = mysql_fetch_array($result)){ echo ("<tr><td>". $rij['ID'] . " </td> " . "<td>" . $rij['naam'] . " " . $rij['telefoon'] . " </td> " . "<td>" . $rij['adres'] . " </td> " . "<td>" . $rij['mobiel`'] . " </td>". "</td></tr>\n "); } ?> But i dont know how to solve it. Code: [Select] <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("root123", $con); $sql= "SELECT * FROM root123 ORDER BY ID;"; $result = mysql_query($sql, $con); $rij = mysql_fetch_array($result); ?> <table width="80%" align="center"> <tr> <th>ID</th> <th width="150px">Naam</th> <th>Gebruikersnaam</th> <th width="300px">Wachtwoord</th> </tr> <?php while ($rij = mysql_fetch_array($result)){ echo ("<tr><td>". $rij['ID'] . " </td> " . "<td>" . $rij['naam'] . " " . $rij['telefoon'] . " </td> " . "<td>" . $rij['adres'] . " </td> " . "<td>" . $rij['mobiel`'] . " </td>". "</td></tr>\n "); } ?> </table> Ive tried to create a function to create a dropdown select box but im getting alot of errors to do with the if statement saying the ($_POST[$name]) value is not set??? function selectBox($name, $firstvalue, $limit, $increment) { // echo "<br>"; // echo $name; // echo "<br>"; // print_r($_POST[$name]); // echo "<br>"; $select ="selected=\"selected\""; $body = "<select name='$name' id='$name' method='POST'> <option value=''>$name</option>"; for ($value = $firstvalue; $value <= $limit; $value += $increment) { $body .= "<option value= '$value' "; if ($_POST["$name"] === $value) { $body .= $select; } $body .= ">$value</option>"; } $body .= "</select>"; // echo $value; // echo $_POST[$name]; // echo $_POST['Width']; return $body; } Any help would be greatly appreciated Hi, I'm an novice/intermediate PHP prgogrammer, so excuse me if this is a stupid question. I have one php program that consist of a few functions. Inside the function taht is the starting point I pickup a few entries from a MYSQL database, and display those in my html page. I wish to dynamically put a link on those entries that makes it possible for the users to click those entries an get directed to other functions inside this very same PHP program WITH a parameter. The parameters that should accompany the url/link into the selected function to be able to select apropriate records from the mysql tables. Any hints and tips for this would be highly appreciated! Kind Regsrds, Terje Hvidsten. Is it possible to achieve a similar thing to this... SELECT FROM db WHERE value!=$array My example code is this: $browser_hide = array("1" => "YandexBot 0.0 for unknown"); $query = "SELECT * FROM hitlist_hits WHERE month='$month' AND browser!='$browser_hide' ORDER by id DESC"; Thanks Guys. In witch conditions you need to transfer the data from a select in array and after that to show from array in content of the page ? How can I select a value in an array but not its key. Hi guyz, please help me I want to get an id from my database, I use a word to search it which is stored as keys of array. But all the keys can't be found in my database, even though it's there. Here some parts of my code Code: [Select] $arrKeys = array_keys($arrResult); foreach($arrKeys as $key) { //if($arrResult[$key]>1) { echo $key/*."=>". $arrResult[$key]*/."<br>"; //$q = mysql_query("select id_katadasar from tb_katadasar where katadasar='".$key."'"); $kon->query("select id_katadasar from tb_katadasar where katadasar='".$key."'"); var_dump($kon->query); //echo "select id_katadasar from tb_katadasar where katadasar='".$key."'"; //$jum = $kon->getJumlah(); //$row = mysql_fetch_array($q, MYSQL_ASSOC) or die(mysql_error() . ''. $q); //echo count($row); if($row = $kon->tampilkan()) //if($row = mysql_fetch_array($q, MYSQL_NUM)) { //$res = $kon->tampilkan(); echo $row[0]; } //else { //echo "tidak ada di db <br><br>"; } } } $kon->query is equal to mysql_query() $kon->tampilkan return $row=mysql_fetch_array(query) Thanks I have a table with a field called "tags" and below is an example of how this field might look like. pensions, employee benefits, group benefits, defined contribution, auto-enrollment I want to perform a query to get this field for all of the rows in my table. However is there a way to stop mysql from returning duplicates. For example if another field read, pensions, employee benefits, group benefits, defined contribution, auto-enrollment, tax Then it would only return "tax" from this field as it's already return pensions, employee benefits etc. Thanks for any help. Hi guys I have a two tables in my mysql branch: id branchname postcode then a user table id username password branch1 branch2 so far I have a select form as below which populates the select form from mysql branch 1<select name='branch1'><p /> <?php $branchdropdown=mysql_query("SELECT id ,branchname, postcode FROM branch"); while($row = mysql_fetch_array($branchdropdown)) { // echo '<option value="' .$row['stationname']. '"></option>' ; echo "<option value=\"".$row['id']."\">".$row['branchname']."\n "; $postcodeone=$row['postcode']; } ?> and then this will be inserted in mysql as below $submit = mysql_query("INSERT INTO users (branch1, branch2) VALUES ($postcodeone, $postcodetwo)"); echo "This entry has been added to our database"; </select> but it only inserts the branch id and not the password, can you please tell me what im doing wrong? |
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