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PHP - Query Multiple Mysql Tables Is Not Working.
Hi guys, I'm trying to query two tables for different data and echo the results that match both tables. Here are the tables I have and the query I'm trying to run.
(Table) qc_reports (Fields) id report_date report_lot_number report_po report_supplier report_buyer report_inspectedby report_pulptemprange report_carrierconditions report_supplierclaim report_carrierclaim report_temprecorder report_temprange_N report_temprange_M report_temprange_B report_suppliercontact report_contactedby report_time report_comments (Table) qc_lots (Fields) id report_id lot_temprange lot_commodity lot_rpcs lot_brand lot_terms lot_cases lot_orgn lot_estnum lot_avgnum This is the query that I'm trying to do. Code: [Select] <?php $sql = "SELECT * FROM qc_reports, qc_lots WHERE "; if (!empty($start_date) and !empty($end_date)) $sql .= " qc_reports.report_date BETWEEN '$start_date' and '$end_date' AND "; if (!empty($search_fronteralot)) $sql .= " qc_reports.report_lot_number = '$search_fronteralot' AND "; if (!empty($search_buyer)) $sql .= " qc_reports.report_buyer = '$search_buyer' AND "; if (!empty($search_supplier)) $sql .= " qc_reports.report_supplier = '$search_supplier' AND "; if (!empty($search_po)) $sql .= " qc_reports.report_po = '$search_po' AND "; if (!empty($search_carrierconditions) and $search_carrierconditions != 'all') $sql .= " qc_reports.report_carrierconditions = '$search_carrierconditions' AND "; if (!empty($search_commodity) and $search_commodity != 'all') $sql .= " qc_lots.lot_commodity = '$search_commodity' AND "; if (!empty($search_inspectedby)) $sql .= " qc_reports.report_inspectedby = '$search_inspectedby' AND "; $sql = substr($sql, 0, -4); $query = mysql_query($sql); $numrows = mysql_num_rows($query); ?> RESULTS - <?php echo $numrows; ?> <hr> <table width='500'><tr><td><b>Date</b></td><td><b>Lot Number</b></td><td><b>PO</b></td><td> </td></tr> <tr> <td> </td> </tr> <?php while ($row = mysql_fetch_assoc($query)) { $id = stripslashes($row['id']); $report_lot_number = stripslashes($row['report_lot_number']); $report_po = stripslashes($row['report_po']); $report_date = stripslashes($row['report_date']); echo "<tr> <td>" . $report_date . "</td></td><td>" . $report_lot_number . "</td><td>" . $report_po . "</td><td><a href='view_report.php?id=" . $id . "'>View</a></td> </tr>"; } echo '</table><br><br><br><hr><br><br>'; } ?> All the variables are passed from a HTML form with $_POST. I need the search to work like this: If there is a value in a form field then the query gets appended with that value but when it gets to the $search_commodity it needs to search the second table (qc_lots) and check for the results. Any results that match have to be matched to the results from the first table (qc_reports) and display (echo) only qc_reports that match to both tables. The only common field is the report_id on the qc_lots table and the id on the qc_reports table. I'm stuck and need some guidance. Can someone help please? Similar Tutorials
Basically I want to add up multiple tables and display a grand total on my page when the radio button is pressed. The radio button has values that connect to my database and the values link to and ID with a price. How can I use this code in order to work out a grand total? Many thanks. I have searched everywhere but found nothing. $query = mysql_query("SELECT COUNT(tickets.id),tickets.id,tickets.status,tickets.date,tickets.question,tickets.title,users.position FROM tickets,users WHERE tickets.id='$existing' users.username='$username'"); How do I get it so it selects data from the table tickets where id='$existing', but I also want to get a user's (who is viewing the ID) position (rank). $username is a session. So, any thoughts on how I would do so? My updated code: <?php session_start(); include("../includes/mysql.php"); include("../includes/config.php"); ?> <!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> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <link href="../style.css" rel="stylesheet" type="text/css" /> <title><?php echo $title; ?></title> </head> <body> <div id="container"> <div id="content"> <div id="left"> <div class="menu"> <?php include("../includes/navigation.php"); ?> <div class="menufooter"></div> </div> <?php include("../includes/menu.php"); ?> </div> <div id="middle"> <?php $existing = $_POST['existing']; $ip = $_SERVER['REMOTE_ADDR']; $username = $_SESSION['user']; $query = mysql_query("SELECT COUNT(tickets.id),tickets.id,tickets.status,tickets.date,tickets.question,tickets.title,users.position FROM tickets,users WHERE tickets.id='$existing' OR users.username='$username'"); $get = mysql_fetch_assoc($query); if(!$existing) { echo ' <div class="post"> <div class="postheader"><h1>Error</h1></div> <div class="postcontent"> <p>You have not enetered in a ticket ID. Please go back and do so.</p> </div> <div class="postfooter"></div> </div> '; } elseif($get['COUNT(id)'] < 1) { echo ' <div class="post"> <div class="postheader"><h1>Error</h1></div> <div class="postcontent"> <p>The ticket ID you are trying to use doesnt exist. Please go back or submit another ticket.</p> </div> <div class="postfooter"></div> </div> '; } elseif($get['tickets.ip']==$ip || $get['user.position'] >= 1) { $status["tickets.status"]["0"] = "Waiting for support..."; $status["tickets.status"]["1"] = "Waiting for user..."; $status["tickets.status"]["2"] = "Ticket Closed..."; $status["tickets.status"]["3"] = "Ticket Opened..."; echo ' <div class="post"> <div class="postheader"><h1>View Ticket Status - ID '. $get["id"] .'</h1></div> <div class="postcontent"> <p>Title: '. $get["tickets.title"] .' - Posted on: '. $get["tickets.date"] .'</p> <p>Ticket Status: '. $status[$get["tickets.status"]] .'</p> <p>Question: '. nl2br($get["tickets.question"]) .'</p> </div> <div class="postfooter"></div> </div> '; } else { echo ' <div class="post"> <div class="postheader"><h1>Error</h1></div> <div class="postcontent"> <p>This is not your ticket. You only have permission to view your tickets. Please go back.</p> </div> <div class="postfooter"></div> </div> '; } ?> </div> </div> </div> </body> </html> Hey, I have a query running inner joins... which currently works, but I need some more information pulled from other tables... Code: [Select] define('WPSC_TABLE_PRODUCT_LIST', "{$wp_table_prefix}wpsc_product_list"); define('WPSC_TABLE_PRODUCTMETA', "{$wp_table_prefix}wpsc_productmeta"); define('WPSC_TABLE_CATREF', "{$wp_table_prefix}wpsc_item_category_assoc"); define('WPSC_TABLE_CATNAME', "{$wp_table_prefix}wpsc_product_categories"); $cf="Linked Products"; $sku = get_post_meta($post->ID, $cf, true); $array = explode(",",$sku); $products = array_count_values($array); foreach($products as $key => $value) { $product = $wpdb->get_row("SELECT meta.product_id AS pid, list.name AS name, list.price AS price, retailer.category_id AS catid FROM ".WPSC_TABLE_PRODUCTMETA." AS meta INNER JOIN ".WPSC_TABLE_PRODUCT_LIST." AS list ON ( list.id = meta.product_id ) INNER JOIN ".WPSC_TABLE_SHOPNAME." AS retailer ON ( retailer.product_id = meta.product_id ) WHERE `meta_key` IN ( 'sku' ) AND `meta_value` IN ( '{$key}' ) ORDER BY list.id DESC"); ?> But I also need the category name - So I have to cross reference the following... wordpdem_wpsc_product_list id name 433 itemone 432 itemtwo 431 itemthree wordpdem_wpsc_item_category_assoc id product_id category_id 1 433 1 2 432 2 3 410 3 wordpdem_wpsc_product_categories id name 1 Bread 2 Fish 3 Snacks I've now written: Code: [Select] SELECT meta.product_id AS pid, list.name AS name, list.price AS price, retailer.category_id AS catid, category.name AS catname FROM wordpdem_wpsc_productmeta AS meta INNER JOIN wordpdem_wpsc_product_list AS list ON ( list.id = meta.product_id ) INNER JOIN wordpdem_wpsc_item_category_assoc AS retailer ON ( retailer.product_id = meta.product_id ) INNER JOIN `wordpdem_wpsc_product_categories``wordpdem_wpsc_product_list` AS category ON ( retailer.category_id = category.id ) WHERE `meta_key` IN ( 'sku' ) AND `meta_value` IN ( '{$key}' ) ORDER BY list.id DESC However it fails on line 12 (the Where statement...) Any ideas what I've done wrong? Hi, i have a database with multiple records. One of those values is an image. When i query the database and i want only the images as a result it gives the result in a field. It then fills up the field horizontal and when there is no more field left it starts a new row. That is what i want. But now i want some extra info next to the picture so i put that in a table with two colums like this: echo '<table border="1" width="400"><td><a href="'.$row['website'].'"> <img src="'.$row['prod_img'].'"> '.$row['prod_name'].' <a href="'.$row['website'].'">'.$row['website'].'</td></table> ';}} But i still want the result to fill up the field and when the field is full i want the next results on a row below this.... Anyone? Below is a page which is supposed to output the name, blog contribution and picture of contributing members of a website. <div id="blog_content" class="" style="height:90%; width:97%; border:5px solid #c0c0c0; background-color: #FFFFFF;"> <!--opens blog content--> <?php //address error handling ini_set ('display_errors', 1); error_reporting (E_ALL & ~E_NOTICE); //include the config file require_once("config.php"); //Define the query. Select all rows from firstname column in members table, title column in blogs table,and entry column in blogs table, sorting in ascneding order by the title entry, knowing that the id column in mebers table is the same as the id column in blogs table. $sql = "SELECT blogs.title,blogs.entry,members.firstname,images.image FROM blogs LEFT JOIN members ON blogs.member_id = members.member_id LEFT JOIN images ON blogs.member_id = images.member_id ORDER BY blogs.title ASC "; $query = mysql_query($sql); if($query !== false && mysql_num_rows($query) > 0) { while(($row = mysql_fetch_assoc($query)) !== false) { echo '<div id="blog_content1" style="float:left; position:relative;bottom:18px;left:13px; background-color: #FFFFFF; height:16.7%; width:100%; border:0px none none;" <!--opens blog_content1 same as main center top 1 and 2 from index page everything scaled down by a factor of 3, heightwise--> <div class="red_bar" style="height:3%; width:100%; border:1px solid #959595;"> <!--a--> <div class="shade1" style="height:5px; width:100%; border:0px none none;"> </div> <div class="shade2" style="height:5px; width:100%; border:0px none none"> </div> <div class="shade3" style="height:5px%; width:100%; border:0px none none"> </div> </div> <!-- closes red bar--> <div class="content" style="height:28.3%; width:100%; border:0px none none;"> <!----> <div class="slideshow" id="keylin" style="float:left; width:20%; border:0px none none;"> <!--a--> <div><img header("Content-type: image/jpeg"); name="" alt="" id="" height="105" width="105" src="$row[image]" /></div> </div> <!-- closes pic--> <div class="content_text" style="float:right; position:relative;top:7px;left:0px; max-height:150px; width:78.5%; border-width:4.5px; border-bottom-style:solid; border-right-style:solid; border-color:#c0c0c0; "> <!--a-->'; echo "<h3>".$row['title']."</h3>"; echo "<p>" .$row['entry']."<br />".$row['firstname']."</p>"; echo '</div> <!-- closes content text--> </div> <!-- closes content--> </div> <!-- closes blog_content1-->'; } } else if($query == false) { echo "<p>Query was not successful because:<strong>".mysql_error()."</strong></p>"; echo "<p>The query being run was \"".$sql."\"</p>"; } else if($query !== false && mysql_num_rows($query) == 0) { echo "<p>The query returned 0 results.</p>"; } mysql_close(); //Close the database connection. ?> </div> <!-- closes blog content--> The select query is designed to retrieve all the blog contributions(represented by the fields blogs.title and blogs.entry) from the database, alongside the contributing member (member.firstname) and the member's picture(images.image), using the member_id column to join the 3 tables involved, and outputs them on the webpage. The title, entry and firstname values are successfully displayed on the resulting page. However, I can't seem to figure out how to get the picture to be displayed. Note that the picture was successfully stored in the database and I was able to view it on a separate page using a simple select query. It is now just a question of how to get it to display on this particularly crowded page. Anyone knows how I can output the picture in the img tag? I tried placing the header("Content-type: image/jpeg"); statement at the top of the php segment, then just right below the select query and finally just right above the img tag, but in every case, I just got a big white blank page starring at me. How and where should I place the header statement? And what else am I to do to get this picture displayed? Any help is appreciated. I have a table that contains the schools in my system: schools id name location ... Then, I have three tables that use the id of this table: schoolAdmins schoolID schoolContests schoolID students schoolID When I go to delete a school from my system, I want to check to see if that school is connected to any of these three other tables first. This is what I tried (but obviously failed because I'm here) where I'm passing the query the $studentID in question: SELECT * FROM schoolAdmins, schoolContests, students WHERE (schoolAdmins.schoolID = $schoolID) OR (schoolContests.schoolID = $schoolID) OR (students.schoolID = $schoolID) I'm really new to the concept of querying multiple tables in a single statement, so I'm just kind of guessing at this point. Thanks in advance. hi everybody simply love this forum because i always get the question perfectly answered. i am here this time with a rather complicated question: i am creating a simple duty plan for different departments for exeample department #1 is "Tech" and department #2 is "Service" different employees working in these departments change their dutys and will are sent to another department or section or go on holidays. this is why i have at least 4 different mysql tables to store the data:- personel(storing the personal information of the workers) dutyplan(storing the week's working daysy of the workers) departments(storing the starting and ending dates of the department change) vacation(stores the start and ending dates of the vacations) in all the tables the common id is the pid which is issued unique to every worker. i am still able to work only with the 1st two tables. i can edit and create new plans for the weeks for employees working in the departments with the following php code(submitting only to edit the plan) if i add a date range for example Monday the 24th of January to Sunday the 30th of January for an employee working in the Tech department to work a few days in service department, it will show the employee in the week on the plan of both departments. if both department heads edit the plan without communicating to each other, and knowing on which date the worker goes out and in to the department, the 1st one will plan him for the whole week on his dutyplan and the second department head will overwrite this plan(if edited) or if creating a new plan(create the duty of the worker also one more time). i want to avoid this confusion and manual work. the same is with the vacation or sickness tables, if the employee has vacation, the program should check and return the selected value in the pulldown menu with the value stored in the vacation or sickness tables appropriate to the date in the plan table. ************************************************************************ form.php: <? require "config.php"; $result=mysql_query("select * from personel, dutyplan, department, vacation WHERE personel.pid = dutyplan.pid and personel.pid = vacation.pid and dp.pid = departments.pid and departments.Section = 'Service' order by dp.id asc"); ?> <? while($row=mysql_fetch_assoc($result)){ ?> // after that i create table, displaying all the days of the week from monday to sunday and use this code <tr> <td><? echo $row['FirstName'] . " " . $row['LastName']; ?>: </td> <td> <select name="monday_<? echo $row['id']; ?>" id="select1"> <option><? echo $row['Mo']; ?></option> <option>Duty</option> <option>Free</option> <option>Vacation</option> //the same way down to sunday, for every day the different options to be selected. it displays the records perfectly, as long as there is no change of department planed and the vacation must also be selected manually. +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ update.php <? if($_POST['Submit']){ require "includes/config.php"; $result=mysql_query("select * from dutyplan, departments, vacation order by dutyplan.id asc"); while($row=mysql_fetch_assoc($result)){ $mo=$_POST["monday_".$row[id]]; $tu=$_POST["tuesday_".$row[id]]; $we=$_POST["wednesday_".$row[id]]; $th=$_POST["thursday_".$row[id]]; $fr=$_POST["friday_".$row[id]]; $sa=$_POST["saturday_".$row[id]]; $su=$_POST["sunday_".$row[id]]; mysql_query("update dp set Mo='$mo', Tu='$tu', We='$we', Th='$th', Fr='$fr', Sa='$sa', Su='$su' where id='$row[id]'"); } echo "Records updated"; } ?> **************************************************** how would you gueys solve this problem? many thanks in advance Hi, So I'm trying to make a car rental project, and I need to show cars of a specific type, available in a specific period. I'm currently using this query: SELECT * FROM car, reservations WHERE car.car_id = reservations.car AND car.type = 1 AND reservations.reserved_from >= "2011-12-22" AND reservations.reserved_to <= "2012-12-20"; So this returns all the cars of the type 1 reserved in 2011-12-22 - 2012-12-20 period. So know that I know which cars are unavailable, how would you go ahead and pick out what's left? I have a photo album style gallery to build and i'm finding it dificult to list all the table names (these are names of photo albums) and then enter the data into a seperate query for each album name (these will change often so i cant keep updating the file as normal. this will then post all the data to the xml file and show the set of photos in the individual albums in a flash file. can anyone help me where im going wrong at all? <?php $dbname = 'cablard'; if (!mysql_connect('localhost', 'cablard', '')) { echo 'Could not connect to mysql'; exit; } $sql = "SHOW TABLES FROM $dbname"; $result = mysql_query($sql); if (!$result) { echo "DB Error, could not list tables\n"; echo 'MySQL Error: ' . mysql_error(); exit; } while ($row = mysql_fetch_row($result)) { echo "Table: {$row[0]}\n"; } mysql_free_result($result); $query = "SELECT * FROM photo ORDER BY id DESC"; $result2 = mysql_query ($query) or die ("Error in query: $query. ".mysql_error()); while ($row = mysql_fetch_array($result2)) { echo " <image> <date>".$row['date']."</date> <title>".$row['title']."</title> <desc>".$row['description']."</desc> <thumb>".$row['thumb']."</thumb> <img>".$row['image']."</img> </image> "; } ?> Thanks James Not sure if this topic goes in here, it is related to PHP but also MySql, so if i'm on the wrong board sorry! What i'm trying to do is search for a keyword in 5 different tables and return the keyword ID from the table that its in The tables i'm trying to search are as follow location state county region continent The "location" table has all the locations i.e cities and each row has the following columns: id | continent_id | country_id | state_id | region_id | city_name The "state" table is set the the following: id | name "county" table : id | name "region" table: id | region and "continent" table id | name The way it works is the can search for any city or state or county or region or continent and ideally it should look into the five different tables and return the id of that table. So if the use searches for United States it will look for United States in all five tables, obviously it would find it in the "country" table so it should return that "id". The results are returned in "json format" Below is the code i have: Code: [Select] <? $input = $_GET['keyword']; $data = array(); /* In this query i'm attempting to search in all databases, but i'm not sure if i'm doing this right. I'm not getting any results so i know something is wrong just don't know how to write it. */ $query = mysql_query("SELECT * FROM locations JOIN states ON states.id = locations.state_id JOIN countries ON countries.id=locations.country_id JOIN regions ON regions.region_id = locations.region_id JOIN continents ON continents.id=locations.continent_id WHERE name LIKE '$input%' OR state LIKE '$input%' OR region LIKE '$input%' OR country LIKE '$input%' OR continent LIKE '$input%'") or die(mysql_error()); /* Here the values are added to to the $json array. The "value" should be the "id" from the table that the keyword matched. The 'name' Should be the name of the actual keyword. Again if they search for United States the "id" will come from the "countries" table and the "value" would come from the "countries" table as the name */ while ($row = mysql_fetch_assoc($query)) { $json = array(); $json['value'] = $row['id']; $json['name'] = $row['name']; $data[] = $json; } header("Content-type: application/json"); echo json_encode($data); ?> Any help would he be appreciated, i don't want people to do it for me, but rather just guide me a little bit. Back with a new problem. I have 8 tables interconnected. Table#1 - Users user_id | name Table#2 - user_categories id | user_id | category_id Table#3 - user_cities id | user_id | city_id Table#4 - user_dates id | user_id | dates_available Table#5 - categories category_id | category_name Table#6 - cities city_id | city_name Table#7 - provinces province_id | province_name Table#8 - categories country_id | country_name Each user will have multiple categories, cities and available dates listed in these tables. I simply want to retrieve and list each user and their data on a page. Here's my query. $url_city = 1;
$url_category = 2;
$url_date = '2021-07-19';
$find_records = $db->prepare("SELECT user_categories.*, categories.*, user_cities.*, cities.*, provinces.*, countries.*, user_dates.*, users.* FROM users
LEFT JOIN user_categories ON users.user_id = user_categories.user_id
LEFT JOIN user_cities ON users.user_id = user_cities.user_id
LEFT JOIN user_dates ON users.user_id = user_dates.user_id
LEFT JOIN categories ON user_categories.category_id = user_categories.category_id
LEFT JOIN cities ON user_cities.city_id = cities.city_id
LEFT JOIN provinces ON cities.province_id = provinces.province_id
LEFT JOIN countries ON provinces.country_id = countries.country_id
WHERE user_cities.city_id = :city_id AND user_categories.category_id = :category_id AND user_dates.date_available = :date_available GROUP BY users.user_id");
$find_records->bindParam(':city_id', $url_city);
$find_records->bindParam(':category_id', $url_category);
$find_records->bindParam(':date_available', $url_date);
$find_records->execute();
$result_records = $find_records->fetchAll(PDO::FETCH_ASSOC);
if(count($result_records) > 0) {
foreach($result_records as $row) {
$user_id = $row['user_id'];
$name = $row['name'];
$country_id = $row['country_id'];
$country_code = $row['country_code'];
$country_name = $row['country_name'];
$province_id = $row['province_id'];
$province_code = $row['province_code'];
$province_name = $row['province_name'];
$city_id = $row['city_id'];
$city_name = $row['city_name'];
$category_id = $row['category_id'];
$category_name = $row['category_name'];
}
}
There are no errors but the above query would only return a single row with only 1 "user" despite having multiple users in the "users" table. If I remove the GROUP BY, then it'll return multiple rows of the same user instead of all the relevant users. So what do you think I am doing wrong with my query? Edited July 20 by imgroootTo accomplish the layout I was looking for, I used the following code: for ($i=1; $i<=7; $i++) { //set day of week for display as title of columns $dow=date("l", strtotime($year.'W'."$weekno"."$i")); echo "<td class=\"wvcolumn\" valign=\"top\"> <table cellpadding=\"0\" cellspacing=\"0\" border=\"0\" align=\"center\" width=\"$daycolumn\"> <tr><th class=\"wvtitle\" valign=\"bottom\" align=\"center\" width=\"$daycolumn\"><a class=\"titledate\" href=\"createticket.php?action=o&month=".date("m", strtotime($year.'W'."$weekno"."$i"))."&date=".date("d", strtotime($year.'W'."$weekno"."$i"))."&year=".date("Y", strtotime($year.'W'."$weekno"."$i"))."\">$dow, ".date("F d", strtotime($year.'W'.$weekno."$i"))."</a> <a onClick=\"window.open('http://www.batchgeo.com')\" title=\"Map $dow's Appointments\" href=\"mapday.php?week=$weekno&year=$year&day=$i\"><img width=\"20\" style=\"border-style: none\" src=\"../images/icon_globe.png\"></a></th></tr>"; if(date("Y-m-d")==date("Y-m-d", strtotime($year.'W'."$weekno"."$i"))) { $query = "SELECT fieldtickets.ticketnumber, fieldtickets.business, title, apptdatetime, DATE_FORMAT(apptdatetime,'%h:%i %p') AS fapptdatetime, status, billstatus, type, assigntech, clients.business FROM fieldtickets LEFT JOIN clients ON fieldtickets.business = clients.id WHERE status='Open' AND (type = 'Field' OR type = 'Phone') AND apptdatetime BETWEEN '".date("Y-m-d", strtotime($year.'W'."$weekno"."$i"))." 00:00:00' AND '".date("Y-m-d", strtotime($year.'W'."$weekno"."$i"))." 23:59:59' UNION SELECT fieldtickets.ticketnumber, fieldtickets.business, title, apptdatetime, DATE_FORMAT(apptdatetime,'%h:%i %p') AS fapptdatetime, status, billstatus, type, assigntech, clients.business FROM fieldtickets LEFT JOIN clients ON fieldtickets.business = clients.id WHERE status = 'Pending' ORDER BY status ASC, apptdatetime ASC"; } else { $query = "SELECT fieldtickets.ticketnumber, fieldtickets.business, title, apptdatetime, DATE_FORMAT(apptdatetime,'%h:%i %p') AS fapptdatetime, status, billstatus, type, assigntech, clients.business FROM fieldtickets LEFT JOIN clients ON fieldtickets.business = clients.id WHERE status='Open' AND (type = 'Field' OR type = 'Phone') AND apptdatetime BETWEEN '".date("Y-m-d", strtotime($year.'W'.$weekno."$i"))." 00:00:00' AND '".date("Y-m-d", strtotime($year.'W'.$weekno."$i"))." 23:59:59' ORDER BY apptdatetime ASC"; } $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ //if ticket is open show appt time and business, if ticket pending display DELIVERY and business name if($row[status]=="Open") { $href="createticket.php?action=e&ticket=".$row[ticketnumber]; $time=$row[fapptdatetime]; $delivery=""; } elseif($row[status]=="Pending") { $href="createticket.php?action=c&ticket=".$row['ticketnumber'].""; $time=""; $delivery="DELIVER"; } echo "<tr><td valign=\"top\"><a title=\"".$row[ticketnumber]." - ".$row[title]."\" class=\"".$row[status]."\" href=\"$href\"><span>$delivery$time ".substr($row['business'], 0, $displaychars)."</span></a></td></tr>\n"; } echo "</table>"; } echo "</td></tr> </table>"; Output is similar to this, and any existing "deliveries" are displayed under the current day. _________________________________________________ ______________ | Monday | Tuesday | Wednesday | Thursday | Friday | |appointments |appointments |appointments |appointments |appointments This works great, but requires 5 separate queries. My database is very small for now, so its not a big deal, but I know this can be done much more efficiently. How can I query the whole week and put appointments in their corresponding tables (days)? Thank you for your help! Ok, so I've spent quite a bit of time piecing together this solution from a variety of sources. As such, I may have something in my code below that doesn't make sense or isn't neccessary. Please let me know if that is the case. I'm creating an administrative form that users will you to add/remove items from a MySQL table that lists open positions for a facility. The foreach loop generates all of the possible job specialties from a table called 'specialty_list'. This table is joined to a second table ('open_positions') that lists any positions that have been selected previously. Where I'm stuck is getting the checkbox to be checked if the facility_ID from the open_positions table matches the $id passed in the URL via ?facility_id=''. Here's where I am so far: $query = "SELECT specialty_list.specialty_displayname , specialty_shortname , open_positions.position , facility_ID FROM specialty_list LEFT OUTER JOIN open_positions ON open_positions.position = specialty_list.specialty_shortname ORDER BY specialty_list.specialty_shortname"; $results = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_assoc($results)) { $positions[$row['specialty_shortname']] = $row['specialty_displayname']; } echo "<form method='POST' action='checkbox.php'>"; foreach($positions as $specialty_shortname => $specialty_displayname) { $facility_ID = $row['facility_ID']; $checked = $facility_ID == $row['facility_ID'] ? ' checked' : ''; echo "<input type='checkbox' name='position[]' value=\"{$specialty_shortname}\"{$checked}> {$specialty_displayname}</input><br/>"; } echo "<input type='hidden' name='facility_ID' value='$id'>"; echo "<input type='submit' value='Submit Checkboxes!'>"; echo "</form>"; Any ideas how to get this working? I feel like I'm very close, but I just can't get it. I also tried starting from scratch with a WHILE statement instead of a FOREACH, but haven't tweaked it enough to prevent duplicate checkboxes. With that in mind, here it is, just in case that's a better direction: $query = "SELECT specialty_list.specialty_displayname , specialty_shortname , open_positions.position , facility_ID FROM specialty_list LEFT OUTER JOIN open_positions ON open_positions.position = specialty_list.specialty_shortname ORDER BY specialty_list.specialty_shortname"; $results = mysql_query($query) or die(mysql_error()); echo "<form method='POST' action='checkbox.php'>"; while($row=mysql_fetch_assoc($results)) { $facility_ID = $row['facility_ID']; $specialty_shortname = $row['specialty_shortname']; $specialty_displayname = $row['specialty_displayname']; if ($facililty_ID==$id) { $checked=' checked'; } echo "<input type='checkbox' name='position[]' value=\"$specialty_shortname\"$checked> $specialty_displayname</input><br/>"; } echo "<input type='hidden' name='facility_ID' value='$id'>"; echo "<input type='submit' value='Submit Checkboxes!'>"; echo "</form>"; I am loading notifications from a database table called "notifications" and I am having a little trouble getting them to order in the correct way. my query I'm using right now: Code: [Select] $query = mysql_query("SELECT B.* FROM (SELECT A.* FROM notifications A WHERE A.user_id='$session' AND A.from_id!='$session' ORDER BY A.id ASC ) AS B ORDER BY B.state ASC LIMIT 7"); this works well as far as showing the unread notifications on top, then the read notifications below, however it's not ordering the two sets by ID (which the id auto increments so the higher id number is the newest) from newest on top to the oldest on bottom, still keeping them separated by the unread on top, read on bottom (column name is state for showing whether they're read or not). The order the notifications are displaying by their ID is: 3 5 2 4 1 when it should be: 5 3 2 4 1 I made a small editing system for my news page, and I need to update three columns within my table "announcements" in the database. I tried a method of updating all of them with one MySQL query instead of using three as it just isn't neat. I've searched several methods via google and I've tried all of them, but just can't seem to get it to work. Is this MySQL query correct? mysql_query("UPDATE announcements SET title = {$title} WHERE id = '$id', content = {$content} WHERE id = '$id', lastmodified = ". date('M-d-Y') ." WHERE id = '$id'"); I want to have a search product feature, but I would like members to be able to search multiple fields in one go i.e. product_code, Product_name in one MySQL query. The thing is, members have to be logged on, so the query must also only show results relating to that specific member, via the session[member_ID], my current query for listing products for that specific member is : Code: [Select] $sql = "SELECT productId, productCode, image, name, price, stock_level FROM product_inventory WHERE memberr_ID = '" . $_SESSION['SESS_mem_ID'] . "; How would I change the above into a search query to search for productcode, productname and still only show results beloging to this member using the session data ? all help appreciated I insert data in mysql table row using multiple method : 1#2#3 . 1 is ID of country, 2 is Id of state, 3 is ID of town . now i have this table for real estate listings. for each list(home) i have country/state/town (1#2#3). in country table i have list of country - in country table i have list of state - in country table i have list of town. i need to The number of houses in country / state / town . my mean is : Code: [Select] USA [ 13 ] <!-- This Is equal of alabama+alaska+arizona --> ----Alabama [8] <!-- This Is equal of Adamsville+Addison+Akron --> -------Adamsville [2] -------Addison[5] -------Akron[1] ......(list of other City) ----Alaska [ 3 ] -------Avondale[3] ......(list of other City) ----Arizona [ 2 ] -------College[2] ......(list of other City) Lisintg Table : Code: [Select] ID -- NAME -- LOCATION -- DATEJOIN -- ACTIVE 1 -- TEST -- 1#2#3 -- 20110101 -- 1 2 -- TEST1 -- 1#2#3 -- 20110101 -- 1 3 -- TEST2 -- 1#3#5 -- 20110101 -- 1 4 -- TEST3 -- 1#7#6 -- 20110101 -- 1 Country Table : Code: [Select] id -- name 1 -- USA stats Table : Code: [Select] id -- countryid -- name 1 -- 1 -- albama 2 -- 1 -- alaska 3 -- 1 -- akron town Table : Code: [Select] id -- countryid -- statsid -- name 1 -- 1 -- 1 -- adamsville 2 -- 1 -- 1 -- addison 3 -- 1 -- 1 -- akron Thanks For Any Help. Hi there, I have this query, which outputs the records properly in PhpMyAdmin. But when using this query inside my PHP code, the result is incomplete. Instead of retrieving the expected 15 results, I only get the first one twice. Also, I don't get any errors. Code: (php) [Select] <?php $get_files = "SELECT `id` FROM `files` WHERE `category` = 'W';"; include_once('connection.inc.php'); $con = mysql_connect(MYSQL_SERVER, MYSQL_USERNAME, MYSQL_PASSWORD) or die ("Connection failed: " . mysql_error()); $dbname = "myDB"; $select_db = mysql_select_db($dbname, $con) or die ("Can't open database: " . mysql_error()); $result_files = mysql_query($get_files) or die ("Query failed: " . mysql_error()); $files = mysql_fetch_array($result_files); $W_files = "'".implode("', '", $files)."'"; // echo $W_files; outputs the first number twice, instead of the expected 15 different ones $get_checklist = "SELECT * FROM `checklist` WHERE `id` IN ($W_files) ORDER BY `id` ASC;"; $result_checklist = mysql_query($get_checklist) or die ("Query failed: " . mysql_error()); // .... code for displaying ?> Hello all, I am using the Lynda.com PHP course to further learn php. On one of the examples, they are using a SQL Query that works for them but is erroring out for me. It is the query below for the $page_set variable, particularly the use of the WHERE subject_id = {$subject["id"]}". Code: [Select] <?php require_once ('includes/conf.php'); ?> <?php include ('includes/header.php'); ?> <?php require_once ('includes/functions.php'); ?> <table id="stucture"> <tr> <td id="navigation"> <ul class="subjects"> <?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>"; } $page_set = mysql_query("SELECT * FROM pages WHERE subject_id = {$subject["id"]}"); // This appears to be the problem area, unless it is fooling me. It works with this exact statement in their code. if (!$page_set) { die ("Database query failed: " . mysql_error()); } echo "<ul class='pages'>"; while ($page = mysql_fetch_array($page_set)) { echo "<li>{$page['menu_name']}</li>"; } echo "</ul>"; ?> </ul> </td> <td id="page"><h2>Content Area</h2> </td> </tr> </table> <?php require ('includes/footer.php'); ?> The error listed is this: Database query failed: 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 Thanks, you all are the best! BtD |
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