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PHP - Multiple Rows Query And Displaying Data
Ok so I am pretty good with mysql and performing queries to get the data that I need. However, I have a question for any of you gurus out there that may be able to help me with an issue that I always run into with certain types of queries. Say I have 2 tables like this:
Galleries id name 1 My Photos Gallery Photos id gallery_id photo 1 1 photo1.jpg 2 1 photo2.jpg 3 1 photo3.jpg Now usually when I pull this info from the database I have to do 2 separate queries in order to get the data and then link them so I would do something like. <?php //make the gallery query $query = "SELECT * FROM galleries"; $results = mysql_query($query); //setup an array for the galleries data $gallery_data = array(); //loop through the galleries for($i=0;$i<mysql_num_rows($results);$i++){ //add the gallery info to the gallery data array $gallery_data[$i] = mysql_fetch_array($results); //make the photos query with the gallery id $photo_query = "SELECT * FROM gallery_photos WHERE gallery_id='".mysql_real_escape_string($gallery_data[$i]['id'])."'"; $photo_results = mysql_query($photo_query); //setup the photos array $photo_data = array(); //put the photos in the array for($n=0;$n<mysql_num_rows($photo_results);$n++){ $photo_data[$n] = mysql_fetch_array($photo_results); } //add the photos to the gallery array $gallery_data[$i]['photos'] = $photo_data; } //now to display it is like this if(is_array($gallery_data)){ foreach($gallery_data as $gallery){ echo $gallery['name']; //show the photos if(is_array($gallery['photos'])){ foreach($gallery['photos'] as $photo){ echo $photo['photo']; } } } } ?> So my question is there a way to get all of this data at one time. I know how to do multiple queries in one and to do joins but they can only return one row as far as I know of. The only other way that I know how to do this is by ordering them by name and selecting them directly from the photos table and then just getting the gallery name like this: SELECT *,(SELECT name FROM galleries WHERE id=gallery_photos.id) AS gallery_name FROM gallery_photos ORDER BY gallery_id Then I could just do one loop and see if the name has changed and if so to output the new name. But I would like to know if there is a way to get a second set of results as an array from a query so I could just select the galleries and photos all in one query. Any help is appreciated. Similar TutorialsThis topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=351561.0 So here is what I a trying to do: Code: [Select] <?php /** * @author * @copyright 2010 */ $res = mysql_query("SELECT * FROM `planetsTable` WHERE `section` = '1' LIMIT 9 ASC"); while($row = mysql_fetch_array($res)){ ?> <li> <span id="sun">Solar System Star</span> <p>The Sun is a star, a hot ball of glowing gases at the heart of our solar system. Its influence extends far beyond the orbits of distant Neptune and Pluto. Without the Sun's intense energy and heat, there would be no life on Earth. And though it is special to us, there are billions of stars like our Sun scattered across the Milky Way galaxy.</p> </li> <li> <span id="mercury"><?php echo $row[0]['name']; ?></span> <p>Sun-scorched Mercury is only slightly larger than Earth's Moon. Like the Moon, Mercury has very little atmosphere to stop impacts and it is covered with craters. Mercury's dayside is super heated by the Sun, but at night temperatures drop hundreds of degrees below freezing. Ice may even exist in craters. Mercury's egg-shaped orbit takes it around the Sun every 88 days.</p> </li> <li> <span id="venus"><?php echo $row[1]['name']; ?></span> <p>Venus is a dim world of intense heat and volcanic activity. Similar in structure and size to Earth, Venus' thick, toxic atmosphere traps heat in a runaway 'greenhouse effect.' The scorched world has temperatures hot enough to melt lead. Glimpses below the clouds reveal volcanoes and deformed mountains. Venus spins slowly in the opposite direction of most planets.</p> </li> <li> <span id="earth"><?php echo $row[3]['name']; ?></span> <p>Earth is an ocean planet. Our home world's abundance of water - and life - makes it unique in our solar system. Other planets, plus a few moons, have ice, atmospheres, seasons and even weather, but only on Earth does the whole complicated mix come together in a way that encourages life - and lots of it.</p> </li> <li> <span id="mars"><?php echo $row[4]['name']; ?></span> <p>Though details of Mars' surface are difficult to see from Earth, telescope observations show seasonally changing features and white patches at the poles. For decades, people speculated that bright and dark areas on Mars were patches of vegetation, that Mars could be a likely place for life-forms, and that water might exist in the polar caps. When the Mariner 4 spacecraft flew by Mars in 1965, many were shocked to see photographs of a bleak, cratered surface. Mars seemed to be a dead planet. Later missions, however, have shown that Mars is a complex member of the solar system and holds many mysteries yet to be solved.</p> </li> <li> <span id="jupiter"><?php echo $row[5]['name']; ?></span> <p>The most massive planet in our solar system, with four large moons and many smaller moons, Jupiter forms a kind of miniature solar system. Jupiter resembles a star in composition. In fact, if it had been about 80 times more massive, it would have become a star rather than a planet.</p> </li> <li> <span id="saturn">Saturn</span> <p>Saturn was the most distant of the five planets known to the ancients. Like Jupiter, Saturn is made mostly of hydrogen and helium. Its volume is 755 times greater than that of Earth. Winds in the upper atmosphere reach 500 meters (1,600 feet) per second in the equatorial region. These super-fast winds, combined with heat rising from within the planet's interior, cause the yellow and gold bands visible in the atmosphere.</p> </li> <li> <span id="uranus">Uranus</span> <p>The first planet found with the aid of a telescope, Uranus was discovered in 1781 by astronomer William Herschel. The seventh planet from the Sun is so distant that it takes 84 years to complete one orbit.</p> </li> <li> <span id="neptune">Neptune</span> <p>Nearly 4.5 billion kilometers (2.8 billion miles) from the Sun, Neptune orbits the Sun once every 165 years. It is invisible to the naked eye because of its extreme distance from Earth. Interestingly, the unusual elliptical orbit of the dwarf planet Pluto brings Pluto inside Neptune's orbit for a 20-year period out of every 248 Earth years</p> </li> <li> <span id="pluto">Pluto</span> <p>Tiny, cold and incredibly distant, Pluto was discovered in 1930 and long considered to be the ninth planet. But after the discoveries of similar intriguing worlds even farther out, Pluto was reclassified as a dwarf planet. This new class of worlds may offer some of the best evidence of the origins of our solar system.</p> </li> </ul> <?} ?> What I need to know is if this is possible and if so how would I got about doing this? Code: [Select] <?php $row['0']['name']; // Main Issue. ?> McBryver Hi, after banging my head against the wall for a while thinking this would be a simple task, I'm discovering that this is more complicated than I thought. Basically what I have is a link table linking together source_id and subject_id. For each subject there are multiple sources associated with each. I had created a basic listing of sources by subject... no problem. I now need a way of having a form to create an ordered list in a user-specified way. In other words, I can currently order by id or alphabetically (subject name lives on a different table), but I need the option of choosing the order in which they display. I added another row to this table called order_by. No problem again, and I can manage all of this in the database, however I want to create a basic form where I can view sources by subject and then enter a number that I can use for sorting. I started off looping through each of the entries and the database (with a where), and creating a foreach like so (with the subject_id being grabbed via GET from the URL on a previous script) Code: [Select] while($row = mysqli_fetch_array($rs)) { //update row order if (isset($_POST['submit'])) { //get variables, and assign order $subject_id = $_GET['subject_id']; $order_by = $_POST['order_by']; $source_id = $row['source_id']; //echo 'Order by entered as ' . $order_by . '<br />'; foreach ($_POST['order_by'] as $order_by) { $qorder = "UPDATE source_subject set order_by = '$order_by' WHERE source_id = '$source_id' AND subject_id = '$subject_id'"; mysqli_query($dbc, $qorder) or die ('could not insert order'); // echo $subject_id . ', ' . $order_by . ', ' . $source_id; // echo '<br />'; } } else { $subject_id = $_GET['subject_id']; $order_by = $row['order_by']; $source_id = $row['source_id']; } And have the line in the form like so: Code: [Select] echo '<input type="text" id="order_by" name="order_by[]" size="1" value="'. $order_by .'"/> (yes I know I didn't escape the input field... it's all stored in an htaccess protected directory; I will clean it up later once I get it to work) This, of course, results in every source_id getting the same "order_by" no matter what I put into each field. I'm thinking that I need to do some sort of foreach where I go through foreach source_id and have it update the "order_by" field for each one, but I must admit I'm not sure how to go about this (the flaws of being self-taught I suppose; I don't have anyone to go to on this). I'm hoping someone here can help? Thanks a ton in advance On my web page, there is a variable called $submission. I would like to display exactly 11 rows from the query below: the row where $submission equals $row["title"], 5 rows above it, and 5 rows below it. All ranked by points descending. How can I do this? Code: [Select] $sqlStr = "SELECT title, points, submissionid FROM submission ORDER BY points DESC"; $result = mysql_query($sqlStr); $arr = array(); $count=1; echo "<table class=\"samplesrec\">"; while ($row = mysql_fetch_array($result)) { echo '<tr >'; echo '<td>'.$count++.'.</td>'; echo '<td class="sitename1">'.$row["title"].'</td>'; echo '<td class="sitename2"><div class="pointlink2">'.number_format($row["points"]).'</div></td>'; echo '</tr>'; } echo "</table>"; Hey, I was wondering if there is a way to pull multiple rows at once using a list of unique identifiers. For example, I want to pull the rows with the IDs of 4,13,91 and 252 I know the WHERE part of this query is incorrect, but I'm putting it to hopefully help you guys understand what I'm looking for. $result = mysql_query("SELECT * FROM $table WHERE id='4' OR '13' OR '91' OR '252'"); while($row = mysql_fetch_array($result)) { echo($row['name']); } Or is the best way simply to do it one query at a time without a while statement? There could be as many as a few dozen records being pulled per page. 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. 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'm trying to write a php page that displays data from a JOIN query for a specific ID table view brandinfo ID, brand, discounttype 1, antioni, no discount brandproducts brandID, producttype, price 1, Tshirt, 20.00 1, Pants, 30.00 1, Shoe, 40.00 the returned result is 1 antioni, no discount, Tshirt, 20.00, 2 antioni, no discount, Pants, 30.00 3 antioni, no discount, Shoe 40.00 The way I want the page to be displayed is ------------------ Antioni (at the top) Table 1. Tshirt 20.00 2. Pants 30.00 3. Shoe 40.00 no discount (at the bottom) ---------------------------- How should I construct the PHP page from the result since they're retrieved as rows? Hi Guys Just need some advice to go in the right direction. I'm working on a csv upload script (part of a bigger thing i'm building), so i read in the csv to a multipdimensional array and then build a query that inputs all rows in one query - i read this is the most efficient way to import multiple rows of data at once(rather than multiple insert statements). Just for illustration here's the code i use to build the query so you understand what i'm on about: Code: [Select] $sql = "INSERT INTO teams (company, teamname, teamnum, amountraised, country, president) VALUES "; // $rows is a count of the rows in the csv for($i=1; $i<$rows; $i++){ $sql.="('{$myarray[$i][0]}','{$myarray[$i][1]}','{$myarray[$i][2]}','{$myarray[$i][3]}','{$myarray[$i][4]}','{$myarray[$i][5]}')"; echo $i . "<br/>"; if($i >= 1 && $i < $rows - 1) { $sql.= ","; } } Anyway, the issue is that one of the fields("teamnum") needs to be unique - so i've set this as unique on the table in mysql. But when i run my query it doesn't import anything if one of the records isn't unique. What i really want is for it to import the ones it can and catch the ones it cant import to present to the user. So my question is - to acheive the above would i need to rewrite the query so that it inserts each row one at a time, instead of all together? Or can someone point me in the right direction for a better solution? Probably something very simple i've missed i am sure... Thanks chaps! This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=344772.0 I am trying to array data to database against same id.
Here is code.
form.php
<form name="users" method="post" action="order_submit.php" enctype="multipart/form-data" onSubmit="return validate();" id="inv_form">
<div class="formSep">
<select name="company" onChange="showSubcat(this);">
<option value="">Company</option>
<?php $s1 = mysql_query("select * from leads where lead_customer='Lead' ") or die (mysql_error());
while($s2 = mysql_fetch_array($s1))
{ ?>
<option value="<?php echo $s2['id']; ?>"><?php echo $s2['company']; ?></option>
<?php
} ?>
</select>
</div>
<div class="formSep">
<table class="table invE_table">
<thead>
<tr>
<th></th>
<th>Item</th>
<th>Unit</th>
<th>Unit Cost ($)</th>
<th>Qty</th>
<th>Tax (%)</th>
<th>Total ($)</th>
</tr>
</thead>
<tbody>
<tr class="inv_row">
<td class="inv_clone_row"><i class="icon-plus inv_clone_btn"></i></td>
<td><input type="text" class="span12" name="invE_item[]" /></td>
<td><input type="text" class="span12" name="invE_description[]" /></td>
<td><input type="text" class="span12 jQinv_item_unit" name="invE_unit_cost[]" /></td>
<td><input type="text" class="span12 jQinv_item_qty" name="invE_qty[]" /></td>
<td><input type="text" class="span12 jQinv_item_tax" name="invE_tax[]" /></td>
<td><input type="text" readonly class="span12 jQinv_item_total" name="invE_total[]" /></td>
</tr>
<tr class="last_row">
<td colspan="5"> </td>
<td colspan="2">
<p class="clearfix">Subtotal: <span class="invE_subtotal">$<span>0.00</span></span></p>
<p>Tax: <span class="invE_tax">$<span>0.00</span></span></p>
<p>Discount: <span class="invE_discount">$<span>0.00</span></span></p>
<p><strong>Balance: <span class="invE_balance">$<span>0.00</span></span></strong></p>
</td>
</tr>
</tbody>
</table>
</div>
Here invE_item[], invE_description[], invE_unit_cost[].... are the array , i mean dynamically one can add as many as items and its details.
In my order_submit.php page
<?php
error_reporting(0);
include("connect.php");
include("admin_auth.php");
if(isset($_POST['save']))
{
$company = $_POST['company'];
$contact_person = $_POST['contact_person'];
$billing = $_POST['billing_address'];
$shipping = $_POST['shipping_address'];
$reference = $_POST['reference'];
$t_c = $_POST['t_c'];
$payment = $_POST['payment'];
$ship_in = $_POST['ship_inst'];
$validity = $_POST['validity'];
$currency = $_POST['currency'];
$order_for = $_POST['order_for'];
$assigned_to = $_POST['assigned_to'];
$item = $_POST['invE_item'];
$unit = $_POST['invE_description'];
$price = $_POST['invE_unit_cost'];
$qty= $_POST['invE_qty'];
$tax = $_POST['invE_tax'];
$total = $_POST['invE_total'];
$sql = mysql_query("insert into orders (order_id, company_id, contact_person, billing_address, shipping_address, reference, t_c, payment, shipping_inst, validity, order_for, currency, assigned_to, last_modified, order_quote) values ('', ".$company.", '".$contact_person."', '".$billing."', '".$shipping."', '".$reference."', '".$t_c."', '".$payment."', '".$ship_in."', ".$validity.", '".$order_for."', '".$currency."', '".$assigned_to."', NOW(), 'Order')");
$last_id = mysql_insert_id();
$msql = "insert into order_line_items (id, order_id, company_id, item, unit, unit_cost, quantity, tax, total) values ('', ".$last_id.", ".$company.", '".$item."', '".$unit."', ".$price.", ".$qty.", ".$tax.", ".$total.")";
$l1 = mysql_query($msql) or die (mysql_error());
}
I want to insert each item in different row with $last_id , as in the attached image .
Please somebody help me in this
Attached Files
db.PNG 11.01KB
4 downloadsThis topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=327017.0 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 stored supplier information in a table and same form user will select multiple products for that supplier and that will be stored in a seperate table against that supplier id. Now when i would like to display all supplier information with muliple products in a single line. Now all the information will be displayed as many time as the products are. How to display all the products in a same line.
Now my display is like this
if(isset($_POST['submit']) && ($_POST['vendor']!='') && ($_POST['item']!=''))
{
$sql="SELECT supplier.id AS sid, supplier.name AS SNAME, supplier.category, supplier.website, supplier.email, supplier.phone, supplier.vat, supplier.pan, supplier_location.id, supplier_location.supplier_id, supplier_location.location, supplier_products.id, supplier_products.supplier_id, supplier_products.product_id, location.loc_id, location.name AS locname, products.product_id, products.name AS pname FROM supplier INNER JOIN supplier_location ON supplier.id = supplier_location.supplier_id INNER JOIN supplier_products ON supplier.id=supplier_products.supplier_id INNER JOIN location ON supplier_location.location = location.loc_id INNER JOIN products ON supplier_products.product_id=products.product_id WHERE supplier.id=".$sup." AND supplier_products.product_id=".$product;
$sql1 = mysql_query($sql) or die(mysql_error());
<table>
<thead><tr>
<th>Vendor ID</th>
<th>Vendor</th>
<th>Category</th>
<th>Website</th>
<th>Email</th>
<th>Phone</th>
<th>Products</th>
<th>Locations</th>
<th>VAT</th>
<th>PAN</th>
</tr>
</thead>
<tbody>
<tr>
<?php
while($row = mysql_fetch_array($sql1))
{
?>
<td><?php echo $row['sid'] ?></td>
<td><?php echo $row['SNAME'] ?></td>
<td><?php echo $row['category'] ?></td>
<td><?php echo $row['website'] ?></td>
<td><?php echo $row['email'] ?></td>
<td><?php echo $row['phone'] ?></td>
<td><?php echo $row['iname']; ?></td>
<td><?php echo $row['locname']; ?></td>
<td><?php echo $row['vat'] ?></td>
<td><?php echo $row['pan'] ?></td>
</tr>
<?php
}
?>
</tbody></table>
}
Ok I'm trying to insert multiple rows by using a while loop but having problems. At the same time, need to open a new mysql connection while running the insert query, close it then open the previous mysql connection. I managed to insert multiple queries before using a loop, but for this time, the loop does not work? I think it is because I am opening another connection... yh that would make sense actually? Here is the code: $users = safe_query("SELECT * FROM ".PREFIX."user"); while($dp=mysql_fetch_array($users)) { $username = $dp['username']; $nickname = $dp['nickname']; $pwd1 = $dp['password']; $mail = $dp['email']; $ip_add = $dp['ip']; $wsID = $dp['userID']; $registerdate = $dp['registerdate']; $birthday = $dp['birthday']; $avatar = $dp['avatar']; $icq = $dp['icq']; $hp = $dp['homepage']; echo $username." = 1 username only? :("; // ----- Forum Bridge user insert ----- $result = safe_query("SELECT * FROM `".PREFIX."forum`"); $ds=mysql_fetch_array($result); $forum_prefix = $ds['prefix']; define(PREFIX_FORUM, $forum_prefix); define(FORUMREG_DEBUG, 0); $con = mysql_connect($ds['host'], $ds['user'], $ds['password']) or system_error('ERROR: Can not connect to MySQL-Server'); $condb = mysql_select_db($ds['db'], $con) or system_error('ERROR: Can not connect to database "'.$ds['db'].'"'); include('../_phpbb_func.php'); $phpbbpass = phpbb_hash($pwd1); $phpbbmailhash = phpbb_email_hash($mail); $phpbbsalt = unique_id(); safe_query("INSERT INTO `".PREFIX_FORUM."users` (`username`, `username_clean`, `user_password`, `user_pass_convert`, `user_email`, `user_email_hash`, `group_id`, `user_type`, `user_regdate`, `user_passchg`, `user_lastvisit`, `user_lastmark`, `user_new`, `user_options`, `user_form_salt`, `user_ip`, `wsID`, `user_birthday`, `user_avatar`, `user_icq`, `user_website`) VALUES ('$username', '$username', '$phpbbpass', '0', '$mail', '$phpbbmailhash', '2', '0', '$registerdate', '$registerdate', '$registerdate', '$registerdate', '1', '230271', '$phpbbsalt', '$ip_add', '$wsID', '$birthday', '$avatar', '$icq', '$hp')"); if (FORUMREG_DEBUG == '1') { echo "<p><b>-- DEBUG -- : User added: ".mysql_affected_rows($con)."<br />"; echo "<br />-- DEBUG -- : Query used: ".end($_mysql_querys)."</b></p><br />"; $result = safe_query("SELECT user_id from ".PREFIX_FORUM."users WHERE username = '$username'"); $phpbbid = mysql_fetch_row($result); safe_query("INSERT INTO `".PREFIX_FORUM."user_group` (`group_id`, `user_id`, `group_leader`, `user_pending`) VALUES ('2', '$phpbbid[0]', '0', '0')"); safe_query("INSERT INTO `".PREFIX_FORUM."user_group` (`group_id`, `user_id`, `group_leader`, `user_pending`) VALUES ('7', '$phpbbid[0]', '0', '0')"); mysql_close($con); } include('../_mysql.php'); mysql_connect($host, $user, $pwd) or system_error('ERROR: Can not connect to MySQL-Server'); mysql_select_db($db) or system_error('ERROR: Can not connect to database "'.$db.'"'); } So I need to be able to insert these rows using the while loop.. how can I do this? I really appreciate any help. I want to echo out a few subcategories per main category. It should be a maximum of 20 entries per column At this stage I have the following... It echo's the results out in 4 columns and not what I want. Any suggestions? $query = "SELECT adsubcat.name AS subname, adsubcat.linkname AS sublink, adcat.clinkname AS clink FROM adsubcat JOIN adcat ON adcat.id=adsubcat.catid WHERE adsubcat.catid='$catid' ORDER BY adsubcat.name ASC"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)) { $count = 0; print '<table style="text-align:left;margin-left:0px;width:943px;">'."\n"; while ($row = mysql_fetch_assoc($result)){ $gal = '<a href="'.$root.'/'.$row['clink'].'/'.$row['sublink'].'">'.$row['subname'].'</a>'; if ($count == 0){ print '<tr>'; } ++$count; print '<td>'.$gal.'</td>'; if ($count == 4){ $count = 0; print'</tr>'."\n"; } } print'</table>'."\n"; } Alright, I need to find a way to display the top ten duplicated rows for a table in my database. I have tried multiple methods but I have failed. Can you please assist me with this problem. I currently run a query in phpmyadmin to get the result, but i cant figure out how to properly code the php script. Hi I'm not at all good at PHP and haven't really played with it for some years, so bare with me if code examples are old I'm looking for a simple solution to count rows (MySQL), without using multiple queries. The below code is quite simple, counting rows where 'Department' is 'Sales': $sql ="SELECT * FROM DB WHERE Department ='Sales'"; $result=mysqli_query($con,$sql); $rowcount=mysqli_num_rows($result); But what If I need to count the rows for the different departments?: $sql ="SELECT * FROM DB WHERE Department ='Sales'"; $result=mysqli_query($con,$sql); $rowcount=mysqli_num_rows($result); $sql ="SELECT * FROM DB WHERE Department ='Support'"; $result=mysqli_query($con,$sql); $rowcount=mysqli_num_rows($result); $sql ="SELECT * FROM DB WHERE Department ='Online'"; $result=mysqli_query($con,$sql); $rowcount=mysqli_num_rows($result); There must be a better / simpler way to this with a single query? Hope for someone to show me the light
PS. Once this is resolved, I'll probably need further help, to accomplish my final result. I have a query that pulls 1 field with 20 rows of data. I need to assign each row to a different variable so that I can then display them in different locations on a page. I cannot make each row of data a different field because of other constraints. My data is very well normalized. I am using mysqli so something like the old mysql_result would be lovely! How can this be done without hitting my database 20 times? Thanks for the help. I have 3 set of pages. The first html page only contains a search textbox and a submit button which then call up the second page search.php which populate html table with search result. The third page is update.php which allows to update the records populated by search.php. The problem is that it when it update the records it automatically updates all rows in the table making all records look the same. I do not know where I have made a mistake please help. When I use phpadmin to update the records no problem as I can specifically input the cf_id number directly. Search.php code Code: [Select] <?php if(isset($_POST['submit'])){ if(isset($_GET['go'])){ if(preg_match("^/[A-Za-z]+/", $_POST['name'])){ $name=$_POST['name']; } } else{ echo "<p>Please enter a search query</p>"; } } //connect to the database $con = mysql_connect("localhost","dbusrn","dbpwd"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("mydb", $con); $name = $_POST['name']; $result = mysql_query ("select * from Customer_Registration where Firstname like '%$name%' or lastname like '%$name%' "); $row = mysql_fetch_row($result); $cf_uid = $row[0]; $Firstname = $row[6]; $lastname = $row[7]; $Address = $row[8]; $Postcode = $row[9]; $Phone = $row[10]; $Email = $row[11]; $Customer_Type = $row[12]; $Computer_type = $row[13]; $Computer_maker = $row[14]; $Model = $row[15]; $OS = $row[16]; $Appointment_date = $row[17]; $Problem = $row[18]; $Solution = $row[19]; $Comment = $row[20]; ?> <form action="updatecus.php" method="post"> <table width="100%" border="2" cellspacing="0" cellpadding="8"> <tr><td width="45%" class="FormText">Customer ID:</td> <td width="55%"><?php echo $row[0];?></td></tr> <tr><td width="45%" class="FormText">First name:</td> <td width="55%"><input name="Firstname" type="text" value="<?php echo $Firstname;?>"?> </td></tr> <tr><td width="45%" class="FormText">Last name:</td> <td width="55%"><input name="lastname" type="text" value="<?php echo $lastname;?>"?> </td></tr> <tr><td width="45%" class="FormText">Address:</td> <td width="55%"><input name="Address" type="text" value="<?php echo $Address; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Postcode:</td> <td width="55%"><input name="Postcode" type="text" value="<?php echo $Postcode; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Phone:</td> <td width="55%"><input name="Phone" type="text" value="<?php echo $Phone; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Email:</td> <td width="55%"><input name="Email" type="text" value="<?php echo $Email; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Customer:</td> <td width="55%"><input name="Customer_Type" type="text" value="<?php echo $Customer_Type; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Computer :</td> <td width="55%"><input name="Computer_type" type="text" value="<?php echo $Computer_type; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Manufactural:</td> <td width="55%"><input name="Computer_maker" type="text" value="<?php echo $Computer_maker; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Model:</td> <td width="55%"><input name="Model" type="text" value="<?php echo $Model; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Operating System:</td> <td width="55%"><input name="OS" type="text" value="<?php echo $OS; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Appointment:</td> <td width="55%"><input name="Appointment_date" type="text" value="<?php echo $Appointment_date; ?>"?> </td></tr> <tr><td width="45%" class="FormText">Problem:</td> <td width="55%"><textarea name="Problem" rows="10" cols="20" ><?php echo $Problem; ?></textarea></td></tr> <tr><td width="45%" class="FormText">Solution:</td> <td width="55%"><textarea name="Solution" rows="10" cols="20" ><?php echo $Solution; ?></textarea></td></tr> <tr><td width="45%" class="FormText">Comment:</td> <td width="55%"><textarea name="Comment" rows="10" cols="20" ><?php echo $Comment; ?></textarea></td></tr> <td width="55%"><input name="submit" value="submit" type="submit" <br> <input type="Submit" value="Cancel"</br></td> </form> <?php ?> update.php code Code: [Select] <?php //connect to the database $con = mysql_connect("localhost","dbusrn","dbpwd"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("mydb", $con); //if (isset($_POST['submit'])) { $cf_id = $_POST['cf_id']; $Firstname = $_POST['Firstname']; $lastname = $_POST['lastname']; $Address = $_POST['Address']; $Postcode = $_POST['Postcode']; $Phone = $_POST['Phone']; $Email = $_POST['Email']; $Customer_Type = $_POST['Customer_Type']; $Computer_type = $_POST['Computer_type']; $Computer_maker = $_POST['Computer_maker']; $Model = $_POST['Model']; $OS = $_POST['OS']; $Appointment_date = $_POST['Appointment_date']; $Problem = $_POST['Problem']; $Solution = $_POST['Solution']; $Comment = $_POST['Comment']; if (isset($_POST['submit'])) { $query ="UPDATE Customer_Registration SET Firstname='$Firstname',lastname='$lastname',Address='$Address',Postcode='$Postcode',Phone='$Phone',Email='$Email',Customer_Type='$Customer_Type',Computer_type='$Computer_type',Computer_maker='$Computer_maker',Model='$Model',OS='$OS', Appointment_date='$Appointment_date',Problem='$Problem',Solution='$Solution',Comment='$Comment' WHERE cf_id=cf_id"or die (mysql_error()); echo $query; mysql_query($query) or die(mysql_error()); //mysql_close($con); echo "<p>Congrats Record Updated</p>"; } ?> Thank you in advance for your help |
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