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PHP - Querying And Displaying Data Using Php And Mysql Using Checkboxes?
I have a table in a mysql database with 5 columns, id, Name, Wifi, Bluetooth, GPS, with rows that are for example 1, Galaxy S2, yes, yes, yes. So basically i want to build a form that has check boxes (3 checkboxes for wifi bluetooth and GPS respectively) that once selected will query the database depending on which check boxes are selected. I have made the form but need to know what to put in the filter.php to make the results be displayed accordingly. Ideally if anyone knows how i would want it so the results were shown on the same page which i believe u need to use ajax to happen but any help on how to show the results will be greatful.
the code for the form i have made is as follows: Code: [Select] <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <title>test</title> </head> <body> <form action="filter.php" method="post"> <INPUT TYPE=CHECKBOX NAME="option[]" VALUE="Wifi" id="r1">Wifi <INPUT TYPE=CHECKBOX NAME="option[]" VALUE="Bluetooth" id="b1">Bluetooth <INPUT TYPE=CHECKBOX NAME="option[]" VALUE="GPS" id="g1">GPS <input type="submit" name="formSubmit" value="Submit" /> </form> </body> </html> im not sure on how to make the filter.php page but i do have this code for displaying the all the data but i want to kno how to make it only display the data from the choices on the checkboxes Code: [Select] <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <title>pls work</title> </head> <body> <?php function h($s) { echo htmlspecialchars($s); } mysql_connect("localhost", "root", "") or die (mysql_error()); mysql_select_db("project") or die (mysql_error()); $result= mysql_query('SELECT * FROM test') or die('Error, query failed'); ?> <?php if (mysql_num_rows($result)==0) { ?> Database is empty <br/> <?php } else { ?> <table> <tr> <th></th> <th>Name</th> <th>Wifi</th> <th>Bluetooth</th> <th>GPS</th> </tr> <?php while ($row= mysql_fetch_assoc($result)) { ?> <tr> <td> <a href="uploaded-images/<?php h($row['Name']); ?>.jpg"> <img src="uploaded-images/<?php h($row['Name']); ?>.jpg" alt="test"/> </a> </td> <td><a href="textonly.html"><?php h($row['Name']); ?></a></td> <td><?php h($row['Wifi']); ?></td> <td><?php h($row['Bluetooth']); ?></td> <td><?php h($row['GPS']); ?></td> </tr> <?php } ?> </table> <?php } ?> </body> </html> Similar TutorialsHey guys have been trying to get this script to work for a while now, i am new to php and mysql so i am sure i am missing something simple. I have DB setup and need to pull data based on the key item code and get the following I want to get the fields item_code description allergy_statement useable_units region_availability order_lead_time ingredients for item_code 12-100 LITERALLY 12-100, no range, but like i said before i am really new to php and mysql. I have 1187 items that when a user clicks a link in search results it takes them to the product details page for that item code All that data is in my database just can't figure out how to get it out of the database. Is this even the right script to achieve that result. here is the code to get the data from database Code: [Select] <?php require_once('includes/mysql_connect_nfacts_ro.php'); $query = "SELECT item_code, description, allergy_statement, useable_units, region_availability, order_lead_time, ingredients " . "FROM products " . "WHERE item_code = '12-100' "; $resuts = mysql_query($query) or die(mysql_error()); ?> And need to display the data like so : Code: [Select] <td width="715" align="center" valign="top"> <h1>Product Details</h1> <h3>DISPLAY description HERE</h3> <table width="420" border="0"> <td class="ingreg"> </td> </table> <h5>Item Number</h5> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY ITEM_CODE HERE</td> </tr> </table> <h3>Ingredients:</h3> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY INGREDIENTS HERE</td> </tr> </table> <h4>Allergy Statement:</h4> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY Allergy Statement HERE</td> </tr> </table> <h4>Useable Units Per Package:</h4> <table width="420" border="0"> <tr> <td class="ingreg">DISPLAY Useable Units Per Package HERE</td> </tr> </table> <h4>Region Availability: </h4> <table width="420" border="0"> <tr> <td class="ingreg">&DISPLAY ITEM_CODE HERE</td> </tr> </table> <h4>Order Lead Time:</h4> <table width="420" border="0"> <tr> <td class="ingreg">&DISPLAY order lead time HERE</td> </tr> </table> <p> </p> <div align="right"></div></td> </tr> </table> how do i get data in database to display where i need it to? Can any one shine some light on this So i pull some records out of a mysql table and i want to display them in 5 even columns. I'm not entirely sure how to do the math & logic to accomplish this. The pull is simple $qry = "SELECT DIST_PART_NUM FROM $tablename"; $sql = mysql_query($qry) or die(mysql_error()); while($res = mysql_fetch_assoc($sql)) { // CREATE 5 even columns here. } so let's say i just retrieved 5,000 part numbers, i'd like to display then in a table of 5 columns with 1000 records per column. This is easy math, but i need the script to automatically figure out the #'s. Also the tricky part is that i dont want to display the part numbers like so 11111 22222 33333 44444 55555 66666 77777 88888 99999 00000 but rather 11111 44444 77777 22222 55555 88888 33333 66666 99999 00000 the remainder if there is one can go in the last column or whatever is easier. I'd tried googling this, but it's not easy to phrase what i'm looking for. Thanks for the help. PS: I'm not looking to copy and paste code, if possible please explain your way so that i can learn the logic. Ok... I need some help - I want to show a players balance in a game beside there name (Balance is in mysql database)I can do that but... - I also want to show if there online or offline at the same time( This is stored in a different database) I have the code which says whether they are online or offline <?PHP // Conect to the Mysql Server $connect = mysql_connect("localhost","scswccla_bukkit","********"); //connect to the database mysql_select_db("scswccla_bukkit"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); //title echo "<font size='100'>NovaCraft Users</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='black'>|Online|<br><font color='green'>$users</font></font><br>"; endwhile; //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 0"); //title echo "<font color='black'>|Offline|</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='red'>$users</font><br>"; endwhile; ?> Here is the page: www.scswc.com/Offline_Users.php displaying that But I want to Create something like this: Nocvacraft Players |Online| Name:Player Balance:$20 |Offline| Name:Player Balance:$15 Here is what I have tried: <?PHP // Conect to the Mysql Server $connect = mysql_connect("localhost","scswccla_bukkit","**********"); //connect to the database mysql_select_db("scswccla_bukkit"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); $query2 =mysql_query("SELECT * FROM iBalances WHERE player = $users"); //title echo "<font size='100'>NovaCraft Users</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query) $rows2 = mysql_fetch_array($query2)): $users = $rows['name']; $balance = $rows2['balance']; echo "<font color='black'>|Online|<br><font color='green'>Name:$usersBalance:$balance</font></font><br>"; endwhile; //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 0"); //title echo "<font color='black'>|Offline|</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='red'>$users</font><br>"; endwhile; ?> I know I am trying to use a variable before it is been set - but if I don't how I have tried this as well... <?PHP // Conect to the Mysql Server $connect = mysql_connect("localhost","scswccla_bukkit","**********"); //connect to the database mysql_select_db("scswccla_bukkit"); //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 1"); //title echo "<font size='100'>NovaCraft Users</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='black'>|Online|<br><font color='green'>Name:$users</font></font><br>"; endwhile; //query the database $query = mysql_query("SELECT * FROM users_online WHERE online = 0"); //title echo "<font color='black'>|Offline|</font><br>"; // fetch the results / convert into an array WHILE($rows = mysql_fetch_array($query)): $users = $rows['name']; echo "<font color='red'>$users</font><br>"; endwhile; $query = mysql_query("SELECT * FROM iBalances WHERE player = $users"); WHILE($rows = mysql_fetch_array($query)): $balance = $rows['balance']; echo "<font color='red'>$users $balance</font><br>"; endwhile; // ?> Can you use variables in mysql_query()?Is that why it isn't working? This is my first php script so if I need to give you more information for you to help me just tell me Thanks Here is database pictures iBalances users_online Hi guys. I'm trying to build a games site for a friend. Currently, the front-end (HTML/CSS) of the site is done. Now, I want to make a way for him to easily add games to the site. Ideally, I'd like to make a database with the following columns: ID, Name, Category, Link, Thumbnail_Link. So, those would be the ID, name of the game, the category, a link to the game, and a link to the 50x50 thumbnail image respectively. Then, using PHP, I'd like to call the first x number (not sure what it will be yet, let's say 50) and make format it as a grid in the following way: There's the thumbnail image followed by the game name, and they're all a clickable link to the game URL. Is this possible? How would I go about doing this? I've already set up a database for a login module to the site, so each page has already opened a connection to the MySQL database. However, I've only ever done basic PHP for mail forms and am otherwise extremely new to it, and am especially new to MySQL. Could anyone walk me through how to do this or even give me a quick example script to work off of? Thanks, any of your time is greatly appreciated! I am querying my database to show the visit statistics for a particular week and it shows the number of visits for the countries, but does not display the country name.
I have proved that the MySQL works by going into phpMyAdmin and pasting the query into SQL query tab, replacing the POST with 1, for week 1.
I can't see why it is not displying the country.
Here is the code:
<?php
include('connect_visits.php');
doDB7();
$WVisit_data="SELECT WeekNo15.WNo, WeekNo15.WCom, Countries.Country, Countries.CID, ctryvisits15.CVisits
FROM ctryvisits15
LEFT JOIN Countries
ON ctryvisits15.country = Countries.CID
LEFT JOIN WeekNo15
ON ctryvisits15.WNo = WeekNo15.WNo
WHERE ctryvisits15.WNo = '{$_POST['WeekNo']}'
ORDER BY ctryvisits15.CVisits DESC";
$WVisit_data_res = mysqli_query($mysqli, $WVisit_data) or die(mysqli_error($mysqli));
$display_block ="
<table width=\"20%\" cellpadding=\"3\" cellspacing=\"1\" border=\"1\" BGCOLOR=\"white\" >
<tr>
<th>Country</th>
<th>Visits</>
</tr>";
while ($WV_info = mysqli_fetch_array($WVisit_data_res)){
$Ctry = $WV_info['country'];
$Visits = $WV_info['CVisits'];
//add to display
$display_block .="
<tr>
<td width=\"10%\" valign=\"top\">".$Ctry."<br/></td>
<td width=\"5%\" valign=\"top\">".$Visits."<br/></td>
";
}
mysqli_free_result($WVisit_data_res);
mysqli_close($mysqli);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<!--
Design by Free CSS Templates
http://www.freecsstemplates.org
Released for free under a Creative Commons Attribution 2.5 License
Name : Yosemite
Description: A two-column, fixed-width design with dark color scheme.
Version : 1.0
Released : 20091106
-->
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta name="keywords" content="" />
<meta name="description" content="" />
<meta http-equiv="content-type" content="text/html; charset=utf-8" />
<title>1066 Cards 4U - Stats for country</title>
<link href="style.css" rel="stylesheet" type="text/css" media="screen" />
</head>
<body>
<div id="wrapper">
<div id="menu">
<ul>
<li class="current_page_item"><a href="index.php">Home</a></li>
<li><a href="Links.html">Links</a></li>
<li><a href="Verse_Menu.html">Verses</a></li>
<li><a href="Techniques.html">Techniques</a></li>
<li><a href="blog.php">Blog</a></li>
<li><a href="Gallery.html">Gallery</a></li>
<li><a href="contact.html">Contact</a></li>
<li><a href="AboutUs.html">About Us</a></li>
<li><a href="stats1.html">Stats</a></li>
</ul>
</div><!-- end #menu -->
<div id="header">
<div id="logo">
<h1><a href="http://www.1066cards4u.co.uk">1066 Cards 4U</a></h1>
</div><!-- end #wrapper -->
</div><!-- end #header -->
<div id="page">
<div id="page-bgtop">
<div id="page-bgbtm">
<div id="content">
<h3>Statistics for Week Commencing <? echo $WkCom; ?> in 2015</h3>
<div id="table">
<?php echo $display_block; ?></div>
</div><!-- end #content -->
</body>
</html>
Can you help please?
Edited by rocky48, 07 January 2015 - 07:33 AM. Hi all, I'm trying to display a form based on a dropd own selection. The drop down is propagated from a database query. I have the drop down list displaying, but when I click submit, I can't get the result to display. Code: [Select] $sql="SELECT id, company_name FROM webprojects ORDER BY company_name ASC"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $id=$row["id"]; $company_name=$row["company_name"]; $options.="<OPTION VALUE=\"$id\">".$company_name; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>PHP Dynamic Drop Down Menu</title> </head> <body> <form id="form1" name="form1" method="post" action="selected.php"> <SELECT NAME=id> <OPTION VALUE=0>Choose <?=$options.="<OPTION VALUE=\"$id\">".$company_name.'</option>';?> </SELECT> <label> <input type="submit" name="submit" id="submit" value="Submit" /> </label> </form> </body> </html> Any help is greatly appreciated Hi There! Hopefully someone can help. I'm looking for some php script that allows me to display mysql table data in alternating row colors (eg 5 columns and 500 rows). I would also like to add pagination and limit number of rows displayed at a time to e.g. 50. If possible I'd like to be able to sort asc/desc using titles(links) at the top of the displayed data. Hopefully someone may have a script like this that also shows the mysql connection code (sample host, pass, user etc...) - a working php file so to speak that I can edit. Would really appreciate the help. If you could recommend some software or a wizard that could generate the php code for me I would really appreciate it too! THANKS! Hello everyone, I begin in everything web related but I have been programming for years. I tried to code something simple : small Mysql DB (works fine) and to begin a search bar to browse data. I adapted a code that I understood provided here : https://www.cloudways.com/blog/live-search-php-mysql-ajax/. Base principle is simple : as you type in your query, it will pass the text to script.js that will forward this request to ajax.php file. In the ajax.php, a javascript function named “fill()” will pass the fetched results. This function will also display the result(s) into “display” div in the “search.php” file. The problem is that when I type anything it displays, below the search bar, at the moment I type a character: Quote
'; //Fetching result from database. while ($Result = MySQLi_fetch_array($ExecQuery)) { ?> ")'>
instead of the actual answer from my database (no error in the browser console). I tested the SQL query + the user I provide and everything seems fine. Any clue what could be the root cause ? I strongly suspect a mistake in the code as I already corrected one (script.js instead of scripts.js) but I really cannot figure out where. Thanks in advance,
problematic code (ajax.php):
<?php
//Including Database configuration file.
include "db.php";
//Getting value of "search" variable from "script.js".
if (isset($_POST['search'])) {
//Search box value assigning to $Name variable.
$Name = $_POST['search'];
//Search query.
$Query = "SELECT Name FROM search WHERE Name LIKE '%$Name%' LIMIT 5";
//Query execution
$ExecQuery = MySQLi_query($con, $Query);
//Creating unordered list to display result.
echo '
<ul>
';
//Fetching result from database.
while ($Result = MySQLi_fetch_array($ExecQuery)) {
?>
<!-- Creating unordered list items.
Calling javascript function named as "fill" found in "script.js" file.
By passing fetched result as parameter. -->
<li onclick='fill("<?php echo $Result['Name']; ?>")'>
<a>
<!-- Assigning searched result in "Search box" in "search.php" file. -->
<?php echo $Result['Name']; ?>
</li></a>
<!-- Below php code is just for closing parenthesis. Don't be confused. -->
<?php
}}
?>
</ul>
Hi, I'm trying to make a dynamic html table to contain the mysql data that is generated via php. I'm trying to display a user's friends in a table of two columns and however many rows, but can't seem to figure out what is needed to make this work. Here's my code as it stands: Code: [Select] <?php //Begin mysql query $sql = "SELECT * FROM friends WHERE username = '{$_GET['username']}' AND status = 'Active' ORDER BY friends_with ASC"; $result = mysql_query($sql); $count = mysql_num_rows($result); $sql_2 = "SELECT * FROM friends WHERE friends_with = '{$_GET['username']}' AND status = 'Active' ORDER BY username ASC"; $result_2 = mysql_query($sql_2); $count_2 = mysql_num_rows($result_2); while ($row = mysql_fetch_array($result)) { echo $row["friendswith"] . "<br>"; } while ($row_2 = mysql_fetch_array($result_2)) { echo $row_2["username"] . "<br>"; } ?> The above simply outputs all records of a user's friends (their usernames) in alphabetical order. The question of how I'd generate a new row each time a certain amount of columns have been met, however, is beyond me. Anyone know of any helpful resources that may solve my problem? Thanks in advance =) This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=334481.0 Hey there... I'm so stuck on this problem I was hoping someone could help... I need to query a MySQL Database using array values... Here is my array ... $query_keywords = Array ( [7] => business [10] => home [11] => depo ) Here is my php ...$query = "SELECT * FROM product WHERE product_name, product_comment IN ($query_keywords)"; $result=mysql_query($query); if(mysql_num_rows($result) > 0) { echo "results"; } else { echo "no results"; } Unfortunately, I get this ... Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in blah\blah\results.php on line 99 Please help me All comments greatly appreciated... I created a simple search box which will query my table and match the input value to one of my columns. two of these columns store comma separated values. if i query a column other than a column which stores my csv i can see my search results. if i query a column which stores my csv i will not see results unless the search value matches the first value within the column. how would i be able to get say the second or third or forth value. here is the code i am using to query the table any help would be appreciated thanks. Code: [Select] $q = $this->db->query("SELECT * FROM table WHERE col1 LIKE 'searchvalue'". " OR col2 LIKE 'searchvalue'". " OR col3 LIKE 'searchvalue'". " OR col4 LIKE 'searchvalue'". " OR FIND_IN_SET('searchvalue', col5) > 0 ". " OR FIND_IN_SET('searchvalue', col6) > 0"); I was fiddling around with php and concerning data in a mysql database. All I was wanting to understand was being able to remove values with cheking multiple items in outputted html list. One that reads a set of values from a database, then displays a check box allowing the user to delete such values, that's it basically, here is my script I used to do it: Code: [Select] <?php function deleteRow($value) { $sql = "DELETE FROM checkbox WHERE value LIKE '$value'"; $result = mysql_query($sql); } if(array_key_exists('delete', $_POST)){ if(isset($_POST['value'])) { $count = count($_POST['value']); $connect = mysql_connect('localhost', 'jeremy', 's56pj989'); if($connect) { $select_db = mysql_select_db('test'); if($select_db) { for($i=0; $i<$count; $i++){ // print 'This is item = '.$_POST['value'][$i]; // outputs fine if(strlen($_POST['value'][$i]) > 0) { deleteRow($_POST['value'][$i]); } } } } } } ?> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html lang="en"> <head> <title></title> </head> <body> <form id="myform" name="myform" action="<?=$_SERVER['PHP_SELF']; ?>" method="post"> <?php if($_SERVER['REMOTE_ADDR'] == '192.168.0.2') { // printf("<p>Yes you have connected to this script successfully!</p>"); $connect = mysql_connect('localhost', 'mydbuser', 'mydatabasepassword'); if($connect) { // printf("<p>You have now connected to the database, well done!</p>"); $select_db = mysql_select_db('test'); if($select_db) { // printf("<p>Selected database now</p>"); $sql = 'SELECT * FROM checkbox'; $result = mysql_query($sql); if($result) { if(mysql_num_rows($result)>0){ while($row = mysql_fetch_assoc($result)) { printf("%s <input type=\"checkbox\" name=\"value[]\" value=\"%s\">\n<br />\n", $row['value'], $row['value']); } } else { printf("<p>No rows in system!</p>"); } } } } } ?> <input type="submit" id="delete" name="delete" value="Delete" /> </form> </body> </html> There must be a better way shouldn't there? I am not too concerned with error checking, ie if result of removing the values doesnt work, since I will just use this as an example of how to do this. Going on from that, how would I allow it to update a set of say changed values, just purely out of interest, pretty happy since it works though this script. But I was also wondering if there's some easy way of updating a set of values perhaps? Hmm good though and I look forward to any replies, Jeremy. Hey everyone. [= I don't have much time to read through all of the forums, because I'm on a time crunch and I'll be putting this certain project behind a couple others while I wait for some answers. This project is confusing me a bit, because I haven't worked much with arrays and the good ol' checkbox, so I need some help getting this to work. This project requires me to set-up a checkbox next to each order that was pulled off a database. You are allowed to select as many checkboxes as you'd like, and then click on the "process receipts" for all of the orders that were selected. I need to be able to have the form send me straight to a new page that pulls a loop of receipts one after another. I have the receipts built, and I can access a receipt one at a time based off the ID of a certain order, and I can easily build a loop that presents the receipts. I'm just having a problem pulling the information off the database based from the checkbox array. If you can give me some ideas on where to start with this, I'd greatly appreciate it. Note: Example is inside a hidden administrative page. Can't really give you something to play with, sorry! QQ Thanks! Alex D.A. Designing having a small issue setting up a search, trying to use check boxes, here is my form code <form action="workorder_index.php" method="POST"> <input type="checkbox" name="wos_0" <? if($wos_0 != NULL) { echo "checked"; }?>>0 - Active <input type="checkbox" name="wos_1" <? if($wos_1 != NULL) { echo "checked"; }?>>1 - Completed awaiting pickup <input type="checkbox" name="wos_2" <? if($wos_2 != NULL) { echo "checked"; }?>>2 - Completed and out the door <input type="checkbox" name="wos_3" <? if($wos_3 != NULL) { echo "checked"; }?>>3 - Waiting call back <input type="checkbox" name="wos_4" <? if($wos_4 != NULL) { echo "checked"; }?>>4 - Waiting for parts<br> <input type="checkbox" name="wos_5" <? if($wos_5 != NULL) { echo "checked"; }?>>5 - Parts in waiting for completion <input type="checkbox" name="wos_6" <? if($wos_6 != NULL) { echo "checked"; }?>>6 - not used <input type="checkbox" name="wos_7" <? if($wos_7 != NULL) { echo "checked"; }?>>7 - Overdue for pickup <input type="checkbox" name="wos_8" <? if($wos_8 != NULL) { echo "checked"; }?>>8 - Scrapped per customer <input type="checkbox" name="wos_9" <? if($wos_9 != NULL) { echo "checked"; }?>>9 - Scrapped by overdue<br><br> <input type="hidden" value="1" name="customsearch"> <input type="Submit" value="Update"> </form> here is my handling code $wos_0=$_POST[wos_0]; $wos_1=$_POST[wos_1]; $wos_2=$_POST[wos_2]; $wos_3=$_POST[wos_3]; $wos_4=$_POST[wos_4]; $wos_5=$_POST[wos_5]; $wos_6=$_POST[wos_6]; $wos_7=$_POST[wos_7]; $wos_8=$_POST[wos_8]; $wos_9=$_POST[wos_9]; if(isset($_POST['customsearch'])) { $str=''; if($wos_0 != NULL) { $str.= "'0'"; } if($wos_1 != NULL) { $str.= ",'1'"; } if($wos_2 != NULL) { $str.= ",'2'"; } if($wos_3 != NULL) { $str.= ",'3'"; } if($wos_4 != NULL) { $str.= ",'4'"; } if($wos_5 != NULL) { $str.= ",'5'"; } if($wos_6 != NULL) { $str.= ",'6'"; } if($wos_7 != NULL) { $str.= ",'7'"; } if($wos_8 != NULL) { $str.= ",'8'"; } if($wos_9 != NULL) { $str.= ",'9'"; } $query="SELECT * FROM workorders a INNER JOIN customers b ON a.customer_id = b.customer_id WHERE a.workorder_status IN (".$str.") ORDER BY a.workorder_id"; } everything works as intended as long as the first check box remains checked, the others can be checked or not, if the first box gets unchecked I get this error Quote SELECT * FROM workorders a INNER JOIN customers b ON a.customer_id = b.customer_id WHERE a.workorder_status IN (,'1','2','3','4','5','6','7','8','9') ORDER BY a.workorder_id Warning: mysql_numrows() expects parameter 1 to be resource, boolean given in workorder_index.php on line 84 basically I can't get the commas right, ...? any help is much appreciated Hey all... I'm running into a problem here. If you take a look at the code for this, you'll see it's to enter "horses" into a "class." It's all fine and dandy except I'm trying to get the way the horses are selected to have a different option. Right now they exist as a drop down/select option where you have to choose one at a time and enter submit. My goal is to have checkboxes that users can toggle and submit them all at once (with the rest of the eligibility checks etc still applicable). I've looked it up and the problem is, each user has a different amount of horses, so I can't account for all of them manually. Could someone help me out? On DaniWeb a user was trying to help me with an implode option, but I have zero idea how to work with that. I will literally need my hand to be held through this one. Thank you in advance - I have been trying to do this myself for a good chunk of the day, but my PHP skills are quite limited. Code: [Select] <?php session_start(); // Maintain session state header("Cache-control: private"); // Fixes IE6's back button problem. $page_title = "Class Information"; include('header.php'); $class_id = $_GET['id']; $enter = $_POST['enter_horse']; $enter_check = $_POST['check']; $horse_id = $_POST['horse_id']; //general show information $result = @mysql_query("SELECT s.show_id, s.player_id, s.type, s.name, DATEDIFF(s.run_date, NOW()) AS datedif, c.class_id, s.entry_fee FROM classes c, shows s WHERE c.class_id='$class_id' AND c.show_id=s.show_id LIMIT 1")or die("Cannot find class! " . mysql_error()); $row = @mysql_fetch_array($result); $show_id = $row['show_id']; $show_name = $row['name']; $runs_in = $row['datedif']; $species = $row['species']; $type = $row['type']; $owner_id = $row['player_id']; if(!$row['class_id']){myError("Invalid class!");include('footer.php');} $entry_fee = $row['entry_fee']; $num_entries = @mysql_num_rows(@mysql_query("SELECT horse_id FROM horses_entered WHERE class_id='$class_id'")); $runs_in = "$runs_in day[s]"; if($enter){ //ensure horse is eligible to enter $good = @mysql_num_rows(@mysql_query("SELECT horse_id FROM horses WHERE horse_id='$horse_id' AND player_id='$player_id' AND age>=2 AND age<=20")); $exists = @mysql_num_rows(@mysql_query("SELECT horse_id FROM horses_entered WHERE horse_id='$horse_id' AND class_id='$class_id' LIMIT 1")); if($my_money < $entry_fee){myError("You cannot afford the entry fee.", 1);} if(!$good){myError("Are you sure you own the horse and it is between 3 and 20 years of age?"); }elseif($exists){myError("That horse is already entered in this class!"); }else{ @mysql_query("INSERT INTO horses_entered(horse_id, class_id) VALUES('$horse_id', '$class_id')")or die("Cannot create entry!"); if($type == 1 AND $entry_fee){@mysql_query("UPDATE players SET money=money+'$entry_fee' WHERE player_id='$owner_id' LIMIT 1")or die("Cannot update player money!"); $points=1; }elseif($type == 2 AND $entry_fee){@mysql_query("UPDATE clubs SET money=money+'$entry_fee' WHERE president='$owner_id' LIMIT 1")or die("Cannot update player money2!"); $points=2;} @mysql_query("UPDATE players SET money=money-'$entry_fee', points=points+'$points' WHERE player_id='$player_id' LIMIT 1")or die("Cannot update player money3! " . @mysql_error()); @mysql_query("UPDATE horses SET points=points+'$points' WHERE horse_id='$horse_id' LIMIT 1")or die("Cannot update horse points!"); myError("Class entered!"); } } //display the show information echo "<table> <tr><td><b>Class:</td><td>#$class_id</td></tr> <tr><td><b>Show:</td><td><a href='shows.php?id=$show_id'>$show_name (#$show_id)</a></td></tr> <tr><td><b>Runs:</td><td>$runs_in</td></tr> <tr><td><b>Entry Fee:</td><td>$$entry_fee</td></tr> <tr><td><b>Total Entrants:</td><td>$num_entries</td></tr> <tr><td valign=top><b>Your Horses:</td><td> <form action='classes.php?id=$class_id' method=POST> <select name='horse_id'> "; $result = @mysql_query("SELECT horse_name, breed, horse_id FROM horses WHERE player_id='$player_id' AND age>2 AND age<=20 ORDER BY horse_name ASC")or die("Cannot find horses! " . mysql_error()); while($row = @mysql_fetch_array($result)): $horse_id = $row['horse_id']; $horse_name = stripslashes($row['horse_name']); $breed = $row['breed']; echo "<option value='$horse_id'>$horse_name (#$horse_id), $breed</option>\n"; $prev_species = $species; endwhile; if(!$horse_id){echo "<option value=0>No eligible horses!";} echo " </select> <input type=submit name='enter_horse' value='Enter Horse!'></td></tr></form> <tr><td valign=top><b>Entrants:</td><td> "; $query = "SELECT h.horse_name, h.horse_id, h.breed FROM horses_entered he LEFT JOIN horses h ON he.horse_id=h.horse_id WHERE he.class_id='$class_id' ORDER BY h.horse_name ASC"; $result = @mysql_query($query)or die(mysql_error()); while($row = @mysql_fetch_array($result)): $name = $row['horse_name']; $aid = $row['horse_id']; $breed = $row['breed']; $page = "horses.php"; echo "<a href='$page?id=$aid'>$name (#$aid)</a>, $breed<br>\n"; endwhile; if(!$aid){echo "<i>No entrants.";} echo "</td></tr> </table>"; include('footer.php'); ?> I currently have a page where I have each row in my table listed using: Code: [Select] $getinfo="SELECT * FROM tablename WHERE Closed='n'"; $result=mysql_query($getinfo); while ($row=mysql_fetch_array($result)){ ... // list of user information } That successfully displays my content. What I am trying to do is have a checkbox next to each user (row) that when I check them and hit submit at the bottom, it will change the value of each row on the 'Closed' column to 'y' so it will not be displayed when the page is refreshed. I want to be able to select multiple items to "close". And I would like it to come back to this page. I think I can use this page as the form action and have the code above my while loop to change the Closed column so it then would not be displayed when the while loop is called... is that right? How do I set this up? Not sure where to start the form tag and where to end it, and not sure how to...well...do any of it. I built a series of insertion forms to put entries into a mysql database. Everything is working fine, except that I just now realized my checkboxes don't put any entry in. It's just leaving it's respective smallint field as a default "0". Being new, I've must've overlooked how to handle these. The form's checkboxes have a bit of code in the value that remembers what users chose in case they have to backtrack and redo their form. How would I solve the problem and keep this code? Here's what all the checkboxes are written like: Code: [Select] <input type="checkbox" name="green_leaf" value="<?php if(isset($_POST['1'])) echo $_POST['1']; ?>" > green Hi people. I have a form which inputs into a database. Here is the code that inserts a yes no option... Code: [Select] <select name = "consent"> <option value = "Yes" <?php if ($_POST['consent'] == 'Yes') { echo 'selected="selected"'; } ?>>Yes</option> <option value = "No" <?php if ($_POST['consent'] == 'No') { echo 'selected="selected"'; } ?>>No</option> </select> However, I have been asked if I can make it a yes or no checkbox instead. Please can you tell me how I need to code it so that the "yes" or "no" is recorded in the DB. At the moment I just have this Code: [Select] <input name="consent" type="checkbox" value="Yes" />Yes<br /> <input name="consent" type="checkbox" value="No" />No<br /> Thanks in advance VinceG I have figured out far enough to loop through my table and return a list of check boxes that will be matched to a product. My issue now, is i cannot figure out the proper way to loop through them after they are selected and return them as checked/unchecked for the given product. Thought there is some MySQL involved, that part isn't the problem, I know how to insert into a database or update...when the product is selected from a drop down, I need to be able to populate the checkboxes. Here is the code i am using to initially display all of the categories to choose from... <ul class="categories"> <?php $query = "SELECT * FROM products"; $result = mysql_query($query) or die(mysql_error()); while ($row = mysql_fetch_assoc($result)) { $category = $row['category']; $catId = $row['id']; echo "<li><input class='catCheck' type='checkbox' name='p_cat[]' value='$catId' /> $category</li>"; } ?> </ul> Thanks guys.... |
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