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PHP - How To Make A Php Form That Queries A Database Based On Check Box Options?
Hi im new to php and I need help making webpage that queries a mysql database based on a 3 check boxes and displays results on the same page or on another page. The table being queried has 4 columns, name, gps, wifi, bluetooth. So for example a row in the table would be like, samsung galaxy s2, yes, yes, yes. The idea is for it to be a website that will display phones according to their features.
So the idea is depending on if the boxes were ticked the samsung galaxy would be displayed as a result. So i need some help understanding how to make this. Some1 gave me the code below in attempt to help me (im not sure it works or not) but im not sure how fully use it, ie what pages i need to make and how i create the connection to the mysql database, and how to use the query that they wrote to display the results thanks code: Code: [Select] <form action="?do=filter" method="post"> <table cellspacing="0" cellpadding="3" border="1"> <tr> <td>GPS<input type="checkbox" name="gps" value="checked"></td> <td>Wifi<input type="checkbox" name="wifi" value="checked"></td> <td>Bluetooth<input type="checkbox" name="bluetooth" value="checked"></td> </tr> <tr><td><input type='submit' name='filter' value='Filter'></td></tr> </table> </form> </html> <?php function filterMe($filter){ if(isset($_POST[$filter])){ return "Yes"; }else{ return "No"; } } if(isset($_POST['filter'])){ echo "Gps - " . filterMe('gps'); echo " Wifi - " . filterMe('wifi'); echo " Bluetooth - " . filterMe('bluetooth'); } ?> All you need to do is use a query something like SELECT name,gps,wifi,bluetooth FROM `product` WHERE `gps`='".filterMe('gps')."' AND `wifi`='".filterMe('wifi')."' AND `bluetooth`='".filterMe('bluetooth')."' Similar TutorialsHow to make a form using HTML/PHP with conditional responses based on zip code? What I want to do exactly... zip codes 91900-92600 go to URL www.example.com after submit. All others go to URL www.example.com/page after submit. Any help would be much appreciated. I am using a captcha like image verification for my form fields. I want to make my check case insensitive: For example, if the security word is 1A3e then 1a3e or 1a3E should work.(Upper Case or lower case does not matter) Here is the check I am doin so far to see if my security word matches. Code: [Select] if(md5($_POST['security_word']).'a4xn' == $_COOKIE['tntcon']) Hi
I have a form that has a drop-down with a few to choose from, unfortunately I don't get results for some due to the query involved.
Some need the AND channel LIKE '%$channel%'"; and some don't and therefore will not get desired results. So I would like to run two queries one with and one without.
$query = "SELECT * FROM asterisk_cdr WHERE calldate BETWEEN '$calldate' AND '$calldate2' AND clid LIKE '%$clid%' AND channel LIKE '%$channel%'";
Thanks
We recently upgraded from PHP4 to PHP5 and the below script that was working perfectly in 4 has completely stopped working and I can't figure out why for the life of me. I'm not an experienced PHP programmer--I've done some forms, but this is the first time I've used a database. What needs to (and was) happen: A user enters their username in the form and gets a readout of their participation so far that month. The problem(s): I know that it's storing the variable 'user' because it echoes it back properly, but the database is no longer allowing me to select the row based on that variable. I know it's not that I can't connect to the database because if instead of '$user' I change the code to a username I know is in there, I get the proper readout. This all started as soon as I transferred over to PHP5--before that, no problems at all. The database information is all correct, I just took it out for privacy's sake. <form id="feedback" method="post" action="index.php"> <input name="user" type="text" value="Enter user name" size="20" maxlength="50" /><br /> <input name="send" id="send" type="submit" value="Submit" /> </form> <?php if (isset($_POST['user'])) { $_session['user'] = $_POST['user']; } ?> <p>You entered your username as: <strong><? echo $_session['user'];?></strong>. If this is not correct or you do not see your information below, please re-enter your username and click Submit again.</p> <?php $con = mysql_connect("database","username","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("database", $con); $result = mysql_query("SELECT * FROM March WHERE Username='$user'") or die ('Error: '.mysql_error ()); while($row = mysql_fetch_array($result)) { echo "<table border='0'>"; echo "<tr>"; echo "<td><strong>Username:</strong> </td><td>" . $row['Username'] . "</td></tr>"; echo "<tr><td><strong>Mar. 2 Discussion:</strong></td><td>" . $row['Mar2Q'] . "</td></tr>"; echo "<tr><td><strong>Mar. 2 Poll:</strong></td><td>" . $row['Mar2P'] . "</td></tr>"; echo "<tr><td><strong>Mar. 9 Discussion:</strong></td><td>" . $row['Mar9Q'] . "</td></tr>"; echo "<tr><td><strong>Mar. 9 Poll:</strong></td><td>" . $row['Mar9P'] . "</td></tr>"; echo "<tr><td><strong>Mar. 16 Discussion:</strong></td><td>" . $row['Mar16Q'] . "</td></tr>"; echo "<tr><td><strong>Mar. 16 Poll:</strong></td><td>" . $row['Mar16P'] . "</td></tr>"; echo "<tr><td><strong>March Participation To-Date:</strong></td><td>" . $row['Participation'] . "</td></tr>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> ANY help would be greatly appreciated! I've got a couple hundred people who use this on a regular basis and are starting to ask why it's not working. hi I have 2 tables (members & friendship) Members table columns: -------------------------------------------- username | password | email | etc .... -------------------------------------------- Person 1 xxxxxx x@x.com person 2 yyyyyy y@y.com Friendship columns: --------------------------------------- username | friend | status | --------------------------------------- person 1 person 2 friends person 2 person 1 friends person 1 person 3 pending I want to query the friendship table: "SELECT friend FROM friendship WHERE username ='$_SESSION['myusername'];' AND status='friends' " (so its basically getting the user names of anyone who is the logged in users friend) And then use the returned user names to select their data from the members table. Any help much appreciated! ^.^ So I'm trying to basically trying to make an "advanced search" function in PHP/mysql that will allow users to search by a number of different options like search by zipcode, username, gender, etc. Now my question is how do I vary my mysql query so that it searches for all these things based on whatever the user inputs? For example, the user might need to search zipcode and username but not gender, but in another case, might need to search username and gender, but not zipcode. Obviously I could just do some if/else statements, but that would be increasingly more difficult as there are more fields. What can I do? I'm new to this form and php/mysql so sorry if this isn't the right place to post this. This is what is in the first dropdown box. Code: [Select] $selValues['john'] = "a, b, c, d, e"; These are the different lists I want it to put in the second drop down box depending on what they choose in the first. Code: [Select] $selValues['list1']= " a1, a2, a3, a4, a5"; $selValues['list2']= " b1, b2, b3, b4, b5"; This is how I made an attempt to make it work before posting here. Along with plently of other ways. Code: [Select] if ($selValues['john']=a) { $chan_input = selectBuilder($selValues['$list1'] }; else if ($selValues['john']=b) { $chan_input = selectBuilder($selValues['$list2'] }; Basicly how I need it to work is there are two drop down boxes. First they will chose between a,b,c,d,e. If they chose a then on the second drop down box I only want them to be able to select a1,a2,a3,a4,a5. I can't seem to figure this out. The queries seem to need to be in the foreach loop. The queries will then work but they update every blog post in my table. I only want it to update the 1 that has the button associated with it. So for instance... only delete the blog post where post_id = ${post['id']} Do I make the queries be outside of the foreach? If I do that then MySQL fails because my foreach is using the $post variable. Code: [Select] <?php if (isset($_POST['approve'])) { $sql = " UPDATE `blog_posts` SET `approved` = 1 WHERE `post_id` = '${post['id']}' "; mysql_query($sql) or die(mysql_error()); } else if (isset($_POST['deny'])) { $sql = " UPDATE `blog_posts` SET `approved` = -1 WHERE `post_id` = '${post['id']}' "; } else if (isset($_POST['delete'])) { mysql_query("DELETE FROM `blog_posts` WHERE `post_id` = {$post['id']}") or die(mysql_error()); } foreach ($posts as $post) { ?> <div class="post" id="post<?php echo $post['id']; ?>"> <form action="blog.php" method="post" id="blogform" class="man"> <fieldset class="mvs buttonfield"> <span class="button"> <label> <input type="submit" id="starttop" name="approve" class="invis dark_grey" value="Approve" /> </label> </span> <span id="smarktop" class="button disabled"> <label> <input type="button" id="marktop" name="deny" class="invis dark_grey" value="Deny" disabled="disabled" /> </label> </span> <span id="sdeletetop" class="button disabled"> <label> <input type="submit" id="deletetop" name="delete" class="invis dark_grey" value="Delete" disabled="disabled" /> </label> </span> </fieldset> </form> </div> <?php } ?> Let me give you a couple of examples.
For the first example, I have a table which includes all the pages I have on a website, and includes information about them such as some directory path, the minimum user access level to view, etc. The only time it will change is if I add a page or modify settings on an existing one. Instead of performing the query each time, maybe I should do it once for each user session and store the results in a session array? But if I do, how do I force a reload should I ever make a change. Or maybe I should do it once for all users, but this probably doesn't make much sense.
For a second example, I have a bunch of sites which use common code. Each has their own subdomain which in turn identifies a primary key which with multiple joins makes each site unique. The user could change the settings, however, will rarely do so. Instead of querying the big query each time, should I store the settings in a session, and only query the database if some part of the session is not set or if the session "last_updated" value is different than that stored in the database? Yes, I will still need a query, but it will be small and hopefully more efficient).
For a third example, I have some JSON data available to the client. Most clients will cache the data, so I shouldn't have much problems there. I could also server cache it, but again, how do I know when to force a reload?
Thank you, and Happy New Year!
I have set up a form page with a select box of colleges to select. I want the "options" in the select box to be values taken from a field called "name" in a table called "colleges" and they should be ordered alphabetically. I also want the default selected option to be "none." I have attached a picture to describe what i want. Please be detailed with the code. I am fairly new to php and mysql. Thank you. Hello, I've created a function whereby I want to return the school_id associated with a particular user. Each user will be associated with exactly one school. My query below works, but the thing that it returns is an array; I need to make use of it as a variable (say $instructor_school_id). Any idea how to make the conversion from a single element array into a non-array variable? Thank you.... function getInstructorSchool($read, $user_id) { $sql = "SELECT school_id FROM users WHERE user_id = $user_id"; return $read->fetchRow($sql); } I have been reading in Larry Ullmans book "Visual QuickPro Guide PHP 6 and MySQL 5" and I find it well-written. Since it is from 2008, it does not contain anything about MySQL PDO, but rather does it in the mysqli_* way. Larry suggest placing the secret database password and more along with a database connection script in a connect.php file, placed above the webroot if possible. Then later, when he is creating queries and executing them in other php files, he includes the connect.php file before making the queries.
Now I know it is very important to be careful with the Error handling, so the script won't output errors, which could reveal something about the database making it less secure. Therefore I am wondering how to structure things when using PDO. I need to write error-handling scripts for the following situations:
a) Connection to the database doesn't succeed
b) The execution of the queries doesn't succeed
c) User input in HTML forms are not appropriate
and probably more. The recommended way of handling errors when using PDO seems to be writing some try-catch code. But then I don't see how I can keep the connection to the database completely inside the connect.php file. Either I will need to use a die() or exit() inside this file or I will need to give up my idea to keep everything which concerns the connection to the database in the file mentioned AND write nested try-catch sentences - first make sure the connection works, then make sure the query will execute properly.
I don't like either of those approaches. Firstly I have been told that using exit() is bad programming and secondly it seems to get more complicated using nested try-catch code and to let database connection take part in diverse php-files.
Maybe somebody have a smart, convenient and secure way to do it?
Erik
As we know the total number of hits to database is proportional to the number of page views if page views * queries per page is more than the capacity of your database, then mysql queries may show problem. is there any way to calculate the database capacity ? I googled someone says 128Mb cache is there for mysql. Hey guys, i'm new to this site and would need some help with coding. So i'm making a car part website which should has brand/model search. It's a dropdown search which will get the brand / model from database and should display all the parts for that exact model, from folder. Database structure which i have is id, master, name. ( Here's some pictures to clear out what i'm doing. http://imgur.com/a/7XwVd ) So the problem is that it does not get the images from folder named ex: *_audi_a3.jpg. And link to codes what have been written already. Parts.php: http://pastebin.com/q6vdypge Update.php: http://pastebin.com/DymhGQ17 Search_images.php: http://pastebin.com/LF5Q0i8f Core.js: http://pastebin.com/bgc0y4TS I don't know what's wrong with it sadly. One thing i noticed when i used firebug it gives error on category when searching error:true. Hope you guys understand what i mean here, and also all help is appreciated. And move this post if it's in wrong place. Thanks! Edited by aeonius, 17 October 2014 - 12:15 PM. I am just new to relational algebra probably a pre-step before learning SQL queries. Can you help me make the expressions of relational algebra expression for each of the following queries. This is the table contained inside a bus driver database.
driver ( driver_id, driver_name, age, rating ); bus ( bus_id, bus_name, color); reserves ( driver_id, bus_id, date);a. Find the names of drivers who have reserved at least three busses. b. Find the names of drivers who have reserved all busses. c. Find the names of drivers who have reserved all busses called Shuttle. d. Find the IDs of drivers whose rating is better than some driver called Paul. I would be grateful if somebody can help me here. This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=347091.0 I was thinking of adding queries using a basic HTML form. To be able to do that I was thinking of having a Query sql table which has all the main create, insert, remove, update, delete and select general syntax into it. The idea is to be able to choose a query based on a HTML select option and add the values to a field, any idea how can you do that dynamically?
Greetings! I'm new in phpmysql. Am making signup form with three users (Admin, office and student), that when student is selected there will be another form that need to fill up and the data is saved in different table. Below are my codes when the data of admin and office are submitted. Code: [Select] mysql_query("INSERT INTO tbl_user (user_Id,user_Password,user_Fname,user_Lname,user_Sex,user_Level,user_Program) VALUES ('$userid','$password','$fname','$lname','$sex','$level','$program')"); these things here are codes for student when submitted Code: [Select] mysql_query("INSERT tbl_studenroll (user_Id, course_Id, ap_Id, se_Sem, se_Yearlevel) VALUES ('$userid','$course', '$acadyear', '$sem', '$year')") or die(mysql_error()); $seID = mysql_query ("SELECT max(se_Id) AS mxSEid FROM tbl_studenroll")or die (mysql_error()); $seID1 = mysql_fetch_assoc($seID); $seID = $seID1['mxSEid']; $offcourse = mysql_query("SELECT office_Id FROM tbl_offcourse WHERE course_Id = '$course'")or die (mysql_error()); while ($row = mysql_fetch_array($offcourse)) { $off = $row['office_Id']; mysql_query("INSERT tbl_data (se_Id,office_Id,user_Id,course_Id) VALUES ('$seID','$off','$userID','$courseID')") or die(mysql_error()); } sEE that I am saving in three tables at the same time when student is selected. How can I do it? I want to save only what for students and for other users..I tried my codes but when I submit data from office, it saves data to studenroll, but in 0 value. hope I am clear, help me guys! thank you in advace. I have a column in my db called 'Sold' which is set to Binary. I need a dropdown in my form, that converst the binary into 'True', 'False' for the options, and also selects the one that is set in the db. This is for an edit form, so will then need to adapt it for a add form. This is my code, which doesnt seem to be working
<?php
$id = $_GET["id"];
$sqlSold = "SELECT Sold FROM Products WHERE idProducts = $id;";
$products = [
'True' => true,
'False' => false
];
?>
<select name="Sold">
<option selected="selected">Choose one</option>
<?php
foreach($products as $item){
echo "<option value='strtolower($item)'>$item</option>";
}
?>
</select>
I understand that is might be something that is already answered and I apologize if it is, I could not find it.
What I need to do is build a simple form that has two options, they will be dropdown options. Dropdown A and Dropdown B then a Submit button. This part I understand in HTML, although it may be easier in php or javascript.
Then I need it to take the two options and create a "if/then" statement that loads a specific pdf that matches the two options selected.
Example.
If someone selects Option 1 from Dropdown A and Option 2 From Dropdown B then it loads 12.pdf If someone selects Option 5 from Dropdown A and Option 3 From Dropdown B then it loads 53.pdf If someone selects Option 2 from Dropdown A and Option 1 From Dropdown B then it loads 21.pdf and so on... It does not have to be the exact thing just some way to take both inputs and have it equal a specific pdf. Here is the form I built but I don't know what to put in the form_action.php file in order to make it work <!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>Untitled Document</title> </head> <body> <center> <h1> Get Directions</h1> <form action="form_action.php" method="get" name="directions" target="_new"> <select name="startpoint" size="1"> <option value="north">North Tower Entrance</option> <option value="south">South Tower Entrance</option> <option value="moba">MOB A Entrance</option></select> -----> <select name="endpoint" size="1"> <option value="onco">Oncology</option> <option value="radio">Radiology</option> <option value="pulm">Pulmanary</option></select> <br /><br /> <input type="submit" value="Submit" /> </form> </center> </body> </html>Any help is appreciated, thanks. |
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