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PHP - Dynamic Drop Down Menu That Filters Database Information
I am having a lot of trouble with this code, and I have no clue how to fix it. Right now, I have a GUI for a fictitious car dealership that has 5 populated drop down menus called Make, Model, Year, Color, and Mileage. What I want the code to do is read the selections made by the user with the drop down menus once the user hits the submit button and then filter the tables that I have in a mysql database to meets the choice requirements of the user.
The code will bring up the GUI, but once I hit the submit button, I get the following errors Please if anyone can help me that would be fantastic. I really have no clue Similar TutorialsHello my problem here is to have a drop down menu which gathers usernames from the database. Then with a click of a button the information for that specific user selected is shown. I'm close code wise but right now it's just showing me all users. I'm displaying the info in text box's to allow an admin to change the info.
<form action="edit.php" method="post">
<select name="username" id="username">
<?php
// Connects to your Database
$con=mysqli_connect('localhost', 'root', '');
/* check connection */
if (mysqli_connect_errno($con)) {
trigger_error('Database connection failed: ' . mysqli_connect_error(), E_USER_ERROR);
}
$query = "SELECT `username` FROM `bencobricks` . `users`";
$result = mysqli_query($con, $query) or trigger_error("Query Failed! SQL: $query - Error: ". mysqli_error($con), E_USER_ERROR);
if($result) {
while($row = mysqli_fetch_assoc($result)) {
//printf ("%s\n %s\n ",
echo "<label for='username'>Username: </label> <option value='$row[username]'>$row[username] </option><br/>";
echo "<br/><br/> ";
}
}
mysqli_close($con);
?>
</select>
<br />
<input name="send" id="send" type="submit" value="Edit User" />
</form>
Then in the edit.php file i have this code:
<?php
// Connects to your Database
$con=mysqli_connect('localhost', 'root', '');
/* check connection */
if (mysqli_connect_errno($con)) {
trigger_error('Database connection failed: ' . mysqli_connect_error(), E_USER_ERROR);
}
if (isset($_POST['send'])) {
$username = $_POST['username'];
$query = "SELECT * FROM `bencobricks` . `users` ";
$result = mysqli_query($con, $query) or trigger_error("Query Failed! SQL: $query - Error: ". mysqli_error($con), E_USER_ERROR);
// Print the result
if($result) {
while ($row = mysqli_fetch_assoc($result)) {
// $query = "SELECT * FROM `bencobricks` . `users` WHERE `username` = username = $row[username]";
printf ("%s\n %s\n %s\n %s\n %s\n %s\n %s\n %s\n %s\n %s\n %s\n",
"<label for='username'>Username: </label> <input type='text' name='username' value='$row[username]'>" . "</input><br/>",
"<label for='email'>Email: </label> <input type='text' name='email' value='$row[email]'>" . "</input><br/>",
"<label for='membership'>Membership: </label> <input type='text' name='membership' value='$row[membership]'>" . "</input><br/>",
"<label for='firstName'>First name: </label> <input type='text' name='firstName' value='$row[firstName]'>" . "</input><br/>",
"<label for='lastName'>Last name: </label> <input type='text' name='lastName' value='$row[lastName]'>" . "</input><br/>",
"<label for='gender'>Gender: </label> <input type='text' name='gender' value='$row[gender]'>" . "</input><br/>",
"<label for='dateOfBirth'>Birthdate: </label> <input type='text' name='dateOfBirth' value='$row[dateOfBirth]'>" . "</input><br/>",
"<label for='date'>Date: </label> <input type='text' name='date' value='$row[date]'>" . "</input><br/>",
"<label for='sets'>Sets: </label> <input type='text' name='sets' value='$row[sets]'>" . "</input><br/>",
"<label for='checkbox'>Checkbox: </label> <input type='text' name='checkbox' value='$row[checkbox]'>" . "</input><br/>",
"<label for='admin'>Admin? </label> <input type='text' name='admin' value='$row[adminFlag]'>" . "</input><BR>",
"<br/><br/> ");
}
}
}
mysqli_close($con);
?>
Any help is greatly appreciated.
hii,everyone... i am new to the forum ..so forgive me if i make some errors in describing my problem i want to create a drop down menu...the values of which i want to retrieve from a table i have created in my database... i have been able to acheive this...but i want a second drop down menu...in which i want the values to get filtered according to the selection of the first drop down menu... i hav a table with the name test...which i am connecting to the first drop down menu i hav created another table with the name testtype...for second menu.. the id of first table is the testid of second... plzz help me with creating the second menu... my code is Code: [Select] <?php $sql_result = "select test_series FROM test"; $result = mysql_query($sql_result); ?> <td>Select Test Series </td> <td> <select name="testseries" id="testseries" )"> <option>Select</option> <?php if(mysql_num_rows($result)) { while($row = mysql_fetch_assoc($result)) { $test = $row['test_series'];?> <option><?php echo $test ?></option> <?php } } else {?> <option>No Names Present</option>"; <?php } ?> </select> </td> </tr> <tr> <?php $testseries = $_POST['testseries']; $query = "select id from test where test_series = '$testseries'"; $resultt = mysql_query($query) or die(mysql_error()); $row = mysql_fetch_array($resultt); $id = $row['id']; $sql_result = "select test_name FROM testtype where testid = $id"; $result = mysql_query($sql_result); ?> <td>Select Test Name </td> <td> <select name="testtype"> <option>Select</option> <?php if(mysql_num_rows($result)) { while($row = mysql_fetch_assoc($result)) { $test2 = $row['test_name'];?> <option><?php echo $test2 ?></option> <?php } } else {?> <option>No Names Present</option>"; <?php } ?> </select> </td> </tr> IN THIS CODE I AM GETTIN AN ERROR Undefined index: testseries in ..addques.php on line 155 WHICH IS THIS ONE $testseries = $_POST['testseries']; i have created a seating system. based on the number of seats booked that value is passed to a page that dynamicly produces a drop down box for each seat booked so the type of seat price can be assigned. this part works fine. the problem i have is how do i code the recieving page not knowing how many values are going to be passed? this is the drop down menu page Thanks Code: [Select] <form action="payment.php?seatco=<?php echo $s_count; ?>" method="post" name="f1"> <div id="tick_select"> <?php for($i=1; $i<=$s_count; $i++){ $tick= mysql_query("select * from ticket_list"); ?> <select name="tick_com<?php echo $i; ?>" id="tick_com" onChange="" title="Choose A Ticket"> <option value="">Select Ticket type</option> <?php while($result= mysql_fetch_assoc($tick)) { ?> <?php echo '<option value="'.$result[ticket_id].'" name="'.$result[price].'">'.$result[type].'</option>';?> <?php }}?> </select> </div> <INPUT TYPE="submit" VALUE="Select"> </form> Hey Guys, I'm building a site and I need a drop down menu created dynamicly.... This is the flat html code Code: [Select] <div class="art-nav"> <div class="l"></div> <div class="r"></div> <ul class="art-menu"> <li><a href="URL 1"><span class="l"></span><span class="r"></span><span class="t">TOP LEVEL</span></a> <ul> <li><a href="URL 1">SUB CAT 1</a></li> <li><a href="URL 2">SUB CAT 2</a></li> <li><a href="URL 3">SUB CAT 3</a></li> </ul> </li> <li class="art-menu-li-separator"><span class="art-menu-separator"></span></li> <li><a href="URL 5"><span class="l"></span><span class="r"></span><span class="t">Contact Us</span></a></li> </ul> </div> This is the code that shows a dyamic code, but not in the menu... <?php wpsc_start_category_query(array('category_group'=> get_option('wpsc_default_category'), 'show_thumbnails'=> get_option('show_category_thumbnails'))); ?> <a href="<?php wpsc_print_category_url();?>" class="wpsc_category_link"><?php wpsc_print_category_name();?></a> | <?php wpsc_print_subcategory(); ?> <?php wpsc_end_category_query(); ?> Hi guys, I'm still not having any luck with coming up with an idea of how to make my multi-level drop down menu work with items from a database. I don't necessarily need code, but an idea of how to make it work, so any ideas are welcome. The table: Column Descriptions: id - The unique id label - The text to display link - The link to be used relation - I cannot remember why I wanted that... just disregard it parent - The parent item's id (0 for no parent) sort - A sort number for the item (ascending) active - 0 for false, 1 for true (not implemented yet, will add a WHERE `active` = 1 into the sql later) depth - How many parent items there are Now, here's what it has to do (the class): Grab items from the database (sorted by depth, parent, and sort) Place the children in a child array of the parents (or however it can be sorted) Implode the array and turn it into an html menu with the below structure Menu HTML Structure Code: [Select] <div id="menu"> <ul> <li><h2>Level 1 A</h2> <ul> <li><a href="?r=level2/a">Level 2 A</a> <ul> <li><a href="?r=level3/a">Level 3 A</a></li> <li><a href="?r=level3/b">Level 3 B</a></li> </ul> </li> </ul> </li> <li><a href="?r=level1/b"><h2>Level 1 B</h2></a></li> </ul> </div> And here is the code I have so far, but I can't figure out how to get the children into the parent array, or at least sort it so that I can achieve the above HTML structure. Thank you in advance for your help, and as I said above, I don't necessarily need code, but an idea of how to make it work, so any ideas are welcome. Hi there. Im a noob to sql and php not sure if this is right place to post, Im trying to get a dynamic drop down menu to show the 1st column in my sql database the column is called cat and holds category info ie audio, internet, music ect. ( i have no idea how to do lol ) it has taken me 2 days to find and edit this the bold and underline'd bit is what im trying to change with the dropdown menu. Or thinking bout it is there a way to do it with the URL. IE.. page name.php?cat=audio ? would that be easer ? is there any security issues with doing it that way ? Code================================== $db_host = '*******'; $db_user = '******'; $db_pwd = '*****; $database = 'nbbcj_co_uk'; $table = 'penapps'; if (!mysql_connect($db_host, $db_user, $db_pwd)) die("Can't connect to database"); if (!mysql_select_db($database)) die("Can't select database"); // sending query $result = mysql_query("SELECT * FROM {$table} WHERE `cat` = 'audio' LIMIT 10 "); if (!$result) { die("Query to show fields from table failed"); } $fields_num = mysql_num_fields($result); //echo "<h1>Table: {$table}</h1>"; echo "<table border='1' width='100%'><tr>"; // printing table headers for($i=0; $i<$fields_num; $i++) { $field = mysql_fetch_field($result); //echo "<td>{$field->name}</td>"; } echo "</tr>\n"; // printing table rows while($row = mysql_fetch_row($result)) { echo "<tr>"; // $row is array... foreach( .. ) puts every element // of $row to $cell variable foreach($row as $cell) echo "<td>$cell</td>"; echo "</tr>\n"; } mysql_free_result($result); ?> code end ====================== the test page can been seen here http://www.nbbcj.co.uk/testd/1/test1.php Any more questions let me know Thanks for any help you can give, a we all have to start some ware lol thanks kaine. Hi there, I have been creating a website which shows products of the companys (www.theadventurestartshere.org) and I have been trying to make some filters for the products using URL Parameters and recordsheet filters... Can someone advise me on the easiest way to do this? As I have created a method to do it with, (check website) but it has to be entered manually and I was hoping there is an easier way? Please Note I like to use drop down boxes for the filters but if there is a way to do the checkbox style you see on say Amazon then that would be brilliant... I use Dreamweaver CS5, and the newest versions of PHP and MySql Many Thanks, Paul Hi all, I'm just wondering if there's an easier way of doing what accomplishing the following: I have a value in my database which represents a selection in a drop down menu, i want to read it from the database and have it automatically selected depending on the stored data. I have the following working but just wondered if there was an easier way to get the same result: Code: [Select] <?php //database connection $query = "SELECT id FROM `tablename` WHERE username='$username'"; $result = mysql_query($query); $row = mysql_fetch_object($result); $id = $row->id; $id = (int)$id; ?> <select name="id"> <option value="">Select your option...</option> <option value="1" <?php if (($id - 1) === 0) { echo 'selected="selected"'; }?>>Selection 1</option> <option value="2"<?php if (($id - 2) === 0) { echo 'selected="selected"'; }?>>Selection 2</option> </select> Sorry if it's not very clear, i'll explain best i can if anyone can help. Thanks I'm trying to make it so that someone can add an item to the shop based upon the current items within the database. How do I go about doing that? I know i have to change <input> to <select>, but beyond that, how do I code it so I can run an array to get all the current items from the table `items` and list them by their `name` field? Thanks in advance! Here is what I have so far: function addshopinv($id) { error_reporting(E_ALL); ini_set('display_errors', 1); if (isset($_POST["submit"])) { extract($_POST); $errors = 0; $errorlist = ""; if ($name == "") { $errors++; $errorlist .= "Name is required.<br />"; } else if ($errors == 0) { $dbh=dbconnect() or die ("Userlist read error: " . mysql_error()."<br>"); mysql_select_db("XXX"); $query = mysql_query("SELECT id FROM items WHERE name='$name'"); while ($result = mysql_fetch_array($query)){ $item_id = $result['id']; $query1 = mysql_query("INSERT INTO sale SET shop_id='$id', item_id='$item_id'"); } admindisplay("Inventory Item Added.","Add New Inventory Item"); } else { admindisplay("<b>Errors:</b><br /><div style=\"color:red;\">$errorlist</div><br />Please go back and try again.", "Add New Item to Shop"); } } $page = <<<END <b><u>Add New Inventory Item</u></b><br /><br /> <form action="admin_panel.php?do=addshopinv:$id" method="post"> <table width="90%"> <tr><td width="20%">Name:</td><td><input type="text" name="name" size="30" maxlength="255" value="" />*255 character max</td></tr> </table> <input type="submit" name="submit" value="Submit" /> <input type="reset" name="reset" value="Reset" /> </form> END; $page = parsetemplate($page, $row); admindisplay($page, "Add New Inventory Item"); } Hello all Ok here is the problem... I want when a user inputs the requested data to the text fields , the script to insert those data in the prope table depending on the choise the user does by choosing one option from the drop down menu. Below is the php code (apparently not working) $editFormAction = $_SERVER['PHP_SELF']; if (isset($_SERVER['QUERY_STRING'])) { $editFormAction .= "?" . htmlentities($_SERVER['QUERY_STRING']); } $site_type = $_REQUEST['category_selection']; if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "link_submission") && ($site_type = "Web_Sites")) { $insertSQL = sprintf("INSERT INTO partner_sites (url, url_title, anchor_text, `description`, webmaster_name, webmaster_email, category) VALUES (%s, %s, %s, %s, %s, %s, %s)", GetSQLValueString($_POST['url_field'], "text"), GetSQLValueString($_POST['title_field'], "text"), GetSQLValueString($_POST['anchor_field'], "text"), GetSQLValueString($_POST['description_field'], "text"), GetSQLValueString($_POST['webmaster_nane_field'], "text"), GetSQLValueString($_POST['webmaster_email_field'], "text"), GetSQLValueString($_POST['category_selection'], "text")); mysql_select_db($database_content_conn, $content_conn); $Result1 = mysql_query($insertSQL, $content_conn) or die(mysql_error()); } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "link_submission") && ($site_type = "Blogs")) { $insertSQL = sprintf("INSERT INTO partner_blogs (url, url_title, anchor_text, `description`, webmaster_name, webmaster_email, category) VALUES (%s, %s, %s, %s, %s, %s, %s)", GetSQLValueString($_POST['url_field'], "text"), GetSQLValueString($_POST['title_field'], "text"), GetSQLValueString($_POST['anchor_field'], "text"), GetSQLValueString($_POST['description_field'], "text"), GetSQLValueString($_POST['webmaster_nane_field'], "text"), GetSQLValueString($_POST['webmaster_email_field'], "text"), GetSQLValueString($_POST['category_selection'], "text")); mysql_select_db($database_content_conn, $content_conn); $Result1 = mysql_query($insertSQL, $content_conn) or die(mysql_error()); } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "link_submission") && ($site_type = "Directories")) { $insertSQL = sprintf("INSERT INTO partner_directories (url, url_title, anchor_text, `description`, webmaster_name, webmaster_email, category) VALUES (%s, %s, %s, %s, %s, %s, %s)", GetSQLValueString($_POST['url_field'], "text"), GetSQLValueString($_POST['title_field'], "text"), GetSQLValueString($_POST['anchor_field'], "text"), GetSQLValueString($_POST['description_field'], "text"), GetSQLValueString($_POST['webmaster_nane_field'], "text"), GetSQLValueString($_POST['webmaster_email_field'], "text"), GetSQLValueString($_POST['category_selection'], "text")); mysql_select_db($database_content_conn, $content_conn); $Result1 = mysql_query($insertSQL, $content_conn) or die(mysql_error()); } And the html form <form action="<?php echo $editFormAction; ?>" method="POST" enctype="multipart/form-data" name="link_submission" id="link_submission"> <table width="630" border="0" align="center" cellpadding="5" cellspacing="5"> <tr> <td width="76">URL:*</td> <td width="519"><label for="url_field"></label> <span id="sprytextfield1"> <label for="url_field"></label> <input name="url_field" type="text" id="url_field" size="50" /> <span class="textfieldRequiredMsg">A value is required.</span><span class="textfieldInvalidFormatMsg">Invalid format.</span></span></td> </tr> <tr> <td>Anchor Text:*</td> <td><label for="anchor_field"><span id="sprytextfield2"> <input type="text" name="anchor_field" id="anchor_field" /> <span class="textfieldRequiredMsg">A value is required.</span></span></label></td> </tr> <tr> <td>URL Title:*</td> <td><label for="title_field"><span id="sprytextfield3"> <input type="text" name="title_field" id="title_field" /> <span class="textfieldRequiredMsg">A value is required.</span></span></label></td> </tr> <tr> <td>Description:*</td> <td><span id="sprytextarea1"> <label for="description_field"></label> <textarea name="description_field" id="description_field" cols="45" rows="3"></textarea> <span id="countsprytextarea1"> </span><span class="textareaRequiredMsg">A value is required.</span><span class="textareaMaxCharsMsg">Exceeded maximum number of characters.</span></span></td> </tr> <tr> <td>Webmaster Name:*</td> <td><label for="textfield2"><span id="sprytextfield4"> <input type="text" name="webmaster_nane_field" id="webmaster_nane_field" /> <span class="textfieldRequiredMsg">A value is required.</span></span></label></td> </tr> <tr> <td>Webmaster E-mail:*</td> <td><label for="textfield3"><span id="sprytextfield5"> <input name="webmaster_email_field" type="text" id="webmaster_email_field" size="40" /> <span class="textfieldRequiredMsg">A value is required.</span><span class="textfieldInvalidFormatMsg">Invalid format.</span></span></label></td> </tr> <tr> <td>Category:*</td> <td><span id="spryselect1"> <label for="category_selection"></label> <select name="category_selection" id="category_selection"> <option>Select An Option</option> <option value="Web_Sites">Web Sites</option> <option value="Blogs">Blogs</option> <option value="Directories">Directories</option> </select> <span class="selectRequiredMsg">Please select an item.</span></span></td> </tr> <tr> <td> </td> <td><label for="select"></label> <input type="submit" name="button" id="button" value="Url Submission" /></td> </tr> </table> <input type="hidden" name="MM_insert" value="link_submission" /> </form> Im begging for your help..... As a complete newbie to php and webdesigning i have a following problem.I would like to retrieve the data from database and display it in a drop down menu.Then i should allow the user to select the values from drop down list along with other details,in other words i have to embed the drop down output as the form input for the user and store the form data in another table.I am running a xampp server and i am using php 5.4 version.Please help.My code is as follows.In this case project_name is displayed as the drop down output.but how do i use the same drop down output as a input in the form. <html> <head></head> <body> <?php error_reporting(E_ALL ^ E_DEPRECATED); include 'connect.php' ; $tbl_name="projects"; $sql="SELECT project_name FROM $tbl_name "; $result=mysql_query($sql); if($result === FALSE) { die(mysql_error()); } ?> <form name="resources" action="hourssubmit.php" method="post" > <?php echo "<select name='project_name'>"; while ($row = mysql_fetch_array($result)) { echo "<option value='" . $row['project_name'] ."'>" . $row['project_name'] ."</option>"; } echo "</select>"; ?> </form> </body> </html> I am currently creating a form and I want to populate a drop down selection menu with data from two fields in a form. For example, I want it to pull the first and last name fields from a database to populate names in a drop down menu in a form. I want the form to submit to the email address of the person selected in the drop down. Is this possible to do? The email is already a field in the record of the person in the database. Can anyone give me some pointers or advice on how I should go about setting up the "Select" box drop down? I am not sure how to code it to do what I am wanting. Any links to relevant help would be appreciated too. Thanks in advance! I have created a drop down list and it does retrieve information from mysql but now I want to use what is been selected to retrieve information. How Do I do this? <?php MYSQL_CONNECT(localhost,'root','') OR DIE("Unable to connect to database"); @mysql_select_db(Examination) or die( "Unable to select database"); $query=("SELECT * FROM subject"); $result=mysql_query($query) or die ("Unable to Make the Query:" . mysql_error() ); echo "<select name=myselect>"; while($row=mysql_fetch_array($result)){ echo "<OPTION VALUE=".$row['Sub_ID'].">".$row['Sub_Name']."</OPTION>"; } echo "</select>"; ?> Hi there I would be most grateful is someone could tell me why my data is not being entered into my database and not responding with an email confirmation link. Thanks in advance. <!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> <? include('config.php'); // table name $tbl_name="temp_members_db"; // Random confirmation code $confirm_code=md5(uniqid(rand())); // values sent from form $name=$_POST['name']; $email=$_POST['email']; $country=$_POST['country']; // Insert data into database $sql="INSERT INTO $tbl_name(confirm_code,name,email,password,country) VALUES('$confirm_code','$name','$email','$password','$country')"; $result=mysql_query($sql); // if suceesfully inserted data into database, send confirmation link to email if($result){ // ---------------- SEND MAIL FORM ---------------- // send e-mail to ... $to=$email; // Your subject $subject="Your confirmation link here"; // From $header="from: your name <your email>"; // Your message $message="Your Comfirmation link \r\n"; $message.="Click on this link to activate your account \r\n"; $message.="http://www.yourweb.com/confirmation.php?passkey=$confirm_code"; // send email $sentmail = mail($to,$subject,$message,$header); } // if not found else { echo "Not found your email in our database"; } // if your email succesfully sent if($sentmail){ echo "Your Confirmation link Has Been Sent To Your Email Address."; } else { echo "Cannot send Confirmation link to your e-mail address"; } ?> </body> </html> hello. please could someone help me with this conundrum im trying to pull my content dynamically. im almost there but im stuck with some code. so i have a layout with 3 columns and in each column i have a place holder. ph01, ph02, ph03 each page has its own id and depending on what the page id is i what to pull different elements into the 3 placeholders. so for this code in the database i have the following: table name: placeholders pages_id = page id ph_number = placeholder number contElements_id = element ids id pages_id ph_number contElements_id 1 1 1 1, 2, 3 2 1 2 4, 5, 6 3 1 3 7, 8, 9 table name: contelements name = name of the element id name 1 E1 2 E2 3 E3 4 E4 5 E5 6 E6 7 E7 8 E8 9 E9 SO, page_ID=1 has 3 placeholders and they should show: ph01 should display = E1, E2, E3 ph02 should display = E4, E5, E6 ph03 should display = E7, E8, E9 this is the code im trying to put together. on the page im using this include function in the left, center and right columns. Code: [Select] $phNo = "1"; echo include_admin_contElements($phNo, $pageID); $phNo = "2"; echo include_admin_contElements($phNo, $pageID); $phNo = "3"; echo include_admin_contElements($phNo, $pageID); now this is where the problem is.... i think in the function file i have this but its a mess.. Code: [Select] function include_admin_contElements($phNo, $pageID){ $PH = Placeholders::find_all(); foreach ($PH as $PHs){ $PHid = $PHs->id; $PHpid = $PHs->pages_id; $PHce = $PHs->contElelments_id; $PHn = $PHs->ph_number; if($pageID == $PHpid){ $CE = Contelements::find_by_PHce($PHce); foreach ($CE as $CEs){ echo $CEid = $CEs->id; echo $CEname = $CEs->name; } } } } - so the idea is i get $phNo, $pageID from the page, - then find all the placeholders in the database Quote Placeholders::find_all(); - then if the $pageID is the same as the $PHpid in the placeholder database Quote if($pageID == $PHpid){ get the content elents for that page. - then i want to get the elements that belong to that page using Quote Contelements::find_by_PHce($PHce); but that does not work Code: [Select] public static function find_by_PHce($PHce=0){ $sql = "SELECT * FROM ".self::$table_name." WHERE id=".$PHce.""; $result_array = self::find_by_sql($sql); return $result_array; } anyway... im stuck. this code doesn't even seperate the 3 different placeholders. it needs a more experienced eye.. its a mess please help thanks ricky I am having trouble pulling a youtube embedded code from my database. Everything else comes out fine however it just doesnt pull anything out where the embedded code is supposed to be. Any ideas I put the code below. All help would be greatly appreciated, also if anyone is feeling generous and would like to help me some more please message me I have a couple other small questions. Code: [Select] <?php //open database $connect = mysql_connect("Database","name","password") or die("Not connected"); mysql_select_db("database") or die("could not log in"); $query = "SELECT * FROM boox ORDER BY date DESC"; $result = mysql_query($query); // Get the page number, if none is set - it is 0 if( isset($_GET['page']) ) { $page =$_GET['page']; } else { $page = 0; } $resultsPerPage = 15; $num = mysql_num_rows($result); // amount of rows $loops = $page*$resultsPerPage; // starting loops at.. while ($loops < $num && $loops < ($page+1)*$resultsPerPage ) { $link = mysql_result($result,$loops,"link"); // get result from the 'Title' field in the table $username = mysql_result($result,$loops,"username"); // get result from the 'Content' field in the table $messsage = mysql_result($result,$loops,"message"); $date = mysql_result($result,$loops,"date"); if ($pagelimit == 0) { $pagelimit == 1; } if ($pagelimit <= 15) // echo stuff here $loopz = $loops + 1; echo "   </br><align='left'><table width='297' height='900' border='1' align='center' bgcolor='#111'> <tr> <td>$loopz. </br> $message </br> Posted By: $username $date </td> </tr> </table></br><br>"; $count++ ; $pagelimit++; $loops++; } if ( $page!=0 ) // Show 'Previous' link { $page--; $prevpage = ($page + 1); echo "<br><br><br><a href='index.php?page=$page'>Previous $prevpage </a>"; $page++; } if ($loops > 5&&($page+1)*$resultPerPage < $num ) // Show 'next' link { $page++; $nextpage = ($page + 1); echo "<a href='index.php?page=$page'> Next $nextpage</a>"; } ?> I have these two files and I need to solve the problem that I need the user data in the role column to be taken when logging in. I have written it so that the data can be extracted from the database, but it does not work for me that it is “forwarded” from process.php to admin.php . Please don’t know what the error is that the admin.php file doesn’t want to load $ _SESSION['Role'] from process.php ? Thanks to everyone for helping and here is the code: process.php
<?php
require_once('connect.php');
session_start();
if(isset($_POST['Login']))
{
if(empty($_POST['Username']) || empty($_POST['Password']))
{
header("location:index.php?Empty= Please Fill in the Blanks");
}
else
{
$query="select * from role_test where Username='".$_POST['Username']."' and Password='".md5($_POST['Password'])."'";
$result=mysqli_query($con,$query);
if(mysqli_fetch_assoc($result))
{
$_SESSION['User']=$_POST['Username'];
while($row = mysqli_fetch_array($result) ){
$_SESSION['Role']=$row['role'];
}
header("location:admin.php");
}
else
{
header("location:index.php?Invalid= Please Enter Correct User Name and Password ");
}
}
}
else
{
echo 'Not Working Now Guys';
}
?>
admin.php
<?php
session_start();
if(!(isset($_SESSION['User'])))
{
header("Location: index.php");
exit(0);
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Role</title>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
</head>
<body>
<?php
$_SESSION['Role']=$role;
echo $role;
?>
</body>
</html>
Hello, I recently came upon this website and I want to implement something similar. The 'remove' box is what I would like to do with PHP. I have no idea where to start. I would like to be able to search through my database, select one of its components and add it to that window. Within that window I would also like to be able to remove it. Does anyone know where I should go from there? here is the site: http://www.sealandserpent.org/schedgen/schedulegenerator.php I am creating a music blogging site however the main page will only show one video the code is below anyone have any ideas? " <?php //open database $connect = mysql_connect("******","username","password") or die("Not connected"); mysql_select_db("collegebooxboox") or die("could not log in"); $query = "SELECT * FROM boox ORDER BY date DESC"; $result = mysql_query($query); // Get the page number, if none is set - it is 0 if( isset($_GET['page']) ) { $page =$_GET['page']; } else { $page = 0; } $resultsPerPage = 15; $num = mysql_num_rows($result); // amount of rows $loops = $page*$resultsPerPage; // starting loops at.. while ($loops < $num && $loops < ($page+1)*$resultsPerPage ) { $link = mysql_result($result,$loops,"link"); // get result from the 'Title' field in the table $username = mysql_result($result,$loops,"username"); // get result from the 'Content' field in the table $messsage = mysql_result($result,$loops,"message"); $date = mysql_result($result,$loops,"date"); if ($pagelimit == 0) { $pagelimit == 1; } if ($pagelimit <= 15) // echo stuff here $loopz = $loops + 1; echo "   </br><align='left'><table width='297' height='900' border='1' align='center' bgcolor='#111'> <tr> <td>$loopz. $link </br> $message </br> Posted By: $username $date </td> </tr> </table></br><br>"; $count++ ; $pagelimit++; $loops++; } if ( $page!=0 ) // Show 'Previous' link { $page--; $prevpage = ($page + 1); echo "<br><br><br><a href='index.php?page=$page'>Previous $prevpage </a>"; $page++; } if ($loops > 5&&($page+1)*$resultPerPage < $num ) // Show 'next' link { $page++; $nextpage = ($page + 1); echo "<a href='index.php?page=$page'> Next $nextpage</a>"; } ?> " |
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