PHP - Assigning Multiple Values To One Variable
Hi folks,
I was wondering how to do this. I want the if statement to detect if the query string has any of these values. so im trying to assign them all to the same variable. However, this code wont work. Whats the trick here? <?php $primary=$_GET['intro']; $primary=$_GET['port']; $primary=$_GET['about']; $primary=$_GET['contact']; if(isset($primary)){ echo "<img src='graphics/left-a.png'>";} else {echo "<img src='graphics/leftb.png'>";}?> Similar Tutorials
Table Issue - Multiple Location Values For User Pushes Values Out Of Row Instead Of Wrapping In Cell
I am facing problem to execute query by assigning NULL value to a variable and then executing query.In MySQL DB four fields Mobile,landline, pincode,dob are set as integer and date(for dob) respectively.The default is set as NULL and NULL option is selected as yes.All these fields are not mandatory.The problem is that when I edit the form my keeping the value as empty in DB these are saved as 0, 0 , 0 & 0000-00-00 inspite of Null. I have tried everything but still the defect persist. Please help me to come out of the problem The code, I have used: <?php //require_once 'includes/config.php'; $dbusername = $_POST['email']; $dbfirstname = $_POST['first_name']; $dblastname = $_POST['last_name']; //$dbmobile_number = $_POST['mobile']; if (isset($_POST['mobile'])) { $dbmobile_number = $_POST['mobile']; } else { $dbmobile_number = "NULL"; } $dblandline_number = $_POST['landline']; $dbdob = $_POST['dob']; if(isset($_POST['is_email'])) { $dbSubscribe_Email_Alert = '1'; } else { $dbSubscribe_Email_Alert = '0'; } if(isset($_POST['is_sms'])) { $dbSubscribe_SMS = 0; } else { $dbSubscribe_SMS = 0; } $dbAddress_firstname = $_POST['shipping_first_name']; $dbAddress_lastname = $_POST['shipping_last_name']; $dbAddress = $_POST['shipping_address']; $dbcity = $_POST['shipping_city']; $dbpincode = $_POST['shipping_pincode']; $dbstate = $_POST['shipping_state']; $dbcountry = $_POST['shipping_country']; echo "Welcome".$dbusername; //if($_POST['btnSave']) //if ($_POST['btnSave']) //{ //echo "Inside query loop"; $connect = mysql_connect("localhost","root","") or die("Couldn't connect!"); mysql_select_db("salebees") or die ("Couldn't find DB"); //$query = mysql_query("SELECT * FROM users WHERE username='$username'"); $query = mysql_query("update users set firstname = '$dbfirstname', lastname = '$dblastname', mobile_number = '$dbmobile_number', landline_number = '$dblandline_number', dob = '$dbdob', Subscribe_Email_Alert = '$dbSubscribe_Email_Alert', Subscribe_SMS = '$dbSubscribe_SMS', Address_firstname = '$dbAddress_firstname', Address_lastname = '$dbAddress_lastname', Address = '$dbAddress', city = '$dbcity', pincode = '$dbpincode', state = '$dbstate', country = '$dbcountry' where username = '$dbusername' "); header("location:my_account.php"); //} //else //{ //die(); //} ?> Hi Guys, I don't know how to do this but basically I want the value of $postcode3 to equal the value of $postcodep1 aswell as $postcodep2 like this... else {$postcode3 = $postcodep1 $postcodep2;} That doesn't work... Say $postcodep1 = LS14 $postcodep2 = 2AB Then $postcode3 = LS14 2AB How do I write this in the above php code?! Thanks, S Hey everyone, I'm new to this forum and I came here looking for some help with PHP. I've just started getting into PHP and you guys are probably a smile ahead. So basically, I have some inputs and want them to be sent to my email in a basic layout shown below. Code: [Select] Inputs: Title = _title Forum Name = _forumname Name = _irlname Age = _age I want the email to be sent to me like: Code: [Select] Email Title --------------- Forum Name: TextGiven Real Life Name: TextGiven Age: TextGiven Here is my current code which titles the email correctly, but I want to figure out how to make $message set to the example email I showed above. Code: [Select] <?php $to = "******@*********.net"; $subject = $_REQUEST['_title']; $headers = "From: ***** Applicant \r\n"; $headers .= "MIME-Version: 1.0\r\n" . "Content-Type: text/html; charset=\"iso-8859-1\"\r\n" . "Content-Transfer-Encoding: 7bit\r\n"; $message = $_REQUEST['_forumname' & '_irlname']; <--- Part I want fixed if(mail($to, $subject, $message, $headers)) { echo 'MAIL SENT'; } else { echo 'MAIL FAILED'; } ?> I have this code, and it works. But I want $message = "" to make the message contain not just the forum name, but as well as the name, age, and all other inputs I want. Hi all, I can't seem to designate an array key by using a variable and I was wondering if this is possible. I'm looking to do something like this: Code: [Select] <?php $key = "apple"; $arr = array($key => "fruit"); ?> Any suggestions would be appreciated! hi there, i'm trying to put a message in the footer of a page which welcomes a person who is logged in with his/her name, using sessions of course; when i place this: Code: [Select] $username = $_SESSION['valid_user']; in the footer before the echo: Code: [Select] echo "You are logged in as $username"; but the session is also needed before the footer to use the username for other things , such as -checking his credit- so if place in footer the footer shows name in browser but checking credit would not happen as the assignment is at the buttom. IF i place the assignment above at the top of the file: everything works for the user and checking credit ..etc...but the footer is not there... cud not put this any clearer..sorry...hope if someone cud help...thanks hi, i am updating records from database using ajax and javascript on php page. The result is displayed inside a div (<div id="show"></div>). Now i want to assign the content of the above div(say y) to php variable for further calculations. How can i assign the value displayed in div tag to a php variable? Thanks. Hi, I have just started creating my first class in php. I'm trying to assign $_SERVER['REMOTE_ADDR']; to a protected variable but I keep getting an error message. I'm still trying to get my head around oop php. My current code is "protected $user_ip = $_SERVER['REMOTE_ADDR'];" and the error message is "Parse error: syntax error, unexpected T_VARIABLE". Hello, I am creating a class which contains one private function that deals with connecting to a SQL Server and executing the sql that is passed in to that function. I have defined a class variable which is assigned the value of sqlsrv_query like so. $this->QueryResult = sqlsrv_query($conn,$sql); I have placed private $QueryResult at the top of my class. The other public methods call this private function to assign the result set to the class variable, the private method returns and then the public methods loop through the results array like so. while($row = sqlsrv_fetch_array($this->QueryResult)) .. but the while loop never gets entered. If I declare everything in the same method then it works, however there are going to be several public methods that will use this private method and I don't want to duplicate all the database coding. Hope someone can help Hello Everyone, I am new to forum and could use some help with some php code that isn't working. I am very new to php/html/javascript and all of what I have learned, I learned from forums like this one so first....thank you! I am trying to assign a value from a php variable to the value of my form element. I'm sure there must be a way to do this but I can't seem to get the syntax right. here is my code... first I set the value of $loginname elsewhere in the script like so... <?php $loginname =strtolower(htmlspecialchars(strip_tags($_GET["loginname"]))); ?> This part works fine.. Then I try to set the value of my hidden text field inside the form to the value of $loginname to be passed to a javascript program. Everything works except that the value passed ends up being <?echo and not the expected user name inside of $loginname. <?php echo '<form name ="currentactivity" Id="currentactivity" action="<?php'.htmlspecialchars($_SERVER['PHP_SELF']).'?>" method="post">'; echo '<fieldset><legend><b>Your Current Activity Information</b></legend>'; echo '<input type="text" name="loginnm" style="visibility: hidden" value="<?php echo $loginname;?>">'; echo "<label for='myactivities'>Activity Name:</label>"; echo "<select name='myactivities' Id='myactivities' onchange=\"showdetails(this.form)\" value=''>"; echo "<option value = 'Select an activity'>Select an activity</option>"; for ($i = 1; $i <=$rowcount; $i++) { echo"<option value=$row[activity_name]>$row[activity_name] </option>"; $row = mysql_fetch_array($result); } echo "</select>"; echo '</fieldset>'; echo '</form>'; ?> Please note..the rest of the code is working perfectly, it is just this one value I can't seem to get. Any help you can give will be greatly appreciated. Okay, really newbie question, but for this code... Code: [Select] <!-- Gender --> <label for="gender">Gender:</label> <select id="gender" name="gender"> <option value="">--</option> <option value="F">Female</option> <option value="M">Male</option> </select> 1.) How do I assign a variable to this? 2.) How do I make this "sticky"? Here is how I have usually done other form types... Code: [Select] <!-- First Name --> <label for="firstName">First Name:</label> <input id="firstName" name="firstName" type="text" maxlength="30" value="<?php if(isset($firstName)){echo htmlentities($firstName, ENT_QUOTES);} ?>" /><!-- Sticky Field --> <?php if (!empty($errors['firstName'])){ echo '<span class="error">' . $errors['firstName'] . '</span>'; } ?> Oh, by the way, at the top of my PHP file I have this code... Code: [Select] if ($_SERVER['REQUEST_METHOD']=='POST'){ // Form was Submitted (Post). // Initialize Errors Array. $errors = array(); // Trim all form data. $trimmed = array_map('trim', $_POST); Thanks, Debbie hey guys, I'm quite new to PHP and i was wondering if someone would be able to help me out with this. The code i have checks the database to see if a user has provided a Vimeo ID. If they haven't, $video_check will equal a line of html that says the user hasn't added any videos to their portfolio. If the a vimeo ID is present, i want $video_check to equal a bunch of html and css with a foreach function inside that is used to display the users videos from vimeo. The foreach function works fine when its not assigned to the $video_check variable. How do i structure it so that it displays correctly? If you click on videos on this page you might get a better idea of what i'm talking about. http://www.myfilmportfolio.ie/profile.php?id=33 and here is the code i'm having the problem with: /////// check if user has provided vimeo id ////////////////////////// $vimeoID = $row["vimeoID"]; $video_check=''; if (empty($vimeoID)){ $video_check = '<h3>'.$firstname .' has not added any videos to their portfolio</h3>'; } else{ $video_check = '<div id="stats"> <div style="clear: both;"></div> </div> <div id="wrapper"> <div id="embed"></div> <div id="thumbs"> <ul> <?php foreach ($videos->video as $video): ?> <li> <h4><?=$video->title?></h4> <a href="<?php echo $video->url ?>"> <img src="<?php echo $video->thumbnail_medium ?>" class="thumb" /></a> <p><?=$video->description?></p> <br /> </li> <?php endforeach ?> </ul> </div> </div>'; } if anyone could help me out id really appreciate it. Cheers, G Hi, My company has 240+ locations and as such some users (general managers) cover multiple sites. When I run a query to pull user information, when the user has multiple sites to his or her name, its adds the second / third sites to the next columns, rather than wrapping it inside the same table cell. It also works the opposite way, if a piece of data is missing in the database and is blank, its pull the following columns in. Both cases mess up the table and formatting. I'm extremely new to any kind of programming and maybe this isn't the forum for this question but figured I'd give it a chance since I'm stuck. The HTML/PHP code is below: <table id="datatables-column-search-select-inputs" class="table table-striped" style="width:100%"> <thead> <tr> <th>ID</th> <th>FirstName</th> <th>LastName</th> <th>Username</th> <th>Phone #</th> <th>Location</th> <th>Title</th> <th>Role</th> <th>Actions</th> </tr> </thead> <tbody> <?php //QUERY TO SELECT ALL USERS FROM DATABASE $query = "SELECT * FROM users"; $select_users = mysqli_query($connection,$query);
// SET VARIABLE TO ARRAY FROM QUERY while($row = mysqli_fetch_assoc($select_users)) { $user_id = $row['user_id']; $user_firstname = $row['user_firstname']; $user_lastname = $row['user_lastname']; $username = $row['username']; $user_phone = $row['user_phone']; $user_image = $row['user_image']; $user_title_id = $row['user_title_id']; $user_role_id = $row['user_role_id'];
// POPULATES DATA INTO THE TABLE echo "<tr>"; echo "<td>{$user_id}</td>"; echo "<td>{$user_firstname}</td>"; echo "<td>{$user_lastname}</td>"; echo "<td>{$username}</td>"; echo "<td>{$user_phone}</td>";
//PULL SITE STATUS BASED ON SITE STATUS ID $query = "SELECT * FROM sites WHERE site_manager_id = {$user_id} "; $select_site = mysqli_query($connection, $query); while($row = mysqli_fetch_assoc($select_site)) { $site_name = $row['site_name']; echo "<td>{$site_name}</td>"; } echo "<td>{$user_title_id}</td>"; echo "<td>{$user_role_id}</td>"; echo "<td class='table-action'> <a href='#'><i class='align-middle' data-feather='edit-2'></i></a> <a href='#'><i class='align-middle' data-feather='trash'></i></a> </td>"; //echo "<td><a href='users.php?source=edit_user&p_id={$user_id}'>Edit</a></td>"; echo "</tr>"; } ?>
<tr> <td>ID</td> <td>FirstName</td> <td>LastName</td> <td>Username</td> <td>Phone #</td> <td>Location</td> <td>Title</td> <td>Role</td> <td class="table-action"> <a href="#"><i class="align-middle" data-feather="edit-2"></i></a> <a href="#"><i class="align-middle" data-feather="trash"></i></a> </td> </tr> </tbody> <tfoot> <tr> <th>ID</th> <th>FirstName</th> <th>LastName</th> <th>Username</th> <th>Phone #</th> <th>Location</th> <th>Title</th> <th>Role</th> </tr> </tfoot> </table>
Hi all, Just curious why this works: Code: [Select] while (($data = fgetcsv($handle, 1000, ",")) !== FALSE){ $import="INSERT into $prodtblname ($csvheaders1) values('$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]')"; } And this does not: $headdata_1 = "'$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]'"; while (($data = fgetcsv($handle, 1000, ",")) !== FALSE){ $import="INSERT into $prodtblname ($csvheaders1) values($headdata_1)"; }it puts $data[#'s] in the database fields instead of the actual data that '$data[0]','$data[1]'... relates to. I wrote a script to create the values in $headdata_1 based on the number of headers in $csvheaders1 but can't seem to get it working in the sql statement. Thanks I get the following error when I try to pass a value to a methiod in a loop: Warning: Attempt to assign property of non-object in /Users/staceyschaller/Sites/dev_zone/ckwv2/classes/class.php on line 670 This one has me very baffled. It will work the first time, and seems to work every other time, so I have no clue what is wrong. Here is the code: This code is part of my "display" class: function display_partner ($type,$loc,$rand=0,$narrow=0) { $this->partners = new partner($this->cxn); $display = ' <div id="cont_info" class="partner-list"> <div> <h3 class="settings">'.ucfirst($loc).' '.ucfirst($type).last_letter($type).'s</h3> </div> <div class="settings-value" style="height:12px;padding:0;margin:0;text-align:right;padding-right:10px;"> <a href="" class="trunc">Add your organization to this list</a></p> </div> <div style="height:2px;padding:0;margin:0;"> <hr class="account" /> </div> '; $ids = $this->partners->get_partners_list($type,$loc,$rand); for ($b=0;$b<sizeof($ids->id);$b++) { $this->partnerID = $ids->id[$b]; $display .= ($narrow)? $this->card_partner_narr():$this->card_partner(); if ($b!=(sizeof($ids->id)-1)) { $display .= '<hr class="account" />'; } } if (sizeof($ids->id)==0) { $display .= '<div style="color:#999999;display:line;text-align:center;height:20px;">No Partners found for '.ucfirst($loc).' '.ucfirst($type).'</div>'; } $display .= ' </div>'; return $display; } function card_partner () { $this->partners->set_partner_id($this->partnerID); $part_info = $this->partners->get_partner_info(); if ($part_info) { $display .= ' <table class="settings"> <tr> '.$this->show_if($part_info['partLogo']['val'],'<td class="settings-value" rowspan="2"><img src="'.LOGO_FOLDER.$part_info['partLogo']['val'].'" '.resize_img(LOGO_FOLDER.$part_info['partLogo']['val'],175).'alt="'.$part_info['partName']['val'].'" /></td>').' <td class="settings-value" colspan="2"><h5>'.$part_info['partName']['val'].'</h5></td> </tr> <tr> <td class="settings-value"> <span style="color:999999;">'.$part_info['partAddress']['val'].'<br /> '.$part_info['partCity']['val'].', '.$part_info['partST']['val'].' '.$part_info['partZIP']['val'].'<br /> '.$part_info['partPhone']['val'].'</span><br /> <a href="'.$this->form->show_href($part_info['partWeb']['val']).'" target="_blank">'.$part_info['partWeb']['val'].'</a> </td> <td class="settings-value">'.$part_info['partInfo']['val'].'</td> </tr> </table> '; } return $display; } This code is part of my "partners" class: function set_partner_id($partID) { echo '<p>partID: '.$partID.' '.gettype($partID).'<br> $this->partner->id: '.$this->partner->id.'</p>'; $this->partner->id = $partID; ///*** ERROR HAPPENS HERE ***/ echo '<p>id set: '.$this->partner->id.'<br> $this->partner->id: '.$this->partner->id.'</p><hr>'; } function get_partner_id() { return $this->partner->id; } // gets user info at login function get_partner_info() { $this->partner = $this->cxn->proc_info('partner','partID',$this->partner->id);//$this->partner->id return $this->partner; } The following is the output generated: partID: 24 string $this->partner->id: id set: 24 $this->partner->id: 24 partID: 26 string $this->partner->id: Warning: Attempt to assign property of non-object in /Users/staceyschaller/Sites/dev_zone/ckwv2/classes/class.php on line 670 id set: $this->partner->id: partID: 17 string $this->partner->id: id set: 17 $this->partner->id: 17 partID: 25 string $this->partner->id: Warning: Attempt to assign property of non-object in /Users/staceyschaller/Sites/dev_zone/ckwv2/classes/class.php on line 670 id set: $this->partner->id: As you can see, the value passes to $this->set_partner_id($partID) each time. It is formatted as a string. When it assigns the value to $this->partner->id, however, sometimes it works, and sometimes it doesn't. It's probably something obvious, but I've racked my brain to see what it is. Any ideas? Hey all, I want to have an object that has a property which is an object containing instances of other objects. I try this: Code: [Select] class Blog extends Posts { public $has_posts; public function __construct($a,$b,$c){ $has_posts = (object) array_merge((array) $a, (array) $b, (array) $c); } } class Posts { public $b; public function __construct($b){ $this->b = $b; } } $post1 = new Posts(1); $post2 = new Posts(2); $post3 = new Posts(3); $blog = new Blog($post1,$post2,$post3); var_dump($blog->has_posts); //null foreach($blog->has_posts as $post){ //Invalid argument supplied for foreach() echo $post->b; } But as you see, has_posts is null, not an object containing other objects. Thanks for response. Is is possible to give multiple values in a single checkbox? i need to have a checkbox that holds multiple values. Help Anyone.. I have a form that has a list of checboxes, each checkbox has multiple values. IE. Checkbox1 has Date 1 and Cost1. Checkbox2 has Date2 and cost2 and so on. This is what I have in form. echo "<input type=checkbox name=service_id[] value=".$id."><label>".$description."</label> Date:<input type=date name=date[] value=".date('Y-m-d').">Date:<input type=number name=cost[] value=".$cost."><br>"; When i run the for loop to insert the values into a table i have this.
foreach($_POST['service_id'] as $key => $value){
The problem is that when I select certain boxes (ie, checkbox #2), it inserts Date1 and Cost1 but does use the correct service id (ie checkbox) I'm sorting out a simple PHP email form, and I would like, when the email arrives in my inbox, for it to display both the email of the sender AND their name as specified in the form on my website. So far I have $from = $_POST['from']; with the 'from' being the client's email address. I would also like their name ('name') to show up next to their email in the header. I have tried such things like $from = $_POST['from' . 'name']; //or $from = $_POST['from']; $_POST['name']; and a few more, but I'm terrible at PHP coding in general. I have searched, but I really am at a loss as to what to search for. Can someone please give me an idea as to what I should do? Thanks, Sacha. i've google but no luck on finding what i need....is there a way in javascript to compare 2 variables ie.
var him = "jack"; var her = "jill"; if (him == her){ alert('same name'); }thank you What I have is a page with a list of cars, each with a checkbox next to them. Code: [Select] <?php $sth = null; $count = 0; $sth = $dbh->prepare("SELECT * FROM garage WHERE warehouse_id = ? ORDER BY car_name ASC"); $sth->execute(array($_POST['ware_id'])); echo "<table>"; echo "<tr>"; echo '<td>Choose Vehicles:</td>'; echo "</tr>"; echo "<tr>"; echo "<td>"; echo "<form action='warehouse_ship_3.php' method='POST'>"; while($row = $sth->fetch()){ echo "<input type='checkbox' name='cars' value='".$row['uci']."' /> ID: ".$row['uci'].", ".$row['car_year']." ".$row['car_name']."<br />"; } echo "<input type='hidden' name='ware_id' value='".$_POST['ware_id']."' />"; echo "<hr />"; echo '<div class="center"><input class="myButton" type="submit" name="submit" value="Next" /></div></form>'; echo "</td>"; echo "</tr>"; echo "</table>"; ?> That works fine. What I'm trying to do with this second page, is select all the car's info from a table called "garage" which stores data for all the cars (car name, year, etc) and display it. UCI stands for Unique car id, every car has a different id. It works when the user only selects one car on the previous page, but if they select two or more cars, only the car with the highest UCI number shows up. How would I work it to display info on every car that they selected? Code: [Select] <?php if(empty($_POST['cars'])) { echo("You didn't select any cars."); } else { $sth = $dbh->prepare("SELECT * FROM `garage` WHERE `uci` = ?"); $sth->execute(array($_POST['cars'])); while($cars = $sth->fetch()){ echo" ".$cars['uci'].", ".$cars['car_year']." ".$cars['car_name']." "; } } ?> |