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PHP - Oop - Echo Inside A Class
This probably doesn't matter much, but I am using a 'Display' class for final output to the browser.
The class will display the obvious HTML header and then display either the full site or the mobile site. It also displays the CSS / JS (which is previously selected in the page-specific controller code as there are variations based on server-side checks). Basically it is the ONLY class that actually needs to send anything to the browser. When I begin output and echo inside the class, that is the end of the script - there is no more server-side code to execute. If I shouldn;t echo inside the class, is it that bad to do so? Similar TutorialsI have an existing instance of my class Database, now I want to call that instance in my Session class, how would I go about doing this? Ok. I know you can pass the object of a class as an argument. Example: class A { function test() { echo "This is TEST from class A"; } } class B { function __construct( $obj ) { $this->a = $obj; } function test() { $this->a->test(); } } Then you could do: $a = new A(); $b = new B($a); Ok so that's one way i know of. I also thought that you could make a method static, and do this: (assuming class A's test is 'static') class B { function test() { A::test(); } } But that is not working. I'd like to know all possible ways of accomplishing this. Any hints are appreciated. thanks Hi people! class FirstOne{ public function FunctionOne($FirstInput){ //do stuff and output value return $value1; } } Then:- class SecondOne{ public function FunctionTwo($AnotherInput){ //do stuff and output value return $value2; } } What I want to know is this, if I want to use FunctionOne() in Class SecondOne do I do it like this:- (Assume as I have instantiated the first class using $Test = new FirstOne(); ) class SecondOne{ function SecondedFunction(){ global $Test; return $Test->FunctionOne(); } public function FunctionTwo($AnotherInput){ //do stuff and output value return $value2; } public function FunctionThree(){ //some code here $this->Test->SecondedFunction();<--I think as I can omit the $this-> reference } } My point is: Do I have to do it this way or is there way of having this done through __construct() that would negate the need for a third party function? I have a version working, I just think that it is a little convoluted in the way as I have done it, so I thought I would ask you guys. Any help/advice is appreciated. Cheers Rw So I need to echo a row from my database with php, but where i need to echo is already inside an echo. This is my part of my code: $con = mysql_connect("$host","$username","$password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("main", $con); $result = mysql_query("SELECT * FROM Vendor"); while($row = mysql_fetch_array($result)) { //I need to echo right here .................. but I get a blank page when I try this. Please Help. echo '<option value=$row['vendor_id']>'; echo $row['vendor_id']; echo '</option>'; } mysql_close($con); Result: A Blank page. Thanks in advance! Is it possible to echo php inside php? Im guessing not because you are already inside php but there must be a way round my problem. What I would like to do is if the url has project = something then echo <?php require "footer.php"; ?> if not dont echo anything. What is the best way to go about doing this ? how can i put an if statement inside an echo this is what i want to do Code: [Select] <?php echo "<li> <a class='thumb' href='../images/Sections/pinkpanthers/" . $year . "/" . $sessions . "/" . $day . "/" . $y . ".jpg' title=''> <img src='../images/Sections/pinkpanthers/" . $year . "/" . $sessions . "/" . $day . "/thumbs/" . $y . ".jpg ' /></a><div class='caption'> <div class='download'><a href='../images/Sections/pinkpanthers/" . $year . "/" . $sessions . "/" . $day . "/big/" . $y . ".jpg ' target='_blank' />Download</a> </div><div class='image-desc'>" . if ($count == 1){echo "<a href='tag.php?tag=$name'" . $name . ">" . $name . "</a>";}if ($count > 1){$z = 0;$w = $count - 1;while ($z <= $w){$p = $multi[$z];echo "<a href='tag.php?tag=$p'" . $p . ">" . $p . "</a>";$z++;}} . "</div></div></li>"; ?> all the code works wright up until i added the if statements in the last little bit Hello, I need to add rel="..." in the first part of the script but it is not working, below is the script: Code: [Select] echo " <a href='products/".$image[$abc]."' target='_blank'><img border=0 src='products/thumbs2/".$image[$abc]."' /></a>"; Thank you, Currently whether data from the queries are found or not, its surrounded inside the div tags... i want to somehow echo the div tags inside the while loop, but only echo once. the confusing part for me is that the queries are connecting to 2 different tables for data.. and both need to be found for the div to be echo. if($item['quantity'] >= 1){ if($item2['type'] == 'weapon'){ code below.... Code: [Select] <div class="g_content"><h3 style="text-align: left"> Weapons</h3><div class="g_text"> <table align='center' cellspacing='10'> <?php $current_col = 1; $max_col = 4; $query = $db->execute("select * from items where `player_id`=?", array($player->id)); while($item = $query->fetchrow()) { $query2 = $db->execute("select * from `blueprint_items` where `id`=?",array($item['item_id'])); $item2 = $query2->fetchrow(); //DISPLAY IF QUANTITY IS 1 OR MORE. if($item['quantity'] >= 1){ if($item2['type'] == 'weapon'){ //Open new row if first column if($current_col==1) { echo "<tr>\n"; } //Display current record echo "<td width='25%'>"; echo "<center><img src=\"{$item2['img']}\" width='80' height='80' style=\"border: 1px solid #CC9900\"></center>"; echo "<center><a href=\"../description.php?id={$item2['id']}\">{$item2['name']}</a> [x".$english_format_number = number_format($item['quantity'])."]</center>"; echo "<center>$".$english_format_number = number_format($item2['value'])."</center>"; echo "<center>[<a href='../item.php?sell=".$item['id']."'>Sell</a>] [<a href='../item.php?market=".$item['id']."'>Market</a>] <br>[<a href='../item.php?send=".$item['id']."'>Send</a>] [<a href='../item.php?equip=".$item['id']."'>Equip</a>]</center><br>"; echo "</td>\n"; //Close row if last column if($current_col==$max_col) { echo "<tr>\n"; $current_col = 0; //<---Changed } $current_col++; } } } //Close last row if needed if ($current_col!=1) { for(; $current_col<=$max_col; $current_col++) { echo "<td> </td>\n"; } } ?> </table> </div></div> Hello guys im new to php i been coding for like a week now, i need some help here i have been stuck for like 6 hours XD,is it possible to write an IF Statement inside an echo? //this is what i want to do // echo a table and a delete button // and if you click on the delete button it echoes out "DELETED" echo " <table width='528px'> <tr> <td> </td> <td> <center><font size='5'>$tittle</font></center><br> </td> </tr> <tr> <td> </td> <td> $message </td> </tr> <tr> <td> </td> <td> <font size='1'>Posted By:<font color='green'>$author</font> on <font color='gray'>$date</font> at <font color='gray'>$time</font></font> <input id='delete' name='delete' type='submit' value='Delete' > if ($_POST['delete']) { echo "DELETED"; } <td> </td> </tr><br><br> </table> "; Im having a problem with getting the quotes correct with this. I can either get the variable to work in the href and img src or in the css. Anyone have any ideas on this. Cuase im really stuck. This allows the $address and $cos to echo in but not the css variables <td><?php echo "<a href='planet_profile.php?planet=$address'> <img src='images/star.jpg' id='$cos' style='position:absolute;' left:'$b_x px;' top:'$b_y px;'></a>"; ?></td> and this allows only the css variables to work <td><?php echo '<a href="planet_profile.php?planet="' . $address . '"> <img src="images/star.jpg" id="' . $cos . '" style="position:absolute; left:' . $b_x . 'px; top:' . $b_y . 'px;"></a>'; ?></td> Hi there i just need to assign a class to my echoed vairable. Code: [Select] echo $test class="somecssclass"; Im sure im missing some braces here or something but i cant seem to find it out. Heres the error: Parse error: syntax error, unexpected T_CLASS, expecting ',' or ';' Please help Thanks Can you echo class constants? This does not work for me. echo “here is a class constant: self::CLASS_CONSTANT”; // using double quotesInstead I *must* concatenate it: echo ‘here is a class constant: ‘ . self::CLASS_CONSTANT;But of course I do not have to do this with a normal variable, which will echo a variable fine: echo “my name is $name_variable”;Am I doing something wrong or is this normal behavior? Edited by E_Leeder, 30 October 2014 - 11:04 AM. Hi im having trouble getting the below code to listen to my css margin settings. Just to clear up the server side of things, can anybody tell me wether or not im using the correct syntax in the code below? Code: [Select] else { echo("<p class=\"passed\">Thankyou for submiting your details, you will be added to our directory shortly</p>"); }} I'm redoing my login script using functions and a basic switch function checking against a $_GET variable (hope you guys know what I'm talking about). What I want to do is create two functions: 1 that displays the login form and 1 that processes the information The form tag would look like this: <form action=\"<?php $_SERVER['PHP_SELF']?>?action=process\" method=\"post\"> </form> Here's my switch statement: Code: [Select] <?php //**************************************** //****************Action****************** //**************************************** switch ($_GET["action"]) { default: case "index": if (!$_SESSION["member"]) { if (!$timeout) { display_form(); } else { echo $timeout_error; } } else { echo "You are already logged in."; } break; case "process": if (!$_SESSION["member"]) { if (!$timeout) { process_form(); } else { echo $timeout_error; } } else { echo "You are already logged in."; } break; } ?> This runs as soon as "login.php" loads. It'll automatically run the commands under case "default" and "index". It'll first check to see if the member's logged in. If not, it'll then check to see if "$timeout" is true ($timeout becomes true if the member has attempted to login 5 times and failed). If not, it'll display the login form, by running "display_form()". Once the form has been filled and submitted, the commands under case "process" will be performed. Again it will first check to see if the member's logged in. If not, it'll check to see if "$timeout" is true. If not, it'll start validating the forms. For the validation, I've created a variable called "$errors_found", and scripted one if statement checking to see if the email and password exist. If so, the variables "$rm_field_un" and "$rm_field_pw" become true, as well as "$errors_found". So, back to submission of the form... If "$errors_found" is true, display the form (when the form is displayed, there will be an if statement within that says "<?php if ($rm_field_un) { echo "Username is wrong."; } ?>", which will be displayed right underneath the username field and label. Same with the password elements). If "$errors_found" is not true, go ahead and register the member. Now, here's where I need help, because I'm really confused as to how to accomplish this. The "display_form()" function will contain a single variable called "$login_form". It will contain a value of the HTML constructed login form, and the function will return the variable. Remember how I stated in the third paragraph up that I will have <php> statements within the form which would display errors if necessary? Well, how do I put those if statements within the HTML, which is contained within the variable? If that last question confuses you, allow me to present you with an instance: Code: [Select] <?php function display_form() { $login_form = " <form name=\"login_form\" method=\"post\" action=\"<?php $_SERVER['PHP_SELF']?>?action=process\"> Username: <input type=\"text\"> <?php if ($rm_field_un) { echo "Username is wrong"; } ?> "; return $login_form; } ?> See what I mean? This code confuses me ALOT. I'm not sure if the <php> tags are needed as it is already contained within existing ones, or what. Can somebody please help me out? Any and all help is much appreciated. =D Hi I am trying to use the nl2br function like this
while ($row = mysqli_fetch_array($query)) {
//May need this later to output pictures
// $imageURL = 'upload/'.rawurlencode($row["filename"]);
echo
" <div class='divTableRow'>
<div class='divTableCell'>{$row['User']} ;</div>
<div class='divTableCell'>nl2br({$row['CommentText']});</div>
</div>
\n";
}
However the output just looks like the attached picture. When I check in the sql db I can see the line breaks when doing a select * from Table ;
hello, im trying to add a hyperlink that launches in a new window to the following (in the last column) any ideas? Code: [Select] echo "<tr>"; echo "<td align='center'>" . $row["ID"] . "</td>"; echo "<td align='center'>" . $row["Name"] . "</td>"; echo "<td align='center'>" . $row["jobNO"] . "</td>"; echo "<td align='center'>" . $epn . "</td>"; echo "<td align='center'>" . $cname . "</td>"; echo "<td align='center'>" . $cadd . "</td>"; //want to add a hyperlink here echo "</tr>"; } echo "</table>"; } Trying to figure out how to make it so name is a link to the profile when its echo anyone know how to do this? im at a huge stand still Code: [Select] <?php $sql = "SELECT name FROM users WHERE DATE_SUB(NOW(),INTERVAL 5 MINUTE) <= lastactive ORDER BY id ASC"; $query = mysql_query($sql) or die(mysql_error()); $count = mysql_num_rows($query); $i = 1; while($row = mysql_fetch_object($query)) { $online_name = htmlspecialchars($row->name); echo '<a href="Inbox.php">"'[$goauld]'</a>"'; ?> Hi I am new to PHP and this is my first post one here so appologies is this questions seems a bit dumb! I have an if clause such that if a button is pressed on my web page then i want to reload the page and include a new form on it. I am having a problem getting the $_SERVER['PHP_SELF'] command to work from iside a echo command. I must not be escaping the code correctly with back slashes: I currently have the line : echo"<form method=\"POST\" action=\"\<?$_SERVER['PHP_SELF']?\>\">"; However this doesnt seem to work as my page just doesnt display in the browser. Any advice is much appreciated. Thanks for taking the time to read. how do i get the image to show as the code i have is echoing what i have been told is binary, not confirmed. and not the images. all images are .jpg Code: [Select] $allowed_types = array('png','jpg','jpeg','gif'); $imgdir = '/home/mysite/ftpfolder'; $imgFilesArray = scandir($imgdir); foreach ($imgFilesArray as $imgkey => $imgvalue) { $imgInfo = pathinfo($imgdir . '/' . $imgvalue); $imgExtension = $imgInfo['extension']; if(in_array($imgExtension, $allowed_types)) { readfile($imgdir . '/' . $imgvalue); //echo '<img src = "'. $imgdir .'/'. $imgInfo['basename'].'" alt="" />'; } } images are FTP'd to the ftpfolder and can not be placed anywhere else but this folder due to security reasons. |
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