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  • How to add a method to an existing class in PHP?

    - by sombe
    I'm using WordPress as a CMS, and I want to extend one of its classes without having to inherit from another class; i.e. I simply want to "add" more methods to that class: class A { function do_a() { echo 'a'; } } then: function insert_this_function_into_class_A() { echo 'b'; } (some way of inserting the latter into A class) and: A::insert_this_function_into_class_A(); # b Is this even possible in tenacious PHP?

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  • PHP: Set variable in to include file in 'if' statement

    - by swiftsly
    I am trying to set a variable in an 'if' statement so that only when the 'if' statement is true, the variable will be set and will be able to be used somewhere else on the page. Here's what I have tried: if ($row[8] == 1) { echo 'Message here'; ob_start(); include('F164.php'); $f164 = ob_get_clean(); } as well as: if ($row[8] == 1) { echo 'Message here'; $f164 = include('F164.php'); }

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  • test remote file if directory

    - by soField
    HOSTNAME=$1 #missing files will be created by chk_dir for i in `cat filesordirectorieslist_of_remoteserver` do isdir=remsh $HOSTNAME "if [ -d $i ]; then echo dir; else echo file; fi" if [ $isdir -eq "dir" ] then remsh $HOSTNAME "ls -d $i | cpio -o" | cpio -id else remsh $HOSTNAME "ls | cpio -o" | cpio -id fi done i need simple solution for checking remote file is directory or file ? thanks

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  • How to display after 4 week date from now?I want to pass an argument.

    - by vinothkumar
    echo date( "F jS, Y" , strtotime("now +3 weeks") ); It gives the result as July 2nd, 2010 . Fine.Now I want to pass the argument like this. The original print_r($originalamount) give the result like this Array ( [0] = 4 Months [1] = 3500 ) My code $text=trim($originalamount[0]); $text1="now +".$text; echo date( "F jS, Y" , strtotime($text1)) ; The out put come like this December 31st, 1969 I dont know why?

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  • MySQLi Insert prepared statement via PHP

    - by Jimmy
    Howdie do, This is my first time dealing with MySQLi inserts. I had always used mysql and directly ran the queries. Apparently, not as secure as MySQLi. Anywho, I'm attempting to pass two variables into the database. For some reason my prepared statement keeps erroring out. I'm not sure if it's a syntax error, but the insert just won't work. I've updated the code to make the variables easier to read Also, the error is specifically, Error preparing statement. I've updated the code, but it's not a pHP error. It's a MySQL error as the script runs but fails at the execution of: if($stmt = $mysqli - prepare("INSERT INTO subnets (id, subnet, mask, sectionId, description, vrfId, masterSubnetId, allowRequests, vlanId, showName, permissions, pingSubnet, isFolder, editDate) VALUES ('', ?, ?, '1', '', '0', '0', '0', '0', '0', '{"3":"1","2":"2"}', '0', '0', 'NULL')")) { I'm going to enable MySQL error checking. I honestly didn't know about that. <?php error_reporting(E_ALL); function InsertIPs($decimnal,$cidr) { error_reporting(E_ALL); $mysqli = new mysqli("localhost","jeremysd_ips","","jeremysd_ips"); if(mysqli_connect_errno()) { echo "Connection Failed: " . mysqli_connect_errno(); exit(); } if($stmt = $mysqli -> prepare("INSERT INTO subnets (id, subnet, mask,sectionId,description,vrfld,masterSubnetId,allowRequests,vlanId,showName,permissions,pingSubnet,isFolder,editDate) VALUES ('',?,?,'1','','0','0','0','0','0', '{'3':'1','2':'2'}', '0', '0', NULL)")) { $stmt-> bind_param('ii',$decimnal,$cidr); if($stmt-> execute()) { echo "Row added successfully\n"; } else { $stmt->error; } $stmt -> close; } } else { echo 'Error preparing statement'; } $mysqli -> close(); } $SUBNETS = array ("2915483648 | 18"); foreach($SUBNETS as $Ip) { list($TempIP,$TempMask) = explode(' | ',$Ip); echo InsertIPs($Tempip,$Tempmask); } ?> This function isn't supposed to return anything. It's just supposed to perform the Insert. Any help would be GREATLY appreciated. I'm just not sure what I'm missing here

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  • MySQL Database Query - Codeigniter

    - by user2450349
    I am building an application with Codeigniter and need some help with a DB query. I have a table called users with the following fields: user_id, user_name, user_password, user_email, user_role, user_manager_id In my app, I pull all records from the user table using the following: function get_clients() { $this->db->select('*'); $this->db->where('user_role', 'client'); $this->db->order_by("user_name", "Asc"); $query = $this->db->get("users"); return $query->result_array(); } This works as expected, however when I display the results in the view, I also want to display a new column called Manager which will display the managers user_name field. The user_manager_id is the id of the user from the same table. Im guessing you can create an outer join on the same table but not sure. In the view, I am displaying the returned info as follows: <table class="table table-striped" id="zero-configuration"> <thead> <tr> <th>Name</th> <th>Email</th> <th>Manager</th> </tr> </thead> <tbody> <?php foreach($clients as $row) { ?> <tr> <td><?php echo $row['user_name']; ?> (<?php echo $row['user_username']; ?>)</td> <td><?php echo $row['user_email']; ?></td> <td><?php echo $row['???']; ?></td> </tr> <?php } ?> </tbody> </table> Any idea of how I can form the query and display the manager name is the view? Example: user_id user_name user_password user_email user_role user_manager_id 1 Ollie adjjk34jcd [email protected] client null 2 James djklsdfsdjk [email protected] client 1 When i query the database, i want to display results like this: Ollie [email protected] James [email protected] Ollie

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  • How to manually create Friendly URLs? (PHP)

    - by Ole Jak
    How to manually create Friendly URLs? (PHP) So I have created simple php file which echos requested string. Now it has form echo.php?string=bla+bla+bla&font=times I want to see it like echo/bla+bla+bla/times How to do such thing (not using external libs)?

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  • php for-loop issue

    - by rajesh1984
    Heya, I have a loop which generates a table code for a specific number of times. What I'm having problems with is to echo a variable inside the loop. The loop runs 10 times, and there are 10 text messages sent to the page, so my problem is how do I get each of the looped tables to echo one of the text messages each time?

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  • Pass PHP variables without being seen when working with a database generated list

    - by Wilcoholic
    Looking for any help regarding the problem. Here's the deal: I have a database that has a teams table and it contains team_id. On one of my pages, I generate a list of links that contain the team_id of the creator in their get URL. I need the team_id on the next page but can't figure out how to pass it through any other way. Using a form and POST isn't an option because this method would only pass through the last links data on the list. Storing in a session isn't an option either because there is no way to discretely pass the the variables I need to a function to set the session variables. I have tried and it can pretty easily be viewed from viewing the source code. So here's some sample code to see exactly what I'm dealing with. <? if(mysql_num_rows($result2)>0){ ?> <a class="fltrt btn btn-danger btn-small" onclick="test()" href="acceptmatch-exec.php?match_id=<?php echo $match_id; ?>&team_id=<?php echo $team_id;?>&action=cancel">Cancel Match</a> <?}else{?> <a class="fltrt btn btn-success btn-small" href="acceptmatch-exec.php?match_id=<?php echo $match_id; ?>&team_id=<?php echo $team_id;?>&action=accept">Accept Match</a> <?} ?> The code above is generated multiple times on a page via a while loop that was excluded. I want to pass the match_id and team_id variables without being seen anywhere. If I made this a form, it wouldn't pass the correct variables unless there is only one result at the time (not likely). I'm sure there has to be an easy method that is eluding me, so please share thoughts on how to solve this. I feel as though I am not explaining it well enough, but it's not really easy to explain. I basically want something that works like GET but acts like POST and can be hidden from people viewing source code or link locations. Thanks

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  • grep value inside a variable pointing to other variable

    - by Joice
    using : ksh *abc = 1 efg = 2 hgd = 3 not known to me * say if i have Value="abc efg hgd" abc efg hgd all contains some value which i dnt know. Now I want to grep the value contained inside abc. like for i in $Value do grep "echo $(($((echo $i | cut -d'|' -f2))))" done this grep should look for the value inside abc efg hgd grep 1 grep 2 grep 3

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  • What's the logic flaw in this conditional?

    - by Scott B
    I've created this code branch so that if the permalink settings do no match at least one of the OR conditions, I can execute the "do something" branch. However, I believe there is a flaw in the logic, since I've set permalinks to /%postname%.html and it still tries echo's true; I believe I need to change the ORs to AND, right? if (get_option('permalink_structure') !== "/%postname%/" || get_option('my_permalinks') !== "/%postname%/" || get_option('permalink_structure') !== "/%postname%.html" || get_option('my_permalinks') !== "/%postname%.html")) { //do something echo "true"; }

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  • My php script only reads the first row from mysql and doesn't check the rest of the rows for matches

    - by RobertH
    I'm trying to write a script for users to register to a club, and it does all the validation stuff properly and works great until it gets to the part where its supposed to check for duplicates. I'm not sure what is going wrong. HELP PLEASE!!! Thank you in Advance, <?php mysql_connect ("sqlhost", "username", "password") or die(mysql_error()); mysql_select_db ("databasename") or die(mysql_error()); $errormsgdb = ""; $errordb = "Sorry but that "; $error1db = "Name"; $error2db = "email"; $error3db = "mobile number"; $errordbe = " is already registered"; $pass1db = "No Matching Name"; $pass2db = "No Matching Email"; $pass3db = "No Matching Mobile"; $errorcount = 0; $qResult = mysql_query ("SELECT * FROM table"); $nRows = mysql_num_rows($qResult); for ($i=1; $i< $nRows+1; $i++){ $result = mysql_query("SELECT id,fname,lname,dob,email,mobile,agree,code,joindate FROM table WHERE fname = '$ffname"); if ($result > 0) { $errorcount = $errorcount++; $passdb = 0; $errormsgdb = $error1db; echo "<div class=\"box red\">$errordb $errormsgdb } else { $pass = 1; $errormsgdb = $pass1db; echo "<div class=\"box green\">$errormsgdb</div><br />"; } //--------------- Check if DB checks returned errors ------------------------------------> if($errorcount <= 0){ $dobp = $_REQUEST['day'].'/'.$_REQUEST['month'].'/'.$_REQUEST['year']; $dob = $_REQUEST['year'].$_REQUEST['month'].$_REQUEST['day']; //header('Location: thankyou.php?ffname='.$ffname.'&flname='.$flname.'&dob='.$dob.'&femail='.$femail.'&fmobile='.$fmobile.'&agree='.$agree.'&code='.$code.'&dobp='.$dobp); echo "<div class='box green'>Form completed! Error Count = $errorcount</div>"; } else { echo "<div class='box red'>There was an Error! Error Count = $errorcount</div>"; } } ?>

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  • Autopopulate from Select box from database

    - by Chris Spalton
    hope you can help, please forgive any poor coding or anytihng, I'm new to this and just hacking my way through to get things to work. That said, on one of my projects I have this code, which successfully populates the dropdown from a database when the page is loaded: <select name="Region" id="Region"> <option value="">-- Select Region --</option> <?php $region=$POST['Region']; if ($region); { $regionquery = "SELECT DISTINCT REGION FROM Sales_Execs "; $regionresult = mysql_query($regionquery); while($row = mysql_fetch_array($regionresult)) { echo "<option value=\"".$row['REGION']."\">".$row['REGION']."</option>\n "; } } ?> <script type="text/javascript"> document.getElementById('Region').value = <?php echo json_encode(trim($_POST['Region']));?>; </script> </select> On my next project that I'm working on now, I need to do the same thing, so I copied the above code amended, and placed in my new project: <select name="Sales_Exec" id="Sales_Exec"> <option value="">-- Select SE --</option> <?php $salesexec=$POST['Sales_Exec']; if ($salesexec); { $salesexecquery = "SELECT DISTINCT Assigned FROM Data "; $salesexecresult = mysql_query($salesexecquery); while($row = mysql_fetch_array($salesexecresult)) { echo "<option value=\"".$row['ASSIGNED']."\">".$row['ASSIGNED']."</option>\n "; } } ?> <script type="text/javascript"> document.getElementById('Sales_Exec').value = <?php echo json_encode(trim($_POST['Sales_Exec']));?>; </script> </select> This second chunk of code doesn't work... and I can't work out why as it seems I've copied it all and amended all the neccersary parts, can anyone spot what is wrong? Thankyou!

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  • PHP MySQL query help

    - by user547794
    Hello, I am trying to use this query to return every instance where the variable $d['userID'] is equal to the User ID in a separate table, and then echo the username tied to that user ID. Here's what I have so far: $uid = $d['userID']; $result = mysql_query("SELECT u.username FROM users u LEFT JOIN comments c ON c.userID = u.id WHERE u.id = $uid;")$row = mysql_fetch_assoc($result); echo $row['username'];

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