jpergega

Hi I have a login form but when I click on the submit button it does nothing. I looked up and some people said is to do with JQuery but I don't know JQuery at all. Could someone please tell me how can I make the submit button work please? Here is my code: <div id="loginbox" style="margin-top:50px;" class="mainbox col-md-6 col-md-offset-3 col-sm-8 col-sm-offset-2"> <div class="panel panel-info" > <div class="panel-heading"> <div class="panel-title">Sign In</div> <div style="float:right; font-size: 80%; position: relative; top:-10px"><a href="#">Forgot password?</a></div> </div> <div style="padding-top:30px" class="panel-body" > <div style="display:none" id="login-alert" class="alert alert-danger col-sm-12"></div> <form id="loginform" class="form-horizontal" action="signup.php" method="POST"> <div style="margin-bottom: 25px" class="input-group"> <span class="input-group-addon"><i class="glyphicon glyphicon-user"></i></span> <input id="login-username" type="text" class="form-control" name="username" value="" placeholder="username"> </div> <div style="margin-bottom: 25px" class="input-group"> <span class="input-group-addon"><i class="glyphicon glyphicon-lock"></i></span> <input id="login-password" type="password" class="form-control" name="password" placeholder="password"> </div> <div style="margin-bottom: 25px" class="input-group"> <span class="input-group-addon"><i class="glyphicon glyphicon-user"></i></span> <select id="type" class="form-control" Placeholder="Select option.."name="permission"> <option value="Student" selected="selected">Student</option> <option value="Staff">Staff</option> <option value="Admin">Administrator</option> </select> </div> <div class="input-group"> <div class="checkbox"> <label> <input id="login-remember" type="checkbox" name="remember" value="1">Remember me</label> </div> </div> <div style="margin-top:10px" class="form-group"> <!-- Button --> <div class="col-sm-12 controls"> <input type="button" name="submit" value="Sign in"> <!--<a id="btn-fblogin" href="" class="btn btn-primary">Login with Facebook</a>--> </div> </div> <div class="form-group"> <div class="col-md-12 control"> <div style="border-top: 1px solid#888; padding-top:15px; font-size:85%" > Don't have an account! <a href="#" onClick="$('#loginbox').hide(); $('#signupbox').show()"> Sign Up Here </a> </div> </div> </div> </form> </div> </div> </div>

Thank you

Hi, I have a table contains 2 foreign keys and I am having trouble inserting a record to this table. Below is my code: $query = " INSERT INTO reply ( foreignKey1_id,foreignKey2_id,message,date_Created ) VALUES ( '$foreignKey1','$foreignkey2', '$msg',NOW() )"; This is the error message: Cannot add or update a child row: a foreign key constraint fails.................... I have used the disable foreign key checks but still getting this error. Is there other way to make this work? Thanks

Hi, I want to add new records to a table and there is a foreign key in this table, so I set the disable foreign key via phpmyadmin export, then I re-imported back to the database. However, I am still getting the cannot add or update a child table message. Please tell me what I'm doing wrong here. Thanks $postNT = " INSERT INTO `tbl_A` ( id, message, date, ) VALUES ( '{$_SESSION['id']}', '{$msg}', NOW())";

It says getImage()

Thanks for this I am using WAMP Server I changed my files' extension to .php and is now giving me error message inside the image frame says: call to undefined function functionName() Here is how I call the function : <img class="img-responsive" src="<?php include 'functions.php'; getImage(); ?>" >

Sorry this is what I got: <a class="thumbnail" href="#"> <img class="img-responsive" src="<?php getImage(); ?>" > </a>

I only can see this error message : Failed to load resource: the server responded with a status of 403 (Forbidden)

Hi I am trying to fetch photos from the database but it doesn't show on the website page: PHP - function getImage(){ $query = "SELECT * FROM imgLib WHERE id=$_SESSION['id']"; $result = mysqli_query($con, $query); while($row = mysqli_fetch_assoc($result)){ echo $row['img']; } } HTML - <img class="img-responsive" src="<?php getImage(); ?>" > Result on Website page - see attachment Thanks

Great thank you is working now

Thanks I found the solution

Hi My query is like this and is saying there is an syntax error near 'cusomter_id')}' what am I doing wrong here? $query = " INSERT INTO `imglib` ( `img_name`, `img_type`, `img_size`, `img`,`customer_id` ) VALUES ( '{$name}', '{$mime}', {$size}, '{$data}','{SELECT(customer_id FROM customer WHERE customer_id='cusomter_id')}' )"; Thanks

Hi My query is like this and is saying there is an syntax error near 'cusomter_id')}' what am I doing wrong here? $query = " INSERT INTO `imglib` ( `img_name`, `img_type`, `img_size`, `img`,`customer_id` ) VALUES ( '{$name}', '{$mime}', {$size}, '{$data}','{SELECT(customer_id FROM customer WHERE customer_id='cusomter_id')}' )"; Thanks

Sorry by the way I am using Chrome

Hi Here is my code but I am not sure why is not working could you please help? Thanks <input id="txt" type="text" name="Name" spellcheck="true" required/>

Right now my timeline chart doesn't show up at all in my web page I am not sure what is wrong with my code is not given me any errors. I think this could be something to do with the date format I set but I am not sure. I double checked the date format and the one I am using it is matching with my database. I would be grateful if you could point out where when wrong. Thanks <?php$sth = "SELECT * FROM goal WHERE account_id =$account_id";$result4 = mysqli_query($con,$sth) or die(mysqli_error());$rows = array();$flag = true;$table = array();$table['cols'] = array(array('type' => 'string', 'label' => 'goalName'),array('type' => 'date', 'label' => 'starts'),array('type' => 'date', 'label' => 'finish'));//$rows = array();while($rows = mysqli_fetch_assoc($result4)) {$temp = array();$starts = date("Y-m-d");$finish = date("Y-m-d");$temp[] = array('v' => (string) $rows['goalName']);$temp[] = array('v' => (string) $rows['starts']);$temp[] = array('v' => (string) $rows['finish']);$rows[] = array('c' => $temp);}$table['rows'] = $rows;$jsonTable = json_encode($table);//echo $jsonTable;?><script type="text/javascript" src="https://www.google.com/jsapi?autoload={'modules': [{'name':'visualization','version':'1','packages':['timeline']}]}"></script><script type="text/javascript">google.setOnLoadCallback(drawChart);function drawChart() {var container = document.getElementById('example3.1');var chart = new google.visualization.Timeline(container);var dataTable = new google.visualization.DataTable();var data = "" google.visualization.DataTable(<?php echo json_encode($table, JSON_NUMERIC_CHECK); ?>);var options = {title: 'Progress',is3D: truevar options = {timeline: { colorByRowLabel: true }};chart.draw(dataTable, options);}</script>

Thank you I did change the operator to if($row['finish'] <= $curdate)

Hi I am trying to insert a query output which displayed in a dropdown box to a table in mysql. Right now the problem is that I am getting this error: Error:Unknown column 'A' in 'where clause' (I think is probably something to do with the name being given to the select and the query output name, but I don't know how to sort this out because I have tweaked around and nothing works. Please help! Thanks ) Here is my code: <select name="staff_id"> <option selected="selected" value ="<?php echo $row["staff_id"];?>"><?php echo $row["staff_id"];?></option> </select> <?php if(isset($_POST['save'])){ $staff_id = $_POST['staff_id']; //inserting data order $order2 = "INSERT INTO goal(staff_id) SELECT staff_id FROM staff WHERE staff_id =$staff_id"; if(!mysqli_query($con,$order2)){ die('Error:'. mysqli_error($con)); } ?>

Hi Here is my tables will you be able to tell me how to display whatever is belongs to this student? Thanks CREATE TABLE userAccount ( account_id SMALLINT NOT NULL AUTO_INCREMENT, username varchar(7) NOT NULL, password varchar(8) NOT NULL, firstname varchar(50) NOT NULL, surname varchar (30) NOT NULL, email varchar(50) NOT NULL, permission ENUM('Student', 'Staff') NOT NULL, PRIMARY KEY (account_id) ) CREATE TABLE staff ( staff_id SMALLINT NOT NULL AUTO_INCREMENT, account_id SMALLINT NOT NULL, PRIMARY KEY (staff_id), FOREIGN KEY (account_id) REFERENCES userAccount(account_id) ) CREATE TABLE IF NOT EXISTS `task` ( `task_id` smallint(6) NOT NULL AUTO_INCREMENT, `taskName` varchar(100) NOT NULL, `description` varchar(300) NOT NULL, `starts` date NOT NULL, `finish` date NOT NULL, `progress` varchar(300) NOT NULL, `status` varchar(10) NOT NULL, `account_id` smallint(6) NOT NULL, `staff_id` varchar(100) DEFAULT NULL, PRIMARY KEY (`task_id`), KEY `account_id` (`account_id`), KEY `staff_id` (`staff_id`) )

Thanks My query is this: <?php if(isset($_SESSION['account_id'])) { $account_id = $_SESSION['account_id']; $query = "SELECT staff.staff_id, staff.account_id, task.task_id, task.staff_id, useraccount.username, useraccount.firstname, useraccount.surname, useraccount.account_id FROM useraccount,staff, task WHERE task.account_id=useraccount.account_id AND task.staff_id=staff.staff_id AND staff.account_id=$account_id"; $result= mysqli_query($con,$query);

Hi I have a form and I only want the user to use it only once. How can I disable this form permanently once been submitted? Thanks <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <label>Unique Ref:<input name="reference" type="text" size="30"/> <br><input type="submit" name="save" value="Save"/> </form>

Hi My system has 2 different accounts staff and student. Both users account details(login, names) are stored in the useraccount table. Student has another table called Task. I logged in as a staff and I want to view a student's tasks by clicking on the view link next to their name. Right now if I click on a student's view link it will just bring up everything in the task table, but that is not what I want it should only display whatever is belongs to this student. I am not sure how to do this could you please help? <?php while($row = mysqli_fetch_array($result)){ ?> <tr> <td><?php echo $row['username'] ; ?></td> <td><?php echo $row['firstname'] ; ?></td> <td><?php echo $row['surname'] ; ?></td> <td><a href="viewStudent.php<?php echo '?task_id='.$row['task.account_id']; ?>">View Account</a></td> <?php } } ?>

Earlier I changed to this $today = date("Y-m-d H:i:s"); Now I have got rid of the time $today = date("Y-m-d"); is working now Thanks

Thanks I read the date function link you provided and I have changed the date format but still nothing happened. Any other suggestions that I am doing wrong?

Hi I have a alert box here that want to pop up when a user is logged into their account. The alert box will tell the user there is a expired task. Here is my code but is not doing anything please could you tell me what is wrong with it? because is not showing any errors. Thanks <?php if(isset($_SESSION['account_id'])) { $account_id = $_SESSION['account_id']; $curdate = date('yyyy/mm/dd'); $query4 = "select finish from task WHERE account_id=$account_id"; $result4= mysqli_query($con,$query4); while($row = mysqli_fetch_array($result4)){ if($row['finish'] == $curdate) { echo '<script language="javascript">alert("Your task had been expired.")</script>'; } } } ?>