Text not allowed in element select in this context - php

I created a forum, where students have to fill out their year of graduation. When I was validating my file, as shown by the comment, the line below produces an error, "Text not allowed in element select in this context." I need to name "year" in the select tag, so I can use the post array to get the year. How can I fix this validation error?
<form method="post">
<div class = "mailInput">
Name: <input type="text" name="name">
<span class="error"> <?php echo $nameErr;?></span></div>
<div class = "mailInput">
Email: <input type="text" name="email">
<span class="error"> <?php echo $emailErr;?></span> </div>
<div class = "mailInput">
Major: <input type="text" name="major">
<span class="error"> <?php echo $majorErr;?></span> </div>
<div class = "mailInput">
Year: <select class="select" name="year"> // this line produces an error
<option value="2017">2017</option>
<option value="2018">2018</option>
<option value="2019">2019</option>
<option value="2020">2020</option>
</select>
</div>
<div class = "mailInput lastInput"> Gender:
<input type="radio" name="gender" value="Female">Female
<input type="radio" name="gender" value="Male">Male
</div>
<span class="error"> <?php echo $genderErr;?></span>
<div class = "submitButton"><input type="submit" name="submit" value="Join">
</div>
<div class="message"> <?php echo $submitMessage;?></div>
</form>
if(isset($_POST['submit']) && !empty($_POST['submit'])) {
if (!empty($_POST["year"])) {
$year = $_POST["year"];
}
}

// this line produces an error will prevent your code from working as it is not the correct way to write comments in HTML code.
The correct way is by using <!-- --> and therefore, in your case, it should have been <!-- this line produces an error -->
// is used to write comments in PHP and JavaScript.
/* */ is used to write comments in CSS.

Related

PHP Form Drop Down menu stills blank after I select an item

I am executing an form based on PHP on Windows, however after I select an item in the drop down menu, the text does not show up (even though it was selected properly, as I can verify in the database after clicking the insert button).
The form code:
<div id="janela_formulario">
<form action="inserir_transportadoras.php" method="post">
<input type="text" name="nometransportadora" placeholder="Nome da Transportadora">
<input type="text" name="endereco" placeholder="Endereço">
<input type="text" name="telefone" placeholder="Telefone">
<input type="text" name="cidade" placeholder="Cidade">
<select name="estados">
<?php
while($linha = mysqli_fetch_assoc($lista_estados)) {
?>
<option value="<?php echo $linha["estadoID"]; ?>">
<?php echo utf8_encode($linha["nome"]); ?>
</option>
<?php
}
?>
</select>
<input type="text" name="cep" placeholder="CEP">
<input type="text" name="cnpj" placeholder="CNPJ">
<input type="submit" value="inserir">
</form>
</div>
The dropdown menu PHP code:
<input type="text" name="cidade" placeholder="Cidade">
<select name="estados">
<?php
while($linha = mysqli_fetch_assoc($lista_estados)) {
?>
<option value="<?php echo $linha["estadoID"]; ?>">
<?php echo utf8_encode($linha["nome"]); ?>
</option>
<?php
}
?>
</select>
How can I solve it?
while($linha = mysqli_fetch_assoc($lista_estados)) {
echo '<option value="'. $linha["estadoID"] .'"> '.utf8_encode($linha["nome"]).' </option>';
}
Use above code instead of while loop
test by inspect HTML code. is this estadoID are printing properly if yes then test you are catching (estados) data properly

Not able to print data from Html in Php file

<body>
<span>
<div class="heading">
<h3><img src="http://zicom.com/img/ilayashop-1482233381.jpg" alt="Zicom Logo" ></h3>
<h1><i>Zicom </i> SMS Application</h1><br>
</span>
</div>
<br>
<div>
<form method='post' action=''>
<label for="fname">Contact Number</label>
<input type="text" id="fname" name="number" placeholder="Contact Number....">
<label for="sms" id="msg">Message</label>
<select id="country" name="message">
<option value="" name="message">Blank</option>
<option value="hello" name="message">Abandoned Call</option>
<option value="" name="message">Audit Call </option>
</select>
<input type="text"id="msg" name="message" placeholder="Type your Message...." >
<input type="submit" value="Send" name="send">
</form>
</div>
<?php
if(isset($_POST)) {
$n = $_POST['number'];
$m = $_POST['message'];
echo "$n $m";
}
?>
</body>
</html>
I have used the same method many times and i have succesfully sent the data to php file, but im unable to do the same. I try to print the content of form, but is not executing.

echo a html page and entering values into a form from php variables

I want to block access to a PHP page.
I'm doing that with this way: If you been logged in, PHP check if exist a cookie, and doing echo the HTML, else it's redirecting you to login page.
Here is the code but when I'm trying to set value attribute equal to a PHP variable, I'm getting back the php code ex.""
The PHP code inside the selection tag, isn't working either!
<?php
if(isset($_COOKIE['User_Email_Cookie'])) {
session_start();
$name =$_SESSION['User_FullName'];
$phone =$_SESSION['User_Phone'];
echo '<!DOCTYPE html>
<html>
<body>
<h1 class="Title">Reserve a table now!</h1>
<center>
<form action="reservation2.php" method="post">
<div class="App">
<div class="User">
<h2 style="text-align:left;"> Contact:</h2>
<input type="text" id="Name" placeholder="Full Name" value="<?php echo $name ?>" required>
<input type="tel" id="Phone" placeholder="Phone" value="<?php echo $phone ?>" required>
</div>
<div class="DatePeople">
<h2> Choose the Date:</h2>
<input type="date" id="Date" name="TableDate">
<select name="Time" class="time">
<option>19:00</option>
<option>19:30</option>
<option>20:00</option>
<option>20:30</option>
<option>21:00 </option>
<option>21:30</option>
<option>22:00</option>
</select>
<h2 style="margin-top:0px;">Choose Table, People: <a target="_blank" href="media/diagram.png"><img src="media/info.png" width="23px"></a></h2>
<select name="TableNum" class="table">
<?php
include \'connectDb.php\'; #Eisagwgi stoixeiwn gia syndesi me ti vasi
$result=mysqli_query($con,"SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.Columns WHERE
TABLE_NAME = \'available\' AND COLUMN_NAME NOT IN (\'Date\', \'Time\')");
while($row = mysqli_fetch_array($result)) {
echo \'<option>\'.$row[0].\'</option>\';
}
?>
</select>
<input type="number" id="seats" name="People" min="2" max="8" value="4" >
</div>
</div>
<div>
<input type="submit" name="Submit" value="Reserve">
<a class="button" href="logout.php">Log out</a>
</div> </center>
</form>
else {
header("location: reservation.php");
}
The issue is that you echo the html, and inside that echo you combine "inner" php tags (value="<?php echo $name ?>" instead of value="' . $name . '" for example).
Change:
echo '<!DOCTYPE html>
To:
?><!DOCTYPE html>
And at the end, where you have:
</form>
Replace it with
</form></body></html><?php
The above code allows you combine html markup, by closing the php tags in the correct place, without you having to echo it with php.
Read the documentation for more details.
Please try this code
<?php
if(isset($_COOKIE['User_Email_Cookie'])) {
session_start();
$name =$_SESSION['User_FullName'];
$phone =$_SESSION['User_Phone'];
?>
<!DOCTYPE html>
<html>
<body>
<h1 class="Title">Reserve a table now!</h1>
<center>
<form action="reservation2.php" method="post">
<div class="App">
<div class="User">
<h2 style="text-align:left;"> Contact:</h2>
<input type="text" id="Name" placeholder="Full Name" value="<?php echo $name ?>" required>
<input type="tel" id="Phone" placeholder="Phone" value="<?php echo $phone ?>" required>
</div>
<div class="DatePeople">
<h2> Choose the Date:</h2>
<input type="date" id="Date" name="TableDate">
<select name="Time" class="time">
<option>19:00</option>
<option>19:30</option>
<option>20:00</option>
<option>20:30</option>
<option>21:00 </option>
<option>21:30</option>
<option>22:00</option>
</select>
<h2 style="margin-top:0px;">Choose Table, People: <a target="_blank" href="media/diagram.png"><img src="media/info.png" width="23px"></a></h2>
<select name="TableNum" class="table">
<?php
include \'connectDb.php\'; #Eisagwgi stoixeiwn gia syndesi me ti vasi
$result=mysqli_query($con,"SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.Columns WHERE
TABLE_NAME = \'available\' AND COLUMN_NAME NOT IN (\'Date\', \'Time\')");
while($row = mysqli_fetch_array($result)) {
echo \'<option>\'.$row[0].\'</option>\';
}
?>
</select>
<input type="number" id="seats" name="People" min="2" max="8" value="4" >
</div>
</div>
<div>
<input type="submit" name="Submit" value="Reserve">
<a class="button" href="logout.php">Log out</a>
</div> </center>
</form>
<?php
else {
header("location: reservation.php");
}
?>

select tag name not working while using it in php

I have a form in html and I am saving it as search.php:
<form name="myform" action="" method="POST" onsubmit="search_clicked(); return false;">
Keyword<input type="text" name="Keyword" id="Keyword" value="XYZ" required/><!-- value is the default name that appears in the text box-->
<br>
Type
<select name="sel" id="sel" class="form-control" onchange="checkcolors(this.value)">
<option selected value="Users">Users</option>
<option value="Pages">Pages</option>
<option value="Events">Events</option>
<option value="Places">Places</option>
<option value="Groups">Groups</option>
</select>
</form>
<div id="loc_dist_displayarea" style="display:none;">
Location<input type="text" name="Location" value="90007" required/> Distance(meters)<input type="text" name="distance" value="10000" required/>
</div>
<br><br>
<input type="submit" name="Search"/>
<input type="submit" value="clear" id="clear" onclick="return clearclicked()"/>
</form>
and my php script is in the same file:
<div id="body_area" style="display:none">
<?php
echo "hi I am searching ";
if($_SERVER["REQUEST_METHOD"]=="POST")
{
//echo "yes value is selected";
//echo $_POST["Keyword"];
if (isset($_POST['sel'])) {
$selectedval= $_POST["sel"];
echo "$selectedval";
}
//echo $_POST["Location"];
}
echo "no value is selected";
?>
</div>
I am not able to display the $_POST['sel'] while $_POST['Keyword'] is echoed.Please help.
First of all, you arent using good programming practices, you use quotation marks (these " and these ') Indiscriminately. You should only alternate between them when you have them nested.
Next, on the action paramenter you should put the name of the file, even if it's the same.

PHP echo ID and Pop Up form

I have a pop-up form when user clicks "update", there should be a pop up form where user should be able to update but the pop up does not appear because the "a href" has "echo" in the id value. See below: echo $orderId
Update
But if I delete the <?php echo $orderId ?>part, for example Update the pop up works but echo wrong ID.
Below is the form
<div class="popup">
<form method = "post" action="">
<div>
<label for="Order ID">Order ID:</label><?php echo $orderId ?>
</div><br>
<div>
<label for="Order Date">Order Date:</label><?php echo $orderDate ?>
</div><br>
<div>
<label for="Order Total">Order Total:</label><?php echo $totalAmount ?>
</div><br>
<div>
<label for="Preparation Status">Preparation Status:</label>
<select name="preparationStatus"/>
<option value="<?php echo $preparationStatus; ?>">Default(<?php echo $preparationStatus; ?>)</option>
<option value="Pending">Pending</option>
<option value="Prepared">Prepared</option>
</select>
</div><br>
<div>
<label for="Payment Status">Payment Status:</label>
<select name="paymentStatus">
<option value="<?php echo $paymentStatus;?>">Default(<?php echo $paymentStatus;?>)</option>
<option value="Pending">Pending</option>
<option value="Paid">Paid</option>
</select>
</div><br>
<input type="submit" id="btn" name="submit" value="Save" />
<a class="close" href="#close"></a>
</form>
It only echoes id="id=1"
How can I make the popup form appear?

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