**I have home.php file and superLec.php file. user can enter studentUniversityId and click on the submit button. particular data i want to show studentName textbox and lecturerName textbox. I want to pass data as a array. How to set array data for textbox value.Please any one help me. **.
home.php
<?php
session_start();
include_once 'include/class.supervisor.php';
$supervisor = new Supervisor();
if (isset($_POST['submit'])) {
extract($_POST);
$autofill = $supervisor->check_auto_fill($studentUniversityId);
/* if ($autofill) {
// Registration Success
header("location:homeLecturer.php");
} else {
// Registration Failed
echo 'Wrong username or password';
}*/
}
?>
<form action="" method="post">
<label for="studentID">Student ID</label>
<input type="text" name="studentUniversityId" id="studentID" placeholder="studentID">
<!--input type="submit" id="autoFill" name="autoFill" value="Auto Fill"><br><br>
<input class="btn" type="submit" name="submit" value="Auto Fill"-->
<label for="studentName">Student Name</label>
<input type="text" id="studentName" name="studentName" value = "<?php echo print_r($autofill);?>" placeholder="Student Name"><br><br>
<label for="lecturerName">Supervisor Name</label>
<input type="text" id="lecturerName" name="lecturerName" value = "<?php echo ($lecturerName) ;?>" placeholder="Supervisor Name"><br><br>
<input type="submit" id="submit" name="submit" value="submit">
</form>
**This is my php file. I want pass array values for the home.php file. can any one help me pls. **
superLe.php
public function check_auto_fill($studentUniversityId){
$query = "SELECT supervisor.lecturerName, supervisee.studentName, supervisee.studentUniversityId FROM supervisee, supervisor WHERE supervisee.lecturerId = supervisor.lecturerId AND supervisee.studentUniversityId = '$studentUniversityId' ";
$result = $this->db->query($query) or die($this->db->error);
$supervisor_data = $result->fetch_array(MYSQLI_ASSOC);
$arr = array("lecturerName", "studentName", "studentName");
return $arr;
}
You have to use JQuery function to set value of form elements.
Post studentUniversityId via ajax get data in response and populate the data to form element.
To create a response with array, you need to extract the fetched data from query and put them into the output array.
I suggest you to use a key/value array.
public function check_auto_fill($studentUniversityId){
$query = "SELECT supervisor.lecturerName, supervisee.studentName, supervisee.studentUniversityId FROM supervisee, supervisor WHERE supervisee.lecturerId = supervisor.lecturerId AND supervisee.studentUniversityId = '$studentUniversityId' ";
$result = $this->db->query($query) or die($this->db->error);
$supervisor_data = $result->fetch_array(MYSQLI_ASSOC);
$arr = array(
'lecturerName' => $supervisor_data[0]['lecturerName'], //the "0" index depends of the query structure, but as I can see, this is a right way
'studentName' => $supervisor_data[0]['studentName'],
)
//$arr = array("lecturerName", "studentName", "studentName");
return $arr;
}
After that in your home.php you can simple extract the data from array and put the elements in the right fields as exampled:
<label for="studentName">Student Name</label>
<input type="text" id="studentName" name="studentName" value = "<?php echo $autofill['studentName']; ?>" placeholder="Student Name"><br><br>
<label for="lecturerName">Supervisor Name</label>
<input type="text" id="lecturerName" name="lecturerName" value = "<?php echo $autofill['lecturerName']; ?>" placeholder="Supervisor Name"><br><br>
Hope it help you.
Marco
From what I see and understand (please excuse if I don't), you want to populate the text fields studentName and lecturerName with their respective values. To me, you code looks correct except one thing. You need to drop the line
$arr = array("lecturerName", "studentName", "studentName");
from superLe.php. The line
$supervisor_data = $result->fetch_array(MYSQLI_ASSOC);
already returns an associative array, which you can use in the variable $autofill in file home.php as $autofill['studentName'] or $autofill['lecturerName'].
Related
i'm doing this exercise to understand te concepts and syntax of php.The error i get is that the array is not being appended by a new entry from a text box, if i hard code the values the program works fine and prints the array values. but as soon as i put the variable it just assign a new value to the [0] index.
here is the code:
<label for="Name">Student Name:
<input type="text" name="StdName" placeholder="Your name">
</label>
<label for="Name">Grade:
<input type="text" name="StdGrade" placeholder="Your Grade">
</label>
<input type="submit" name="submit"value="submit">
</form>
<?php
// Associative array new
$studentName = $_POST['StdName'];
$studentGrade = $_POST['StdGrade'];
$classGrades = Array();
$classGrades['name'][] = $studentName;
$classGrades['grade'][] = $studentGrade;
echo "<br> <br> Your name is: $studentName and your grade is: $studentGrade <br> <br>";
foreach($classGrades as $key=>$value){
echo join($value,' <br>');
}
?>```
Your loop is wrong. You have separate name and grade arrays, so that's what you need to loop over:
foreach ($classGrades['name'] as $index => $name) {
$grade = $classGrades['name'][$index];
echo "$name<br>$grade<br>";
}
But it would be better if you didn't create separate array and kept the name and grade together in a single associative array:
$classGrades[] = ['name' => $studentName, 'grade' => $studentGrade];
Then your loop would look like:
foreach ($classGrades as $grade) {
echo $grade['name'] . "<br>" . $grade['grade'] . "<br>";
}
If you want the variable to persist between form submissions, you need to use a session variable.
Put
<?php
session_start();
at the beginning of the script. Then use $_SESSION['classGrades'] instead of $classGrades, and initialize it like this:
if (!isset($_SESSION['classGrades'])) {
$_SESSION['classGrades'] = array();
}
Basically you need to store your values, in some database.
Here is a simple start using a json file named students.json as database. That file will be created along your php file in the same folder, if not existing.
First you was missing the form elements and the method (by default it uses GET, thus this was not working).
Testing with isset() avoids populating the database with empty values, at page load or refreshs.
This is not very secure as such, you will have to work on that later on. Hint: What if a user insert a quote ' or " in his name?
<form method="post">
<label for="Name">Student Name:
<input type="text" name="StdName" placeholder="Your name">
</label>
<label for="Name">Grade:
<input type="text" name="StdGrade" placeholder="Your Grade">
</label>
<input type="submit" name="submit"value="submit">
</form>
<?php
if (!is_file("students.json")){
file_put_contents("students.json", json_encode([]) );
}
$classGrades = json_decode(file_get_contents("students.json"),true);
if (isset($_POST['StdName']) && isset($_POST['StdGrade'])){
var_dump($classGrades);
$studentName = $_POST['StdName'];
$studentGrade = $_POST['StdGrade'];
$classGrades['name'][] = $studentName;
$classGrades['grade'][] = $studentGrade;
echo "<br> <br> Your name is: $studentName and your grade is: $studentGrade <br> <br>";
file_put_contents("students.json",json_encode($classGrades));
}
echo "<table><tr><th>Name</th><th>Grade</th></tr>";
for ($i = 0;$i < count($classGrades['name']);$i++){
echo "<tr><td>".$classGrades['name'][$i] . "</td><td> " .$classGrades['grade'][$i] . "</td></tr>";
}
echo "</table>";
I pretty newbie to php and javascript but more i am curious about PHP. I want to add elements into a empty array every time i add a new element trough a input form, and after that i want those elements to be displayed in to the browser .The code i use is this
<form action="index.php" method="POST">
<input type="text" name="name" placeholder="name"><br><br>
<input type="submit" name="submit" value="enter"><br><br>
</form>
<?php
if (isset($_POST['submit'])) {
$_SESSION['names']=array();
$names=$_SESSION['names'];
$name=$_POST['name'];
array_push($names,$name);
for($i=0;$i<count($names);$i++){
echo $names[$i];
}
};
How could i achieve to display every element inside the array that i add trough the input field in php?
You overwrite the value every time because you wipe out the array on every page load: $_SESSION['names']=array();. Instead check to see if that session variable exists (and is an array) first and, if it doesn't, then create it. Otherwise just append to that array.
<form action="index.php" method="POST">
<input type="text" name="name" placeholder="name"><br><br>
<input type="submit" name="submit" value="enter"><br><br>
</form>
<?php
session_start();
if (isset($_POST['submit'])) {
if (!isset($_SESSION['names']) || !is_array($_SESSION['names'])) {
$_SESSION['names'] = array();
}
$name = $_POST['name'];
$_SESSION['names'][] = $name;
$num_names = count($_SESSION['names']);
for($i=0;$i<$num_names;$i++){
echo $_SESSION['names'][$i];
}
};
When user inputs text in 'ctext' field and press accept, I want to fill the value=" " field with user input, i achieved this but it fills all the value fields of same name in the page, how can i achieve it for different value of different ctext input? Anyone please give me solution with example, Many thanks
<?php
$connect = mysqli_connect('localhost', 'root', 'root123', 'font');
$query = 'SELECT * FROM pens ORDER by id ASC';
$result = mysqli_query($connect, $query);
if($result):
if(mysqli_num_rows($result)>0):
$i=0;
while( $pen = mysqli_fetch_assoc($result) ):
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF'];?>?action=add&id=<?php echo $pen['id']; ?>">
<div class="name pen-<?php echo $pen['id']; ?>">
<input type="text" name="ctext[]" class="form-control" placeholder="Type your text here" value="<?php $ctext = false; if(isset($_POST['ctext'])){ $ctext = $_POST['ctext']; } echo $ctext[$i]; ?>"></input>
<input type="hidden" name="id" value="<?php $pen['id']?>"></input>
</div>
<div class="btn-custom">
<input type="submit" name="add_to_cart" class="btn btn-block" value="Accept"></input>
</div>
</form>
<?php
$i++;
endwhile;
endif;
endif;
?>
I hope I understand what you want. You want to access the ctext for each individual $pen when printing the corresponding form.
You just need to name your <input> with a unique name and then access that value when printing. A possible solution is this:
<input type="text" name="ctext[<?php echo $pen['id']; ?>]" class="form-control" placeholder="Type your text here" value="<?php $ctext = ''; if(isset($_POST['ctext'][$pen['id']])){ $ctext = $_POST['ctext'][$pen['id']]; } echo $ctext; ?>"></input>
What does it do?
name="ctext[<?php echo $pen['id']; ?>]" ensures a unique name for each $pen. For a $pen with id 1 this will result in name="ctext[1]".
if(isset($_POST['ctext'][$pen['id']])){ $ctext = $_POST['ctext'][$pen['id']]; } uses $pen['id'] to look up the corresponding value in $_POST['ctext'].
By the way, when outputting user input you should always escape it, e.g. with htmlspecialchars. This will look like this: echo htmlspecialchars($ctext); That way malicious input like "><script>alert('Hello!')</script> won't execute the javascript.
Update: as requested a solution using session to store data:
<?php
$connect = mysqli_connect('localhost', 'root', 'root123', 'font');
$query = 'SELECT * FROM pens ORDER by id ASC';
$result = mysqli_query($connect, $query);
if($result):
if(mysqli_num_rows($result)>0):
session_start();
if (isset($_POST['ctext'])) {
$_SESSION['ctext'][$_POST['id']] = $_POST['ctext'];
}
while( $pen = mysqli_fetch_assoc($result) ):
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF'];?>?action=add&id=<?php echo $pen['id']; ?>">
<div class="name pen-<?php echo $pen['id']; ?>">
<input type="text" name="ctext" class="form-control" placeholder="Type your text here" value="<?php $ctext = ''; if(isset($_SESSION['ctext'][$pen['id']])){ $ctext = $_SESSION['ctext'][$pen['id']]; } echo htmlspecialchars($ctext); ?>"></input>
<input type="hidden" name="id" value="<?php echo $pen['id']?>"></input>
</div>
<div class="btn-custom">
<input type="submit" name="add_to_cart" class="btn btn-block" value="Accept"></input>
</div>
</form>
<?php
endwhile;
endif;
endif;
Note: I removed the now unnecessary counter $i. The session handling is mainly done before the while loop (start a session and store POST data). During output the values from the session are used. The name of the input is not an array anymore.
Change name of an input to an array.like this . When you submit the form you will get these values as an array. Give it a try
<input type="text" name="ctext[]" class="form-control" placeholder="Type your text here"></input>
I guess your code is misleading you, your form is in while loop so once any of the ctext input is filled your variable $_POST['ctext'] is set on server side and according to your code it sets all the value of ctext once accept is pressed.
You can have different names as a solution or an array indexing in input field name=“ctext[]” to avoid this.
I want to pass two variable from POST, one is the text I write and the other one is the result of a query with I already have.But for some reason I am not getting the variable values. Can you help me?
This is my first page:
<form method="post" action="EliminarGrupos.php">
<label for="nomegrupo"><b>Editar nome do grupo 1 :</label</b><br>
<?php
while ($row = mysqli_fetch_array($result66)){
$result = $row['titulogrupo'];
$_POST['nomegrupo'] = $result; //saving first variable
?>
<input type="text" placeholder="<?php echo $result?>" name="grupo1" id="velhas"></td> //saving second variable
<?php } ?>
<input type="submit" name="submit_x" data-inline="true" value="Submeter">
</form>
This is my second page where I want the variables to appear
$variable = $_POST['nomegrupo'];
$variable2 = $_POST['grupo1'];
The placeholder attribute is for display purposes only. You need to set the value attribute to have it sent to the server.
To send a second value, just use a second <input> element. If you don't want it visible, set type attribute to hidden.
In addition, you are expecting an associative array from mysqli_fetch_array() which is not going to happen. Your HTML had a number of errors in it, which I think I've fixed below. You always need to escape output with htmlspecialchars(). You should separate your HTML and your PHP as much as possible.
<?php
$row = mysqli_fetch_assoc($result66);
$titulogrupo = htmlspecialchars($row["titulogrupo"]);
?>
<form method="post" action="EliminarGrupos.php">
<label for="velhas"><b>Editar nome do grupo 1 :</b></label><br/>
<input type="text" placeholder="" name="grupo1" id="velhas"/>
<input type="hidden" name="nomegrupo" value="<?=$titulogrupo?>"/>
<button type="submit" name="submit_x" data-inline="true">Submeter</button>
</form>
You get the $_POST data from the form submission, specfically from the name attributes. This is what gives the $_POST its information, which it retrieves from value, not placeholder, as you have it now.
<input name="grupo1" value="one"> will make $_POST['grupo1'] equal to one.
You also shouldn't set the $_POST variable on page 1 as you are currently doing, and should make the unchanged variable from the database call a hidden field:
Page 1:
<form method="post" action="EliminarGrupos.php">
<label for="nomegrupo"><b>Editar nome do grupo 1 :</label>
<?php
while ($row = mysqli_fetch_array($result66)){
$result = $row['titulogrupo'];
?>
<input type="text" value="<?php echo $result; ?>" name="grupo1" id="grupo1">
<input type="hidden" value="<?php echo $result; ?>" name="titlogrupo" id="titlogrupo">
<?php } ?>
<input type="submit" name="submit_x" data-inline="true" value="Submeter">
</form>
Page 2:
$variable1 = $_POST['titulogrupo']; // $row['titulogrupo']
$variable2 = $_POST['grupo1']; // Form input
Hope this helps! :)
I'm working on a formular, but for the moment I just want to insert into an array my elements (I have books and authors).
I can display my books with author (name + surname) with the foreach, but I can't add more elements.
Here is the code with the form.
<H1>Exercice 2</H1>
<form method="POST">
<label for"code" >Number :</label>
<input id="code" name="code" type="number" />
<label for"title">Title :</label>
<input id="title" name="title" type="text" />
<label for"author" >Author :</label>
<input id="author" name="author" type="text" />
<button type="input" type="submit">Ok</button>
$title = $_POST['title'];
$code = $_POST['code'];
$author = $_POST['author'];
$book = array();
$book['code'] = 123;
$book['title'] = "Legendes";
$book['author'] = array("David", "Gemmel");
foreach($book as $value){
$book['key'] = $value;
var_dump($book);
if (is_array($value)) {
foreach($value as $otherValue) {
echo($otherValue);
}
} else {
echo($value);
}
}
I did some searcch, but I don't think it works, it's using the array_push() method with the POST, but I don't know where I can manipulate my form into the array.
If you want some details, I'll be happy to do that =) I'm working on it, if i have some news, you will know =)
Have a nice day =)
1) Assignments are in reverse. Correct way:
$myVar = $myValue
2) You need to set the name attribute in your inputs in order to be sent:
<input id="code" type="number" name="code" />
Then you can access them like:
$_POST['code']
3) To add an element by key in an array, use:
$array['key'] = $value;
Your Exercise 2 have some mistakes :
First, your HTML inputs must have the name attribute to be retrieved by post:
<h1>Exercice 2</h1>
<form method="post">
<label>
<input name="code" type="number" />
</label>
<button type="submit">Ok</button>
</form>
With PHP, you can access to any input value using the name:
$code = $_POST['code'];
Now, I think you want to "add" several books using this HTML form without a storage system. The problem is you can not do this if for every a new request since all the elements you have in your array will be deleted each time you run a new post request. To keep this information you need to use some persistent storage system as a database or others.
Since you seem to want to keep the information for each book together, you need to use a multidimensional array - hence, you'll need to redo the whole thing. Here's a suggestion:
Form:
<h2>Exercice 2</h2>
<form method="post">
<label for"code">Number :</label>
<input id="code" name="code" type="number">
<label for"title">Title :</label>
<input id="title" name="title" type="text">
<label for"author-firstname">Author First Name:</label>
<input id="author-firstname" name="author-firstname" type="text">
<label for "author-lastname">Author Last Name:</label>
<input id="author-lastname" name="author-lastname" type="text">
<input type="submit" name="submit_book" value="Ok">
</form>
Fixed the name-problems, changed the heading (you never, ever use H1 for a form, H1 is strictly used for the site-wide heading/logo/name of site). Also changed the button into a simple input type="submit".
$title = $_POST['title'];
$code = $_POST['code'];
$author = $_POST['author'];
$book = []; // changed this to modern PHP version array assignment
$book[0]['code'] = 123;
$book[0]['title'] = "Legendes";
$book[0]['author-firstname'] = "David";
$book[0]['author-lastname'] = "Gemmel"; // no reason to assign a separate array for first and last name, just use two array-keys
for ($c = 0; $c <= count($book); $c++) { //changed this to a for, counting the amount of entries in the $book array
echo 'Title: '.$book[$c]['title'];
echo 'Author: '.$book[$c]['author-firstname'].' '.$book[$c]['author-lastname'];
} // the content should probably be wrapped in a container of some sort, probably a <li> (and then a <ul>-list declared before the for-loop)
Now. None of this has anything to do with putting stuff INTO the array. That would be something like this (there isn't even a point of assigning the $_POST-variables for the code you posted. But, you can do something like this:
if (isset($_POST['submit_book'])) {
$title = $_POST['title'];
$code = $_POST['code'];
$author-firstname = $_POST['author-firstname'];
$author-lastname = $_POST['author-lastname'];
// however, if all you're doing is putting this into the array, no need to assigne the $_POST to variables, you can just do this:
$temp_array = ['code'=>$_POST['code'],'title'=>$_POST['title'],'author-firstname'=>$_POST['author-firstname'],'author-lastname'=>$_POST['author-lastname']];
$book[] = $temp_array;
}
So, that would replace the assigned variables at the beginning of your code.