I am building a code that shows up a random question everytime you refresh the page or answer the question on screen. This is the code I have on my conexao.php file to select a random question:
$research = mysqli_query ($connection, "SELECT * FROM enquete ORDER BY RAND() LIMIT 1");
$result = mysqli_fetch_array($research);
$info = $result;
The code above is working fine to select a random question from my "enquete" table.
The problem I have is when the question is answered. This is the code I have that displays the question and presents 2 submit buttons (YES/NO):
<?php
include 'conexao.php';
$contador = 0;
do{
echo $info['pergunta'];
if(isset($_POST['sim'])){
mysqli_query($connection,"UPDATE enquete SET sim = sim + 1 WHERE pergunta = '".$info['pergunta']."'") ;
echo "<br/>".$info['sim']."people who answered YES <br/>";
echo $info['nao']." people who answered NO <br/>";
}
$contador++;
} while ($contador < 1);
?>
<html>
<head>
<link rel="stylesheet" type="text/css" href="estilo.css">
<link async href="http://fonts.googleapis.com/css?family=Aclonica"
rel="stylesheet" type="text/css"/>
</head>
<div class="wrapper">
<form action="" method="post">
<input type="submit" class="botao-css" value="Sim" name="sim"/><br/>
<input type="submit" class="botao-css" value="Não" name="votarnao"/><br/>
</form>
</div>
</html>
The problem I have is the following: The code above seems to be ok to add 1 to an answered question (YES in the example above), however instead of adding 1 to the current random question shown on screen it adds 1 to the next question which will be shown.
For example:
Question 1: The sky is blue? (if YES is hit then it adds 1 to the next question "sim" row instead of adding to this one. )
Question 2: Do you drink water? (it has already gotten 1 because of the previous question and if I think yes to this one the it adds 1 to the next question "yes" row)
Of course what I intend to do is to have the system to add 1 to "yes" row of the current question and not to the one that is next.
If someone could help me giving tips on this I would be really glad.
First I think you need to add a field in your form to identify which question is relevant for the submission:
<form action="" method="post">
<input type="hidden" name="pergunta" value="<?= $info['pergunta'] ?>" />
<input type="submit" class="botao-css" value="Sim" name="sim"/><br/>
<input type="submit" class="botao-css" value="Não" name="votarnao"/><br/>
</form>
Then you need to use this instead, but to prevent SQL injections, you should make sure it can only be an integer:
mysqli_query($connection,"UPDATE enquete SET sim = sim + 1 WHERE pergunta = ".(int) $_POST['pergunta']) ;
$info will be the information about the next question obtained from conexao.php. $_POST will be the data from the last submission. So it makes sense to not use $info at all if you're referring to the last question.
Note: if pergunta isn't an integer, you should use a prepared statement to protect against SQL injections. (maybe replace pergunta with id in this case)
Thank you it worked. I used post and an int value (the ID of the rows as suggested). Here is how I put the code to work.
Updated form adding this line as suggested:
`<input type="hidden" name="pergunta" value="<?= $info['id'] ?>" />`
And updated the query to this:
`mysqli_query($connection,"UPDATE enquete SET sim = sim + 1 WHERE id = ".(int) $_POST['pergunta']) ;`
Now it updates the question on screen. Thank you.
Related
First of all I'll be sincere, I'm a student and I've been asked to do a task that seems impossible to me. I don't like asking questions because generally speaking I've always been able to fix my coding issues just by searching and learning, but this is the first time I've ever been on this possition.
I need to create a php file that contains a form with two inputs that the user fills. Once he clicks submit the website will show on top of it the two values. Till here I haven't had an issue, but here's the problem, the next time the user sends another submission, instead of clearing the last 2 values and showing 2 new ones, now there needs to be 4 values showing.
I know this is possible to do through JSON, the use of sessions, Ajax, hidden inputs or using another file (this last one is what I would decide to use if I could), but the teacher says we gotta do it on the same html file without the use of any of the methods listed earlier. He says it can be done through an Array that stores the data, but as I'll show in my example, when I do that the moment the user clicks submit the array values are erased and created from zero. I know the most logical thing to do is asking him, but I've already done that 4 times and he literally refuses to help me, so I really don't know what to do, other than asking here. I should point out that the answer has to be server side, because the subject is "Server-Side Programming".
Thank you for your help and sorry beforehand because I'm sure this will end up being a stupid question that can be easily answered.
For the sake of simplicity I erased everything that has to do with formatting. This is the code:
<?php
if (isset($_POST['activity']) && isset($_POST['time'])){
$agenda = array();
$activity = $_POST['activity'];
$time = $_POST['time'];
$text = $activity." ".$time;
array_push($agenda, $text);
foreach ($agenda as $arrayData){
print implode('", "', $agenda);
}
}
?>
<html>
<head>
</head>
<body>
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="POST">
<label for="Activity">Activity</label><br>
<input name= "activity" type="text"><br><br>
<label for="Time">Time</label><br>
<input name= "time" type="time"><br><br>
<input type="submit">
</form>
</body>
</html>
Your question was not very clear to be honest but I might have gotten something going for you.
<?php
$formaction = $_SERVER['PHP_SELF'];
if (isset($_POST['activity']) && isset($_POST['time'])){
$agenda = array();
//if the parameter was passed in the action url
if(isset($_GET['agenda'])) {
$agenda = explode(", ", $_GET['agenda']);
}
//set activity time
$text = $_POST['activity']." ".$_POST['time'];
//push into existing array the new values
array_push($agenda, $text);
//print everything
print implode(", ", $agenda);
//update the form action variable
$formaction = $_SERVER['PHP_SELF'] . "?agenda=" . implode(", ", $agenda);
}
?>
<html>
<head>
</head>
<body>
<form action="<?php echo $formaction; ?>" method="POST">
<label for="Activity">Activity</label><br>
<input name= "activity" type="text"><br><br>
<label for="Time">Time</label><br>
<input name= "time" type="time"><br><br>
<input type="submit">
</form>
</body>
</html>
SUMMARY
Since you cant save the posted values into SESSION vars or HIDDEN input, the next best thing would be to append the previous results of the posted form into the form's action url.
When the form is posted, we verify if the query string agenda exists, if it does we explode it into an array called $agenda. We then concatenate the $_POST['activity'] and $_POST['time'] values and push it to the $agenda array. We then PRINT the array $agenda and update the $formaction variable to contain the new values that were added to the array.
In the HTML section we then set the <form action="" to be <form action="<?php echo $formaction; ?>
The newbie is back with another question. Any help would be much appreciated. Suppose we have got a form in which we have written down the name of a user and in front of which there is an input box in which we can allocate a grade to the mentioned user. Within this scenario, everything is clear. We have a form with the name of user (it's 'id' as the value) and another variable, that is the grade' which are posted to the php-action-page. Hence, in the php-action-page, I get two variables, one is the id of the user and the other allocated grade, through POST. Here, everything is clear and the process easy, since I have got just two defined variables. Now, suppose that we are inserting a list of users from our 'Users' table into the form dynamically. We fill our form with for example 10 users grabbed from the database. In front of them there are input boxes for the 'grade' to be inserted into. So far, everything is fine. The problem, though, lies in the next stage. The problem is I don't know how to ask php-action-page to do the insert, that is insert the grade in the database for specific users as long as there are posted variables of users. Here I have tens of users and tens of dynamic variables. And if the question is a little bit vague, please do excuse me; yet, do your best to get me free from this condition of bafflement. Many thanks.
Here comes some bits of the code to make the problem a little more clear.
I start with the following code:
<?php
require_once ('../inc/takein.php');
$pd = new dbase();
$students = $pd->run_a_query('SELECT * from `checking`');
Here I am including the database and other necessary files. Then I run a query to fetch a list of my students from the table. So far, everything is fine. The next line of action which makes me perplexed is the following code.
Before having a look at the code may you please look at the html design in the following picture:
Final Design
I totally have no idea about it being wrong or correct. You might help with this bit as well.
<form action="grades.php" method="post">
<table class="table table-bordered font-label" id="item_table">
<?php
foreach ($students as $student) {
echo '<tr>';
echo '<td>'.$student['name'].'</td>';
echo '<td><input type="text" name="grade[]" class="form-control omit-radius-input"></td>';
echo '<input type="hidden" name="id[]" value="'.$student['id'].'">';
echo '<tr>';
}
?>
</table>
<input type="submit" name="dispatched" class="btn btn-green">
</form>
Here, I am putting the information in a table within the form element. As you can see in the above picture, I am getting four students from the database. Now I want to send these students back to the database along with their newly set grades. What I want to be posted here is the student id and their grades.
Then, the following is the last part of the code which is left incomplete because I couldn't make any senses how to do it.
if (isset($_POST['dispatched'])) {
$id[] = $_POST['id'];
$grade[] = $_POST['grade'];
// what to do now???!!!
foreach(...HOW TO DO THE 'FOREACH') {
...
}
}
May you please help me insert my student grades. Many thanks in advance.
Simply name your variables as arrays - if your form looks like this
<form method="POST">
<input type="text" name="grade[12]">
<br>
<input type="text" name="grade[15]">
<br>
<input type="text" name="grade[7]">
<br>
<input type="text" name="grade[21]">
<!-- continue here -->
</form>
then in your PHP code you will access the grades like this
if(is_array($_POST['grade'])) foreach($_POST['grade'] as $id => $value)
{
// REPLACE INTO user_grades(user_id, grade) VALUES($id, $value)
}
UPDATE
You should also put the ID of your students in the name of the INPUT field - otherwise you won't know for which student is the given grade.
<?php
foreach ($students as $student) {
echo '<tr>';
echo '<td>'.$student['name'].'</td>';
echo '<td><input type="text" name="grade['.$student['id'].']" class="form-control omit-radius-input" value="'.$student['current_grade'].'"></td>';
echo '<tr>';
}
?>
The foreach is shown above in my original answer.
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I am having some serious trouble for some reason with creating a basic PhpMyAdmin database and making a page with a scoreboard. There will be a game that you play, and when you're done you can input a score and your name using basic form elements. When you press submit the page will reload and you will be able to see the top ten scores, ranked by highest first.
My issue is that I have no idea where to start with this. I have just started Php and don't wish for anything crazy. I have the ~/php/db_connect.php set up correctly already; I just need to make the function work.
How do you recommend I go through with this? Example code is extremely helpful.
I know the first response is "what have you tried?" and I haven't tried much.
This is what I have right now:
// define variables and set to $name = $myArray[0];
$babyinfo = fgets($myfile);
$myfile = scoreboard-dk;
$myArray = explode(',', );
$score = $myArray[1];
$name = $myArray[2];
$insertStmt = "INSERT INTO scoreboard-dk ('score','name') VALUES ('$score','$name')";
// Inserting Babynames into database
$db->query($insertStmt);
?>
<form action=" $db;?>" method="post">
Name: <input type="text" name="name" value=" echo $name;?>" required><br>
Score: <input type="text" name="score-dk" value=" echo $score;?>" required><br>
<input class="btn btn-primary" type="submit">
</form>
<tr> <th scope=row> echo $i;?></th> <td> echo $score;?></td> <td> echo $name;?></td> <td> echo $votes;?></td> </tr>
Thanks in advance.
Ok. First some mistakes you made:
$myfile = scoreboard-dk; isn't working. This way it would be a constant. You need the "$" or quotation marks if it should be a string.
$myArray = explode(',', ); I don't know what you want to do? The second argument is missing. This statement won't work. Second argument has to be a string.
You have to properly escape the query before executing the statement.
You can do this by replacing the following line before building the string:
$score = $db->real_escape_string($myArray[1]);
$name = $db->real_escape_string($myArray[2]);
Furthermore are you sure you use the correct indices for the array access? Counting starts with 0, not with 1.
You can't use PHP code without the opening tags. I thought that you cut that away at the start of the file. You always have to open PHP code blocks with
Perhaps you should search for example code elsewhere. I think stack is more for specific questions. But the code actually shows that you lack of some basic knowledge ... no offense.
The reason people can't help you is that your question is way too broad and everyone will have a different approach about how to implement it.
That being said, here is the pseudo code I would use to implement this. It can be done in a single file. Good luck!
File: score_keeper.php
<?php
error_msg = array
if (form submitted)
$name = name from form
$score = score from form
// Do validation to ensure name and score is as expected.
if name is empty
error_msg[] = 'Name cannot be empty'
if score is not numeric
error_msg[] = 'Score must be numeric'
if empty(error_msg)
// INSERT
// Make sure you use parameterized queries
SQL = INSERT into table (name, score) VALUE (?, ?)
end-if
end-if
// READ top 10
SQL = SELECT name, score FROM table WHERE ...
if !empty(error_msg)
show error_msg
?>
<form method="post">
<input name="name">
<input name="score">
</form>
HTML table
<?php
// output top 10 results
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Closed 8 years ago.
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sorry for simple questions but i tried it too long and cant do this.
So the question is this : I have test with three radio options ,and
for each option being pressed i want that option to increase on 1 or
whatever number i want and then add the count of each radio button
being pressed to the database ??? I have a mysql table named 'test '
with three elements : 'gr' - varchar - 255 - utf8general_ci, 're' -
int - 10 , 'ye' - int - 10 . So When someones press for example radio
button with value green , i want that 'gr' in my sql table 'test '
increase on 1.
PHP Code:
<?php
$a=mysql_connect('xxxx','xxxx','xxxx');
$b=mysql_select_db('xxxx',$a);
$red=1;
$green=1;
$yellow=1;
if ($_POST){
if ($_POST['color']=='green'){
$green+=1;
$query='UPDATE test SET gr='.$green.'';
mysql_query($query);
}
if ($_POST['color']=='red'){
$red+=1;
$query='UPDATE test SET re='.$red.'';
mysql_query($query);
}
if ($_POST['color']=='yellow'){
$yellow+=1;
$query='UPDATE test SET ye='.$yellow.'';
mysql_query($query);
}
}
?>
Here is my html code:
<form action="indexni.php" method="POST ">
<input type="radio" name="color" value="green" />Green<br>
<input type="radio" name="color" value="red" />Red<br>
<input type="radio" name="color" value="yellow" />Yellow<br>
<input type ="submit" />
</form>
Your mistake is 'space'. See on this method="POST ". Change on this method="POST" and it is work for you.
But you are setting those $red, $green, $blue counter to 1 every time?! That way every time you start again from 1. You should just increase current value:
$query='UPDATE test SET ye=ye+1';
But if you don't have different table for every color....and I hope you don't...then you have to distinct somehow rows..... You need WHERE condition.
How your table structure looks like anyway?
Let's say you have table test with fields color and clicks and let's say you have 3 rows of that table where color field has values (per row) red, green, yellow.
In that case you can have mySql query that will increase color counter like this:
$query='UPDATE test SET clicks = clicks+1 WHERE color = "yellow"';
Something like that.
I have worked on this for a few days now.
This snippet finds the correct row(s) from the database. It works fine. However, when the Button is clicked, I only want for the comment it is next to targeted, not all the comments on that page. It's hard to explain.. Let me show you image.
What I am doing currently is selecting the database rows and outputting them for debugging, instead of removing them (I don't want to go around deleting sections of my database with broken code)
$db is mysqli connection to the database
Snippet of the while loop:
TL/TR
I want to do the following with this in the end:
Remove the comment from the database.
Give the feedback to an admin.
This may sound like a noob question, but please help me.
Many thanks!
Just can't wrap my head around this, made like 6 pages single-handed and now I am stuck in a thing this "easy" for a d**n week and i still can't get it right.
EDIT
Latest snippet (Still doing it, What the hell!?!)
$query = $db->query("SELECT * FROM comments WHERE post_id='$id'");
while($row = $query->fetch_object()){
echo "<h5>".$row->name."</h5>","<br>";
$strip_comment = strip_tags($row->comment);
$delComment = $row->comment_id;
$strip_comment_shlashes = stripslashes($strip_comment);
echo "<blockquote>".$strip_comment_shlashes,"<br><br></blockquote>";
//button stuff
if($is_admin){
$query1 = $db->prepare("SELECT comment_id FROM comments WHERE comment_id = '$delComment'");
$query1->execute();
$query1->bind_result($commId);
while($row2 = $query1->fetch()):
?>
<form action="<?php echo $_SERVER['PHP_SELF']."?id=$id"?>" method="post">
<input type="submit" class="closeButton" name="deleteComment" value="<?php echo $commId; ?>" />
</form>
<?php
if(isset($_POST['deleteComment'])&& $is_admin){
if($is_admin && $commId){
echo "Comment ID <b>$commId</b> Removed";
}
}
endwhile;
}
Edit: NON-OBJECT error..
$delComment_2 = $_POST['deleteComment'];
$query2 = $db->prepare("SELECT * FROM comments WHERE comment_id='$delComment_2'");
$query2->execute();
Edit 2: (31/10)
Please, Could someone fix this snippet and post it? I don't usually ask for working snippets, but this one is driving me crazy. I just am too noob to understand how this goes. Thanks.
Change the input to this:
<input type="submit" class="closeButton" name="deleteComment" value="$comment_id_here" />
And then just use the comment id to delete a specific one from the database. (check $_POST['deleteComment'])
Of course, there are numerous other ways to do this - but the point is the same: you need to pass the comment id (not the post id) from your page to the query string. Whatever way you use to accomplish that is up to you, I just gave you an example.