Hello everyone and thanks for your time in advance!
I am trying to finish a project for my university and I am missing something really small.. So basically I have a variable in which I have set a specific value and as a front end user(as the Administrator of that page) I want to make a change on that value. It's like the following.. I have a specific number set and I want to be able to edit/update the value of it from a text box, or somehow from the browser.
Lets say I have this as source code:
$row = 0;
$totalpeoplenumber = 50;
$peoplenumber = 0;
$sql="SELECT peoplenumber FROM bookingform";
$result = mysqli_query($con, $sql) or die ("ERROR 01".mysqli_error($con));
foreach($result as $result_data => $result_row)
{
echo $result_data['peoplenumber'];
$peoplenumber += $result_row['peoplenumber'];
}
echo ('<h2>Seats Availability</h2>');
echo ("<br>");
echo ('<b>Reserved Seats number: </b>');
echo $peoplenumber."<br>";
echo ("<br>");
$totalpeoplenumber = $totalpeoplenumber - $peoplenumber;
echo "<b>The number of available seats is: $totalpeoplenumber</b> </br></br>";
echo "<button>Edit total seats</button>";
So here I need to somehow change the value of $totalpeoplenumber as the administrator of the webpage and not as a programmer.
Thanks a lot.
For sake of simplicity you can do something like:
if(isset($_GET['totalpeoplenumber')){
$totalpeoplenumber = $_GET['totalpeoplenumber '];
}
Then in your browser all you have to do is use the address: localhost/myAdminPage.php?totalpeoplenumber=60
it will set the total number of people to 60.
Note that you don't have to edit the URL itself, you can use a form where the submit sends to the same page (in that case you should use POST instead of GET):
<form action="myAdminPage.php" method="GET">
Insert the totalpeoplenumber: <input type="text" name="totalpeoplenumber"/>
<input type="submit" value="Send"/>
</form>
This propably will do for the form. In your script all you have to do is put the following code:
if(isset($_POST['totalpeoplenumber'])){
$totalpeoplenumber = $_POST['totalpeoplenumber'];
}
done :)
Related
I have a site where an admin can enter exam marks for papers which are part of an exam.
I am on the final part of actually giving a user their mark. But I just can't seem to do it.
So far what I have done is:
Allow the admin to view all of the exams, click on a specific exam and view the papers for that exam, then click on a paper and view all of the people who took that paper.
Then, click on a user and enter their marks and feedback. This is the part which I cannot do. I have pasted my code below along with what I am trying to do but it just is not working, any help would be great!
So, along with their mark and feedback I am also inserting into the marks table the, paperID and the examID.
CODE:
<?php
$epsID = $_GET['epsID'];
$sql = "SELECT * FROM ExamPaperStudent WHERE epsID = '$epsID'";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result))
{
$epID = $row ['epID'];
$sID = $row ['sID'];
echo "<p><form>";
echo "<b>Mark: <input type=text name=mark></b><br>";
echo "<b>Feedback: <input type=text name=feedback></b><br>";
echo $epID;
echo $sID;
echo "</form>";
echo "<a href='insertmark.php?epsID=". $row['epsID']."'>Add Data</a>";
}
?>
INSERTMARK.php code: (At this form I already know the exam/paper ID, which I also am trying to insert (along with marks/feedback).
CODE:
$mark = $_POST["mark"];
$feedback = $_POST["feedback"];
if(isset($_REQUEST['submit']))
{
$sql = mysql_query("insert INTO exammarks (mark, feedback, epID, atID) values ('$mark', '$feedback', '$epID', '$sID')");
$result = mysql_query($result);
}
epsID = exampaperstudent
epID = exampaper
sID = student
your form is wrong .
change this
echo "<p><form>";
to
echo "<p><form action='INSERTMARK.php' method='POST' name='myform'>";
OMG everything is wrong inside your inputs. i just give one and you correct the others
echo "<b>Mark: <input type='text' name='mark'></b><br>";
^----^------^----^---//use single quotes around here
didnt you miss submit button ?
Ok, I haven't done much of this sort of stuff, so I am clueless right now.
On the first page you hit the form submit that generates a bunch of information/stuff and displays it underneath submit button, but I don't know how to take the displayed information and use it on the next page I will show some of my code. btw I know the code is bad, just ignore that fact.
<form name="input" action="slaymonster.php" method="post" id="id">
<div align="center">
<input name="Submit" id="Submit" type="submit" class="button" value="Explore Map!"/>
</div>
</form>
if (isset($_POST['Submit'])) {
include 'includes/mapstuff.php';
// So here we pick a random row from the table pokemon notice the order by rand
$sql23 = "SELECT * FROM map1pokemon ORDER BY RAND() LIMIT 1;";
// We then check for errors
$result23 = mysql_query($sql23) or die(mysql_error());
// we then make the result into a virable called battle_get23
$battle_get23 = mysql_fetch_array($result23);
$sql2 = "SELECT * FROM pokemon WHERE name='".$battle_get23['pokemon']."'";
$result2 = mysql_query($sql2) or die(mysql_error());
$battle_get2 = mysql_fetch_array($result2);
// Now we need to make sure the image is safe be for we use it
$pic2= mysql_real_escape_string($battle_get2['pic']);
$pic = strip_tags($pic2);
include 'includes/maptypes.php';
?>
<form name="inputt" action="" method="post">
<div align="center">
<input type="submit" class="catch" value="Catch Pokemon" name="catch">
</div>
</form>
<p></p>
<?php
echo "You have just found a " ;
echo $randomview97[0];
echo " ";
echo $battle_get23['pokemon'];
$_SESSION['pokemon'] = $battle_get23['pokemon'];
$_SESSION['type'] = $randomview97[0];
$_SESSION['pic'] = $battle_get2;
$_SESSION['money'] = $randomview2[0];
$_SESSION['level'] = $randomview3[0];
$_SESSION['ticket'] = $randomview4;
?>
<p></p>
<?php
echo "You have gained ".$randomview3[0]." levels" ;
echo " ";
?>
<p></p>
<?php
echo "You have received $".$randomview2[0]."" ;
echo " ";
?>
<p></p>
<?php
echo "</center>";
}
?>
it displays the pokemon's picture it's name, type,amount of money you got ect...
I need all that information to be useable on the next page.
Any help is appreciated :)
At the top of your PHP code, be sure to include session_start();
You are already using session variables, so you should refer here to see what a PHP session is: PHP session_start() - Manual. It makes sure to do exactly what you are asking for (someone may point out that in certain cases session_start(); is not necessary, but for your purposes, while learning, stick to the Manual for best practices)
This information will be usable on the next 'page', just as the manual describes, and will be available, until you call something like session_destroy().
If you want to pass the information from one page to another. You have to put the result inside the form tag. Then it is possible to pass the information to another page. Or you can put it on the session and get information from any page.
you got my point? If you explain what you want to do. Then I will do something for you.
Been stuck on this bit of code for a while now. What I'm doing in 'do_assignments.php' (below) is creating a loop that outputs each question in my database and every possible answer associated with that question.
<form method="post" action="user_storeresults.php"> $query = "SELECT * FROM questions, question_choices WHERE q_topic = 'PhoneGap' AND visible = 1 AND q_type = 'Multiple' AND questions.q_id = question_choices.question_id";
$result=mysql_query($query);
while($row = mysql_fetch_array($result)){
if($row['q_id'] != #$qid){
echo "<h4><strong>" . $row['q_string'] . "</strong></h4>";
#$qid = $row['q_id'];
}
echo "<input name='".$row['choice_id']."' value='".$row['correct_choice']."' type='radio' />";
echo $row['choice_string'] . "<br/>";
}
?>
<p><input type="submit" name="submit" value="Submit"><p>
</form>
<?php
The value of each radio button contains data from a field in 'correct_choice' in my database, where there is wrong answers (0) and a correct answer (1) associated with each question.
All of this works well, but because i'm trying to submit a post from a radio button that doesn't have a specific name or value (only what it retrieves from the database), i can't quite figure out how i would be able to retrieve these values in the next page 'user_storeresults.php'...
Any help would be much appreciated. If i'm being a bit too vague then say, and ill try and clear it up as much as possible.
I have the following query that I ran on my database to remove some data:
delete subscriber, subscription from subscriber,subscription where subscription.status = 0 and subscription.snid=subscriber.snid;
But I now need to make the a php function that runs when I press a button called clean
then print out all the subscriber data that was deleted.
Not quitesure where to start with this.
this is my html so far:
<form id="form1" name="form1" method="post" action="">
Clean subscribers:
<input type="submit" name="clean" id="clean" value="Clean" />
</form>
Any help or advice with this is very much appreciated.
C
You'll need the button to submit a form to a handler page, the handler page would then run the query, and collect+print the data.
If you don't want to refresh the page (or have your users diverted into another page), you'll want to use Ajax.
That's where you start.
Is abvious you made no effort! but I will answer you anyway.
<?php
$con = mysql_connect("serverUrl","login","password");
mysql_select_db("dbName", $con);
$result = mysql_query("SELECT * FROM subscriber, subscription where subscription.status = 0 and subscription.snid=subscriber.snid;");
while($row = mysql_fetch_array($result))
{
echo $row['subscriber.name']; //assuming you have a field {name} in your table
echo "<br />";
}
mysql_query("delete subscriber, subscription from subscriber,subscription where subscription.status = 0 and subscription.snid=subscriber.snid;");
?>
First you'll need to select the data you're about to delete.
Then you'll need to delete it and return the selected rows.
$rows = array();
mysql_connect(...);
$res = mysql_query(...select query here...);
while($row=mysql_fetch_assoc($res)) {
$rows[] = $row;
}
$res = mysql_query(...delete query here...);
return $rows;
You might not want to totally delete the subscriber. If I were you I would include a field named "deleted" or something along those lines, indicating whether or not the subscriber has been deleted. Then query according to whether or not that field is true or false.
<?php $name=$_POST['name']; ?>
<form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input type="text" name="name">
<input type="submit" value="GO" name="submit">
</form>
<?php
include ('db.php');
if(isset($_POST['submit']))
{
mysql_query ("INSERT INTO example (name) VALUES('$name')") or die(mysql_error());
}
if (!isset($_GET['startrow']) or !is_numeric($_GET['startrow'])) {
$startrow = 0;
}
else {
$startrow = (int)$_GET['startrow'];
}
$query = "SELECT * FROM example ORDER BY id DESC LIMIT $startrow, 20";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
echo "<li>";
echo $row['name'] ." "." <a href= 'like.php?quote=" . urlencode( $row['name'] ) . "'>Click Here</a>";
echo "</li>";
}
echo 'Next';
?>
I want to make my page links hidden , how can i make then hidden so that a user cant edit it.
2nd question,
currently i am showing total 10 records on each page and then a next page button , but the next button is keep showing even when there is no more records...! how to remove a next page button when records ended. ??
line number 28 is the link to pages which can be easyily editable by any user, i wnat to make them secure (using ID)
and line 35 is n'next' page link , this link should not be appear when number of records ended
I can't think of a reason why you really should hide the page numbers from the user in the link. Keeping them in the query string as $_GET variables is probably the most common practice i know of in this specific case of paging.
I would do validation on the numebrs being recieved in the $_GET variables, since this could often lead to SQL Injection and other problems... Make sure it's a number, possibly divisible by 10 (if that's how you like the site to be), perhaps not bigger than a certain defined number, etc...
If you REALLY still don't agree, and you want to hide it, then you could always do that by saving cookie on the user's computer (this is still exposed to user in some way) or save the page number in the session (although this seems like a big waste of server resources to me!).
About your second question - There are so many possibilities to this...
Here's one way :
Create an sql query that queries how many rows are there to your table.
Let's say the number is 55. You put that into a hidden value.
If you're displaying 10 items on a page then you know the last page is number 6 (showing items 50-55, if you start counting at page number 1).
Simple php check when page loads: if ($_GET['page'] == 5) then you don't display the next button.
something like this (skipping out validation checks and the sql query) :
<input type="hidden" value="<?php echo $itemCount;?>">
<?php
if ($_GET['page'] < ($itemCount \ 10))
{
echo "<a href=\"items.php?page=".($_GET['page']+1)."\">";
}
?>
Using this, I would add a check to make sure the user doesn't enter a number bigger than this number as well, and if they do, just redirect them to the last number they can.