dropdown from MySQL database - php

Having solved the populating the dropdown from the array, the next phase of my project is pulling data from a database to allow users to put limits on the data they want to see.
Again, I'm having trouble populating the dropdown. This time however, the problem seems to be that when I put the HTML interspersed with the PHP, the whole thing stops- the form loads, but nothing goes into the drop down.
Here's the code.
//Step One: Query the DB, and get a list of monthly reports
$sql = "SELECT Report_Text, Report_Date FROM ADM_8_Reports_List";
$result = $conn->query($sql);
//print_r ($result);
$reports_in_db = mysqli_fetch_all($result,MYSQLI_ASSOC);
if (sizeof($reports_in_db) > 0)
{
print_r ($reports_in_db);
echo "\n";
//Now, like the front page, create a form and populate it.
?>
Choose a Monthly Report <select name = "monthly_report">
<?php
foreach($reports_in_db as $row)
{
echo 'option value"' . $row ["Report_Date"] . '">' . $$row ["Report_Date"] . " " . $row ["Report_Text"] . '</option>';
}
}
else
{
echo "I'm sorry, there's no data here. Please start again\n";
} ?>
</select>
<input type="submit" name="monthly report" value="Retrieve Report"">
the output
what it looks like is happening is that the code is freezing where I'm interpolating the HTML and PHP, and actually creating the dropdown.
basically on the previous page, the user can choose one of 4 things to do. On this page, there's an if statement based on $_POST from the previous page. Depending on the choice, the databased will be polled to get, in this case, a list of all the monthly reports. I've copied exactly the structure of what I did on the first page (where I was going from an associative array I hard-defined). I can't figure out what's wrong with the declaration of the dropdown, but that's where the thing's .. well stopping.

I am really annoyed by this.
I forgot '<'when declaring the option. And didn't see it until I posted the question here.
Sorry.

Related

php mysql show result one after another

I wanna build a presence check for our choir in the style of tinder but not as complex.
The database contains names and file paths of pictures of the members. When you click on the "present" or "not present" button, the next picture and name should be shown. In the background, the database table should be updated with true/false for presence. (this will be done later)
My problem is that it almost works, but instead of showing one member, it shows all members with their pictures in one single page.
I understand that I could fire with Javascript to continue and paused php-function but I don't get the clue how.
I tried "break" in the php and call the function again but that didn't work.
<?php
$conn = new mysqli(myServer, myUser, myPass, myDbName);
$sql = "SELECT * FROM mitglieder";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "<img class='pic' src='" .$row["folder"]. "/" .$row["img"]. "'><br>" ;
echo "<div id='name'>" .$row["vorname"]. " " .$row["name"]. "</div> <br>";
echo "<img src=''img.png' id='present' onclick='isPresent()'>";
}
} else {
echo "0 results";
}
$conn->close();
?>
<html>
<script>
$( document ).ready(function() {
console.log("Ready");
);
</html>`
You can use php function
mysqli_fetch_all()
assign it on the variable outside the while loop and loop or access the indexes in your code.
For Example:
$data = mysqli_fetch_all();
echo $data[0]['name'];
foreach($data as $item)
{
echo $item['name'];
}
You need a way to establish a "state" between your web page and the PHP backend so that you can step through the data. I suggest something like this:
Use an auto-increment integer primary key for the database. That way you can access the data in index order. Let's name the column id
Have your JS code send a form variable - named something like LAST_ID to the PHP in your get. i.e http://someurl.com/script.php?LAST_ID=0
On your first call to the server, send LAST_ID = 0
In your PHP code, fetch the value like this: $id = $_GET('LAST_ID');
Change your SQL query to use the value to fetch the next member like this:
$sql = sprintf("SELECT * FROM mitglieder where id > %d limit 1", $id); That will get the next member from the DB and return only 1 row (or nothing at the end of data).
Make sure to return the id as part of the form data back to the page and then set LAST_ID to that value on the next call.
You can use a HTTP POST with a form variable to the server call that sets that member id to present (maybe a different script or added to your same PHP script with a test for POST vs GET). I suggest a child table for that indexed on id and date.
I hope that puts you in a good direction. Good luck

Create a table with blank squares for missing mysql data

I've written a PHP script that can populate a table in a particular way so that multiple events (or no events) can be put in one square in an HTML - similar to the layout a calendar would have. But, there's a problem, the while statement I created to fill in squares in the table when there is no data doesn't detect when there is data, and fills the entire table with empty squares. This is what the output looks like (The page is styled using Bootstrap 3). From the mysql data I have provided, these events should be in the square at {Period 1, Monday}.
Here is my data in a mysql database; mysql data
Here is a snippet of the part of the page related to this table;
<?php
$query = "SELECT * FROM configtimetabletwo WHERE term = ".$term." AND week = ".$week." ORDER BY period, day LIMIT 100;";
$results = mysqli_query($conn, $query);
$pp=1; //The current y value of the table
$pd=0; //The current x value of the table
echo '<tr><td>';
while($row = mysqli_fetch_row($results)) {
while((pd!=$row[3] or $pp!=$row[4]) and $pp<6){ //This while statement fills in empty squares and numbers each row.
if($pd==0) {
echo $pp."</td><td>";
$pd++;
}
elseif($pd<5){
echo "</td><td>";
$pd++;
}
else {
echo "</td></tr><tr><td>";
$pd=0;
$pp++;
}
}
echo '<a href="?edit='.$row[0].'" class="label label-default">';
echo $row[5].' '.$row[6].' - '.$row[7]."</a><br>";
}
echo "</td></tr></table>"
?>
I haven't been able to figure out why this happens so far, thanks in advance to anyone who has any idea what's going on.
In the comments below my question, pavlovich pointed out the error. In this case, it was simply an issue of forgetting to use a $ to reference a variable. It would seem that this doesn't throw an error in a while statement like it would elsewhere.

Trying to copy data to a table with checkboxes, a form, and php

My webpage is pulling data from two tables - applications and archiveapps - and displays them using the following:
$sql = "SELECT * FROM applications";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "Lots of info, including some html thrown in for style."
I want users to be able to click a checkbox listed next to each row and then hit an "Archive Selections" button that then moves all the selected entries from the applications table to the archiveapps one.
So far I've tried a form with it's code half in the echo above (so that a checkbox would go by each listed row from the $sql query) with the submit button outside (so that there would only be one "Archive Selections" button) but I'm sure this isn't proper syntax.
To actually move the data, I had this for the checkbox:
<form method='post' action=''><input type='checkbox' name='archname' value=".$row["charname"].">
(The above was inside an echo statement, so I assume it was able to pull the $row["charname"] no problem, but am unsure how to verify that.)
A little further down the page is the submit button and </form>.
And then I've got the function I want it to run when the submit button is clicked, to check if boxes have been selected and copy them into the archiveapps table. I'm sure there's something I'm missing here, probably in referencing what exactly is selected, and then copying that row's data to the other table... but honestly I just don't know enough about php to know what I'm missing.
if(!empty($_POST['archname'])) {
foreach($_POST['archname'] as $check) {
function archiveapp() {
$sqli="INSERT INTO archiveapps SELECT * FROM applications";
if ($conn->query($sqli) === TRUE) {
echo "<i>Archived</i>";
} else {
echo "Error: " . $sqli . "<br>" . $conn->error;
}}}}
Most of this has just been gathered from google searches and kind of mushed together, so I'm sure there are a lot of things done wrong. Any pointers or advice would be greatly appreciated!
Oh and my end goal is to copy the data to the archiveapps table and then delete it from the applications table, but for now I've just been focusing on the copying part, since I assume deleting a row will be fairly simple? Either way it's not the priority for this question.
Thanks in advance for any help!
Use a tool like FireBug to see what you are actually posting to server. You have multiple checkboxes with the same name, so you are sending only 1 value (the last checked, I think).
You need to send all of them, so use brackets after the name:
<!-- this is just an example -->
<input type="checkbox" name="archname[]" value="1">
<input type="checkbox" name="archname[]" value="2" checked>
<input type="checkbox" name="archname[]" value="3" checked>
Then on PHP, inside $_POST['archname'] you will get this:
Array(2, 3)
After this, you can do a foreach(...) loop like you are already doing, inserting only the ID you get from the $_POST variable:
foreach($_POST['archname'] as $id){
$sqli = "INSERT INTO archiveapps (id) VALUES (".(int)$id.")";
}

How to use a series of drop down list to display data from mysql database

needing some advice.
I am wanting to include 4 drop down lists on a website, which all contain data from different fields in a mysql table.
I then want to be able to press a submit button and display the required data on a webpage.
I am having trouble with knowing what programming language to use and also finding it difficult to find any tutorials. Any help would be greatly appreciated.
Thanks
You mean a HTML dropdown list or just a select dropdown in a form?
If you mean a select dropdown in a form you could do something like this:
<?PHP
$query = mysql_query("SELECT value FROM table ORDER BY value ASC") or die(mysql_error());
$result = mysql_num_rows($result);
// If no results have been found or when table is empty?
if ($result == 0) {
echo 'No results have been found.';
} else {
// Display form
echo '<form name="form" method="post" action="your_result.php">';
echo '<select name="list" id="lists">';
// Fetch results from database and list in the select box
while ($fetch = mysql_fetch_assoc($query)) {
echo '<option id="'.$fetch['value'].'">'.$fetch['value'].'</option>';
}
echo '</select>';
echo '</form>';
}
?>
And then in your_result.php you should fetch the data from your MySQL database based on value from the (when you fetch use mysql_real_escape_string):
<?PHP $_POST['list']; ?>
You could also do everything in just one file, but thats up to you. Try to use google, there are dozens of tutorials out there.

Knowing which <li> item is selected using PHP and mySQL

I have 2 questions. One is if there is any error in this code.
The 2nd question I want to ask is how do you know which <li> item is selected. Right now, the code performs a search in mySQL for all rows that matches city, language and level and returns the results in a list item.
I want it so that when the user clicks on anyone of the list items, it will goes into another page displaying a more detail description by querying the selected list item.
I have a guess, which is for step 2, I also grab the ID (primary key) for each row and somehow keep that stored within the list but not echo.. Would I need to wrap <a> in <form action="XX.php" method="get">?
<?php
//1. Define variables
$find_language = $_GET['find_language'];
$find_level = $_GET['find_level'];
$find_city = $_GET['find_city'];
//2. Perform database query
$results = mysql_query("
SELECT name, city, language, level, language_learn, learn_level FROM user
WHERE city='{$find_city}' && language='{$find_language}' && level='{$find_level}'", $connection)
or die("Database query failed: ". mysql_error());
//3. Use returned data
while ($row = mysql_fetch_array($results)){
echo "<li>";
echo "<a href=\"#result_detail\" data-transition=\"flow\">";
echo "<h3>".$row["name"]."</h3>";
echo "<p>Lives in ".$row["city"]."</p>";
echo "<p>Knows ".$row["level"]." ".$row["language"]."</p>";
echo "<p>Wants to learn ".$row["learn_level"]." ".$row["language_learn"]."</p>";
echo "</a>";
echo "</li>";}
?>
Your code looks alright from what I can see. Does it not work?
One way would be to get the ID through your SQL-query as well and then add the id to the href in your link. Then you can fetch it through the querystring on the other page, to display the proper post depending on which li-element the user clicked on:
echo "<a href=\"details.php?id=". $row["id"] ."\" data-transition=\"flow\">";
Not sure how you mean "somehow keep that stored within the list but not echo" - it wouldn't be possible to store anything in the list, if it is not echoed, as it then wouldn't be sent to the client. You can of course store the id in a data-attribute on the li-element, which won't be displayed to the user. It will however be visible through the source code! Don't know why that should be a problem though?

Categories