PHP - Echo Style Sheet Or Else Help!

Hello, well I guess my question is pretty clear.
How would I be able to read data out of an Excel sheet?

F.E. to allow my users to simply upload an .xls file to help them avoid the coding and such.

Of course, I know how to create the tables etc, the main problem is just getting that data. Also, I am uncertain if uploading an .xls file is a lot harder/totally done differently then images, and if so.. what is the best way to do this?

P.s. I do want all the fancy equations to be included, just normally entered data (which would just follow some sort of a protocol).

Thanks in advance.

Hello to all

Here is the headache:

I want to make a website. In this website there is a table with multiple columns and rows. The cells of the table are editable by the user on the web. After filling the cells with numbers, the user clicks a button and the table is sent to an existing (but blank) excel spreadsheet.

I have the website and i am trying to understand the code beneath this. This is for my graduation paperwork and the schedule is very tight.

If you can teach me how to this I would be most grateful:

How to make a php script that transfers all the data in a table to an excel sheet? I am using ODBC connection and sql language like this one:

<?php
$ligacao=odbc_pconnect("test","","") or die ("error");
$sql="INSERT INTO [sheet$](F1) VALUES ('value1')";
$resultado=odbc_exec($ligacao,$sql);
echo $resultado;
odbc_close($ligacao);
?>

Thank you a lot!
 

I have a PHP code to download one mysql table to excel sheet.what i want to do is download 2 tables in to same excel sheet.those tables are should be in different excel worksheet.please help me. this code work fine.it's download one table

<?php
ob_start();
session_start();
include('dbconnection.php');
$usr= $_SESSION['fname'];

header('Content-Type: text/csv');
header('Content-Disposition: attachment;filename=exported-data.csv');

$select_table=mysql_query("select * from regfarmer WHERE ffname='$usr'");

$rows = mysql_fetch_assoc($select_table);

if ($rows)
{
getcsv(array_keys($rows));
}
while($rows)
{
getcsv($rows);
$rows = mysql_fetch_assoc($select_table);
}

function getcsv($no_of_field_names)
{
$separate = '';

foreach ($no_of_field_names as $field_name)
{
if (preg_match('/\\r|\\n|,|"/', $field_name))
{
$field_name = '' . str_replace('', $field_name) . '';  
}
echo $separate . $field_name;

$separate = ',';
}

echo "\r\n";
}
?>

So I need to echo a row from my database with php, but where i need to echo is already inside an echo. This is my part of my code:


$con = mysql_connect("$host","$username","$password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("main", $con);

$result = mysql_query("SELECT * FROM Vendor");

while($row = mysql_fetch_array($result))
  {
//I need to echo right here  .................. but I get a blank page when I try this. Please Help.
  echo '<option value=$row['vendor_id']>'; echo $row['vendor_id']; echo '</option>';
  }

mysql_close($con);

Result: A Blank page.
Thanks in advance!

I have a log system that allows 10 logs on each side(Left and right). I am trying to make it so that the left side has the 10 most recent logs, then the right as the next 10. Any ideas?

onkeydown="this.style.fontStyle='normal';"

How can I add color: #000; to this?

Thanks,  

Hi   I try to echo out random lines of a html file and want after submit password to whole content of the same html file. I have two Problems. 1st Problem When I echo out the random lines of the html file I don't get just the text but the code of the html file as well. I don't want that. I just want the text. How to do that?

for($x = 1;$x<=40;$x++) {
		$lines = file("$filename.html");
        echo $lines[rand(0, count($lines)-1)]."<br>";
        }

I tried instead of "file("$filename.html");" "readfile("$filename.html");" But then I get the random lines plus the whole content. Is there anything else I can use instead of file so that I get the random lines of text  without the html code?P.S file_get_contents doesn't work either have tried that one.     2nd Problem:   As you could see in my first problem I have a file called $filename.html. After I submit the value of a password I want the whole content. But it is like the program did forget what $filename.html is. How can I make the program remember what $filename.html is? Or with other words how to get the whole content of the html file?   My code:

if($_POST['submitPasswordIT']){
if ($_POST['passIT']== $password ){
$my_file = file_get_contents("$filename.html");
echo $my_file;
}
else{
echo "You entered wrong password";
}
}

If the password isn't correct I get: You entered wrong password. If the password is correct I get nothing. I probably need to create a path to the file "$filename.html", but I don't know exactly how to do that.   

Background:
I'm comparing 2 styles of Ajax:
1.) "jquery style"
2.) "ActiveXObject Microsoft.XMLHTTP style"

Question:
Is one better (faster, more cross-browser compliant) than the other?

My experience:
Both seem equally fast. The Microsoft style is a bit longer, but I don't have to load jquery.js to my page!

Code Examples:

Jquery style on my PHP page:

function getInfo(ProductNumber){
$.ajax({
url:'Ajax-PHP-Page.php?ProductNumber='+ProductNumber,
success: function(html) {
document.getElementById("my_div").value = '';
document.getElementById("my_div").value = html;
}
});
}

Microsoft style on my PHP page:

function getInfo(ProductNumber) {
        if (window.XMLHttpRequest) {
            // code for IE7+, Firefox, Chrome, Opera, Safari
            xmlhttp = new XMLHttpRequest();
        } else {
            // code for IE6, IE5
            xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
        }
        xmlhttp.onreadystatechange = function() {
            if (this.readyState == 4 && this.status == 200) {
                document.getElementById("my_div").value = this.responseText;
            }
        };
    xmlhttp.open("GET","Ajax-PHP-Page.php?ProductNumber="+ProductNumber,true);
  xmlhttp.send();
}

Thank you!!

Edited May 5, 2020 by StevenOliver

Hello All,

I need some help and hopefully will respond. I'm trying to put some information to the second row of the excel sheet once extracted. Can somebody help me with this script? I'm trying to move the address line to the second row. Thanks.

For some reason, I can only get 1 row to echo out when there are actually multiple rows in the database that should be echoing.

Here is what I have:


$result = mysql_query("SELECT * FROM boards WHERE boardname='$board' ORDER BY id LIMIT 10");
$post = mysql_fetch_assoc($result);
echo stripslashes($post['message']) . "<br>\n" . " --- ".stripslashes($post['username']). " on ".stripslashes($post['time']) ."\n<hr width=90%>\n";


It pulls the one record great, but it only shows one record... I want to keep it to 10 records, thus the LIMIT in there (which I think I did right...), but it won't even show the ones in there right now (under 10, so that's not an issue yet).

Code: [Select]
<?php
$sum_num = mysql_query("SELECT SUM(likes) as totallikes FROM facebook");
echo $sum_num;

?>


i got this result Resource id #27
how to get a number only?
 

What is the correct way to write an if/else statement within an echo? I need to alter this so that I can query to see what data is found and if not correct not to echo the rest of the statement.

Code: [Select]
echo '<td class="productbox">
<h1>' . $product_title . '</h1>
</td>';

So from the above code which is echoed within  the single quotes, what is the correct way to include an if else check on a value from the database.

I know how its done, but just want to save time and write it the correct way within this.

Hi all,

I have a page which simply pulls info from a database by id:
<?php
include ('connect.php');
$id = $_GET['id'];

$query = mysql_query("SELECT * FROM JOBS WHERE id=$id");
if (!$query) {
     echo "Could not run query: " . mysql_error();
     exit;
}
$row = mysql_fetch_row($query);
{
echo "<body><h3>" . $row[1]. "</h3>";
echo "<h4>" . $row[2] . "</h4>";
echo "<h4>" . $row[3] . "</h4>";
echo "<h5>Duties & Responsibilities:</h5><ul>";
echo "<li><strong>" . $row[4] . "</strong>" . $row[5] . "</li>";
echo "<li><strong>" . $row[6] . "</strong>" . $row[7] . "</li>";
echo "<li><strong>" . $row[8] . "</strong>" . $row[9] . "</li>";
echo "<li><strong>" . $row[10] . "</strong>" . $row[11] . "</li>";
}
?>

However in some cases the rows in the database may only contain data upto row 6 for example, how would I go about coding this so that it only displays rows that exist. If row 6 exists then 7 always will too as the information is connected. I am manually added this stuff into phpmyadmin as I do not need a form as once it is complete then it will not need to be added to.

Also row 8 and 9 may contain data but 6 and 7 may not

Many Thanks

How do I echo the day from the db? It just displays day and I want it to display the day that was set.


<select name="date_of_birth" id="date_of_birth">
<option value="">Day</option>
<?php 

$isset = isset($_POST['date_of_birth']);

for($day = 1; $day <= 31; ++$day)
{
echo '<option';

if ($isset && $_POST['date_of_birth'] === $day)
{
echo ' selected="selected"';
}

echo ">${day}</option>\n\t\t\t\t\t\t";
}

?>
</select>