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PHP - Help Creating A Vertical Table Chart

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Help Creating A Vertical Table Chart	View Content

I am trying to create something similar to a bar chart, using a html table based on the values stored in an array. However, I cannot seem to get the next column to begin after printing the number at the bottom. I can get it going horizontal, but I would like to create a vertical version. If anyone can provide me with help I would greatly appreciate it.

Code: [Select]

<?php

$month = date("n");     //numeric representation of month (1-12)
$year = date("Y");      //4 digit representation of year
$day = date("d");   //numeric representation of the day of the month

// constant variable to determine how many days alive according to the current date
define('DAYS_ALIVE',(($year - 1902) * 365.24) + (($month - 1) * 30.5 )+ ($day - 1));
$divident = 100000; // variable for modulus
$divider = 10000; // variable to shrink the divident

for($i = 0; $i <= 4; $i++) // for loop to fill the array with days alive
{
$daysArray[$i]= floor((DAYS_ALIVE % $divident)/$divider); //calculates the single digit to store in array

$divident= $divident/10; //after every iteration the divident is decreased
$divider= $divider/10; //after every iteration the divider is decreased
}

echo "<table border='2' width='100%'>";

$arraySize = count($daysArray); //variable for the array size

for($k=0;$k<=$arraySize-1;$k++)
{

echo "<tr>";

for($j=1;$j<=$daysArray[$k];$j++)
{

echo "<td> * </td>";

}

echo "<td>$daysArray[$k]</td>";

}

echo "</table>";
?>

What I want is something like this.

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Moved: Vertical-align='top' Ignored In Table

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This topic has been moved to HTML Help.

http://www.phpfreaks.com/forums/index.php?topic=326428.0

Creating Table From String

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I have a string (Product1-Product2-Product3#2-4-1) where Product1-Product2-Product3 are products and 2-4-1 are quantities. I am struggling to make a table out of it where products and quantities are shown in rows.

here is what i tried.

Code: [Select]
list($products, $quantities) = explode("#", $specs); //to separate Quantities from products

<?php foreach(preg_split("/-/", $products) as $product_list) { ?>

<tr>
<td> <?php echo $product_list; ?></td><? } ?> <?php foreach(preg_split("/-/", $quantities) as $qtylist) { ?> <td></td>
<td><?php echo $qtylist; ?></td> <?php } ?>
</tr>

<?php //} ?>
</table>

I know something is wrong with this and its giving me unexpected reasult

Please help me create a table out of the above string.

Many Thanks
-Max

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Hi there, trying to create a function more out of lazyness an timesaving than anything
im sure it has to do with something combining the use of INSERT and x++ in an array, but im failing at figuring out the right way of doing it to achieve the right combination of effects.

in simple

need to create a 600x600 grid of co-ordinates where x starts at 1, ending in 600 with y starting at 1 and ending in 600

so basically a cubed set of entrys, (360,000 rows) not even sure thats possible lol..

the table only needs to contain the colums of id, x , y , ownerid (where id is auto inc)

any suggestions, as you can tell its a pain-staking long case of affairs to create with a manual VERY long sql statement, trying to figure out a better way of doing this.
help please?

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        //echo $fabricsql;
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        $i=0;
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Hi guys, i cannot figure out what is wrong with this and I hope you could help figure it out?

This is the code I am using -
Code: [Select]

$create_cart_table = "CREATE TABLE `". $string ."` (cart_id int NOT NULL AUTO_INCREMENT, PRIMARY KEY(cart_id), item_name VARCHAR(1000) NOT NULL, item_id VARCHAR(30) NOT NULL, price VARCHAR(30) NOT NULL, date_time TIMESTAMP(100) NOT NULL CURRENT_TIMESTAMP, quantity VARCHAR(100) NOT NULL)";
$set_cart_table = mysql_query($create_cart_table);

$string is defined above and there are no database connection warnings etc coming up and I am lost as to how the heck this isn't working? It was working for about an hour when I first made the script but now it is not. Please help?

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Windows 7, Apache 2.2.21 / PHP 5.3.5. What does the below error mean? I am new and fighting through this particular piece of code.

Thank you in advance for your help.

Warning: mysql() expects parameter 3 to be resource, string given in C:\website\do_createtable.php on line 3

<?
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-Thanks!

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HI. Im new to database. I didnt know how to use bridge table. In my bridge table i have SubjectID and studentID, PK from student table and subject table. In my subject table have SubjectID, SubjectName and SubjectLecturer while in my student table have studentID,,studentName,IC,studentFac,studentPhone, studentEmail,pswd,studentAddress. I didnt know where is the wrong part because when i click my form which is enroll button, this enroll page blank Can someone check my coding ? Help ? <!DOCTYPE HTML> <html> <head> <?php
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echo "Login fail Try again <a href='http://bytes.com/'>here</a>";
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FROM Subject
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ok, I'm not sure I know how to ask this question, but I've created a table that allows the user to add rows. The problem is, upon submit only the last row creates variables. How can I make sure every cell in every row sets a different variable?

This is going to be a purchase order system.

Here is my code:
Code: [Select]
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Insert Table Row using DOM</title>
<script language="javascript">
function addRow()
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var tbody = document.getElementById("po_table").getElementsByTagName("tbody")[0];
var row = document.createElement("TR");
var cell1 = document.createElement("TD");
var cell2 = document.createElement("TD");
var cell3 = document.createElement("TD");
var inp1 = document.createElement("INPUT");
inp1.setAttribute("type","text");
inp1.setAttribute("name","quantity");
cell1.appendChild(inp1);
var inp2 = document.createElement("INPUT");
inp2.setAttribute("type","text");
inp2.setAttribute("name","description");
cell2.appendChild(inp2);
var inp3 = document.createElement("INPUT");
inp3.setAttribute("type","text");
inp3.setAttribute("name","unit_cost");
cell3.appendChild(inp3);
row.appendChild(cell1);
row.appendChild(cell2);
row.appendChild(cell3);
tbody.appendChild(row);
//alert(row.innerHTML);
}
</script>
</head>
<body>
<form method="post" action="review_po.php">
<table id="po_table" border="1">
<tbody>
<tr>
<th bgcolor="#c1c1c1">Quantity</th>
<th bgcolor="#c1c1c1">Description</th>
<th bgcolor="#c1c1c1">Unit Cost</th>
</tr>
<tr>
<td><input type=text name="quantity"></td>
<td><input type=text name="description"></td>
<td><input type=text name="unit_cost"></td>

</tr>
</tbody>
<tfoot>
<tr>
<th bgcolor="#e0e0e0">1</th>
<th bgcolor="#e0e0e0">Shipping</th>
<td><input type=text name="shipping"></td>

</tr>
<tr>
<th bgcolor="#e0e0e0">1</th>
<th bgcolor="#e0e0e0">Tax</th>
<td><input type=text name="tax"></td>

</tr>
</tfoot>
</table>
<input type="button" value="Insert Row" onClick="addRow();">
<input type="submit" value="Review" name="review_po">
</form>

</body>
</html>
right now it goes to this page:

Code: [Select]
<?php

//convert to variables
$quantity = $_POST['quantity'];
$description = $_POST['description'];
$unit_cost = $_POST['unit_cost'];
$extended_cost = $_POST['extended_cost'];
$shipping = $_POST['shipping'];
$tax = $_POST['tax'];

?>

<?php echo $quantity; ?><br>
<?php echo $description; ?><br>
<?php echo $unit_cost; ?><br>
<?php echo $extended_cost; ?><br>
<?php echo $shipping; ?><br>
<?php echo $tax; ?><br>

here is the page:
http://po.shopphgmag.com/create_po.html

Thanks for looking

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Need a little help with this code. I'm running Apache 2.21 / PHP 5.3.5 on Windows. This is a V6 application. I am running Windows 7. I am new, and trying to learn. What am I doing wrong in line 8 with the 'if' statement? The idea is to create a table in mySQL using a form interface.

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Parse error: syntax error, unexpected '{' in C:\website\do_createtable.php on line 8
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Hello I've registered here last March and I'm just lurking around reading threads about php.

Yesterday I've started to self studied php and today I've test some basic codes and I'm having difficulties on creating a table rows and columns.

Here's my code.

Code: [Select]
<?php

echo "<table width=\"100\" border=\"1\">";

$text = 'php';
$mac_var = 1;
$mac_plus = 4;

while ($mac_var < $mac_plus)
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echo "<tr>";
{
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echo "<td>". $text . "</td>";
$mac_var++;
}
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$mac_var++;
}

echo "</table>";

?>

what I want is, it should have 3 columns and 3 rows.

But I'm getting only 3 columns and 1 row only.

I wonder what is the problem?

Sorry for my ignorance as I'm on the 1st basic phase of using php codes. TIA!

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PHP code:

Code: [Select]
<?
//login variable
$username = $_POST ['username'];
$password = $_POST ['password'];
$database = $_POST ['database'];
//table variable
$table = $_POST ['table'];
//stats variables
$name = $_POST ['name'];
$force = $_POST ['force'];
$armor = $_POST ['armor'];
$strength = $_POST ['strength'];
$value = $_POST ['value'];

mysql_connect ("localhost","$username","$password");
@mysql_select_db ($database) or die ("no database name");

$query = "INSERT INTO `$database`.`$table` ('name', 'force', 'armor', 'strength', 'value')
VALUES ('$name', '$force', '$armor', '$strength', '$value')";

mysql_query ($query);

if (mysql_query($query))
{
die('Error: ' . mysql_error());
}
echo "1 record added<br>";

echo "<br>variable $name test";
echo "<br>variable $force test";
echo "<br>variable $armor test";
echo "<br>variable $strength test";
echo "<br>variable $value test";

mysql_close();

?>

HTML files code:

Code: [Select]
<form action = "datainput.php" method = "post">

<div align="center">Item Submission Form</div><br><br>

<br>
<div align="center"><input type="text" name="table" value="head neck back or chest" size="32" maxlength="32"></div>
<br>
<br>
<div align="center"><input type="text" name="name" value="Type Name Here" size="32" maxlength="32"></div>
<br>
<br>
<div align="center"><input type="text" name="force" value="Type force number" size="32" maxlength="32"></div>
<br>
<br>
<div align="center"><input type="text" name="armor" value="Type armor value" size="32" maxlength="32"></div>
<br>
<br>
<div align="center"><input type="text" name="strength" value="Type strength value" size="32" maxlength="32"></div>
<br>
<br>
<div align="center"><input type="text" name="value" value="Type cost Here" size="32" maxlength="32"></div>
<br>
<br>
<br>
<div align="center"><input type="text" name="username" value="username here" size="32" maxlength="32"></div>
<br>
<div align="center"><input type="text" name="password" value="password here" size="32" maxlength="32"></div>
<br>
<div align="center"><input type="text" name="database" value="database Here" size="32" maxlength="32"></div>
<br>

<div align="center"><input type="submit" name="Submit_button" value="Submit"></div>

</form>

What is funky is the page seems to work, it says it writes a record but the record never appears in the database.
I did verify MIME types were correct for HTML to be allowed to write as php if needed, any idea why it wouldn't create a record?

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	$getdata = curl_exec($curl); 
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        'y' => $user->Amount);
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print_r($dataPoints);
echo "</pre>";
echo json_encode($dataPoints, JSON_FORCE_OBJECT);
?>

<!DOCTYPE HTML>
<div class="container">
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  <div>
    <canvas id="myChart"></canvas>
  </div>
</div>
<script src="https://cdn.jsdelivr.net/npm/chart.js@2.8.0"></script>
<script>
var ctx = document.getElementById("myChart").getContext('2d');
var myChart = new Chart(ctx, {
  type: 'pie',
  data: {
    labels: "{label}",
    datasets: [{
      backgroundColor: [
        "#2ecc71",
        "#3498db",
        "#95a5a6",
        "#9b59b6",
        "#f1c40f",
        "#e74c3c",
        "#34495e"
      ],
      data: <?php echo json_encode($dataPoints, JSON_NUMERIC_CHECK); ?>
    }]
  }
});
</script>
<style>
.container {
  width: 80%;
  margin: 15px auto;
}
</style>

what is wrong? I can no longer see the chart.

Json

{
  "result": [
    {
      "ID": 1,
      "Users": [
        {
          "UserObject": {
            "UserName": "User1",
            "-": {
              "ID": 1
            },
            "0": "0"
          },
          "User": "User1",
          "Amount": 57654456
        },
        {
          "UserObject": {
            "UserName": "User2",
            "-": {
              "ID": 1
            },
            "0": "0"
          },
          "User": "User2",
          "Amount": 57654456
        },
        {
          "UserObject": {
            "UserName": "User3",
            "-": {
              "ID": 1
            },
            "0": "0"
          },
          "User": "User3",
          "Amount": 57654456
        }
      ],
      "Reached": false,
      "IsActive": true
    }
  ],
  "error": false,
  "version": 1
}

Pie Chart Statistics

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Hello,

In a MySQL table I have the following rows:
Item name Item price Milk 1.05 Butter 1.50 Lettuce 0.70 Butter 1.55 Bread 0.65 Chicken 3.50 Milk 1.00 Coca Cola 1.20 Apples 0.90 Toothpaste 1.00 Oranges 0.80 Milk 1.00
What I'm looking to do is collect all of the milks, all of the butters, etc and give a $total_price_spent_on value for each product, so for milk we'd have 3.05, for oranges we'd have 0.80, and so on until all items on the list have been totalled. Then I want to select the most expensive 3 totals, display them in a pie chart, and all other items to be listed under an "other" label.

Any pointers as to how to go about this?

Many thanks