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CroNiX

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Posts posted by CroNiX

    form post with array

    in PHP Coding Help

    Posted

    The form controls need to be located within the form open (<form>) and close (</form>) tags.



    
<form action = "test.php" method = "post">
    
<select name="Words">
    
<?php
    
FOREACH($ddArray1 AS $word){
    
    PRINT '<option value="'.$word.'">'.$word.'</option>';
    
}
    
?>
    
</select>
    
</form>

    • Like 1

    swopping description and date in mysql database

    in PHP Coding Help

    Posted

    It's being swapped because that's what you are telling it to do.

    INSERT INTO event_calendar (title,event_date,description)
    VALUES (:title, :description, :event_date)')

    Your insert is saying (in order), title, event_date, description

     

    but your VALUES (in order) are :title, :description, :event_date.

     

    Swap the :description and :event_date placeholders in the VALUE.

    Limit While PDO MySQL Results

    in PHP Coding Help

    Posted · Edited by CroNiX

    Barand has a better solution, but mine was something like:

    $count = 0; //keep track of where we are in the array
$separate = 4; //number of entries before new ul
echo '<ul>';
while($row = $results->fetch(PDO::FETCH_ASSOC))
{
  //If we hit $separate by checking modulus operator, close ul and start a new one
  if ($x%$separate == 0)
  {
    echo '</ul><ul>';
  }

  //output li's
  echo '<li>Mileage: '.number_format($row["Mileage"]).'</li>';
  echo '<li>Engine size: '.$row["EngineSize"].'cc</li>';

  //increment counter
  $count++;
}
echo '</ul>';

    I would like to read a dir of images and save the imagefile name to mysql

    in PHP Coding Help

    Posted

    Since $files is an array, $i should start at 0, not 1 in the for() loop. You're skipping the first image.

     

    Please also note that the mysql extension is deprecated, meaning it will no longer be available in php in the future. If you want to write code that will work on future versions of php without having to rewrite all of your queries, it's highly suggested you use the PDO or mysqli extension.

    Update Query Help

    in PHP Coding Help

    Posted · Edited by CroNiX

    It's most likely on your server but just not enabled. Look in your php.ini in the extensions section and look for

    ;extension=php_mysqli.so (or .dll if windows)

    remove the ; from the front of the line and restart apache.

    Display data in table format from two tables

    in MySQL Help

    Posted

    I believe your data is correct. An order can have many products, correct? What you'd need to do is in your loop when you are outputting the data, check to see if the order_id is the same as the last order_id, if not display the order_id, if so it's part of the same order and output a   in the <td>.

     

    You'd probably want to ORDER your query by (orders.order_id, orders.company_id, orders.order_for)

    some pseudocode:

    $last_order_id = null; //store the last order_id
while($row = $query->fetch_array())
{
  //check to see if this is a new order or part of the same one
  if ($row['order_id'] != $last_order_id)
  {
    //nope, it's a new order, store as $last_order_id and output the order_id
    $last_order_id = $row['order_id'];
    echo '<td>' . $row['order_id'] . '</td>';
  }
  else
  {
    //it's part of the same order, output a space for this td
    echo '<td> </td>';
  }
}

    You'd need to do the same thing for anything else you don't want to repeat

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