PHP - Add Yes To New Field In Database

Here's the scenario:

I have an output (list of fields from a database in a table format). It is a list of items in my game, but I want to make the name be hyperlinked to the information from the database (name, attributes, price, etc.).

How would I create a hyperlink so that when I do click on the item's name, it will open up the item's information to make it editable?

(I have a form that creates the item already, but I want to make it have that information pre-populated for that item so if it needs edited, it could be at that point.)

Thanks!

Hopefully I explained that ok...

Hi,

I need to make it so the items on my page are only displayed if the value in their database column is set to 1 (and 0 for no display). And I should be able to change those values with a checkbox toggle on my admin panel.

Ideally I wouldn't have to click submit either.

Table:

<table id="datatables" class="table table-striped table-no-bordered table-hover" cellspacing="0" width="100%" style="width:100%">
                      <thead>
                        <tr>
                          <th>ID</th>
                          <th>Item name</th>
                          <th>Item group</th>
                          <th>Actions</th>
                          <th>Display</th>
                        </tr>
                      </thead>
                      <tbody>
                        <?php
                        if($result -> num_rows > 0){
                          while($row = $result-> fetch_assoc()){
                            echo "<tr>";
                            echo "<td>" . $row["id"] ."</td>";
                            echo "<td>" . $row["item_name"]. "</td>";
                            echo "<td>" . $row["group_name"] ."</td>";
                            ?>
                            <?php
                            require_once "deleteitem.php";
                            ?>
                            <td>
                              <a href="edititem.php?edit=<?php echo $row['id']; ?>" name="edit"><i class="material-icons">edit</i></a>
                              <a href="?delete=<?php echo $row['id']; ?>" name="delete"><i class="material-icons">delete</i></a>
                            </td>
                            <td>
                            <form action="displayitem.php" method="POST">
                              <div class="togglebutton">
                                <label>
                                  <input type="checkbox" name="display">
                                  <span class="toggle"></span>
                                </label> 
                              </div>
                              </form>
                            </td>
                            <?php
                            echo "</tr>";
                        }
                          }
                        ?>
                      </tbody>
                    </table>

https://i.imgur.com/Qyz6s8v.png

In displayitem.php

<?php
if(isset($_POST['display'])){
    echo '<p>'.$_POST['display'].'</p>';
	// queries etc here
}
?> 

I'm not getting the value here unless I add a submit button.

Guessing jQuery / AJAX is the way to go but can't seem to find any good tutorials on it, are there any other suitable approaches to this?

 

 

 

 

I need some help with updating the database table with the correct selection from that box.
I have a dropdown box. This is actually a list from the images in a curtain directory.
here the code I used:

<?php
$dhandle = opendir('../images/Vessels/');
$files = array();
if ($dhandle) {
   // loop through all of the files
   while (false !== ($fname = readdir($dhandle))) {
      if (($fname != '.') && ($fname != '..') &&
          ($fname != basename($_SERVER['PHP_SELF']))) {
          $files[] = (is_dir( "./$fname" )) ? "(Dir) {$fname}" : $fname; }
            }
   closedir($dhandle);
}


So when inserting a new record I can use the dropdownbox to populate the correct field.

Database:
Id, Vesselname, Image, more, andmore

There is a section where an authorized user can upload an image to the directory.
This image will then be available in dropdown when the need to add a new vessel (in this case)
The problem that I have is, when I later what to update the vessel, I want that same dropdownbox
to be showing the correct imagename as "selected".
How can I do that?

This is what I have been expirementing with sofar:

<?php require_once('../Connections/bgf_db.php'); ?>
<?php
$dhandle = opendir('../images/Vessels/');
$files = array();
$selectedFile = "mare.jpg";
if ($dhandle) {
   // loop through all of the files
   while (false !== ($fname = readdir($dhandle))) {
      if (($fname != '.') && ($fname != '..') &&
          ($fname != basename($_SERVER['PHP_SELF']))) {
          $files[] = (is_dir( "./$fname" )) ? "(Dir) {$fname}" : $fname; }
      }
   closedir($dhandle);
}
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  if (PHP_VERSION < 6) {
    $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
  }
  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);
  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? doubleval($theValue) : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}
$MyFile_Recordset1 = "no-picture.jpg";
if (isset($selectedFile)) {
  $MyFile_Recordset1 = $selectedFile;
}
mysql_select_db($database_bgf_db, $bgf_db);
$query_Recordset1 = sprintf("SELECT vessels.Name, vessels.Image FROM vessels WHERE vessels.Image = %s", GetSQLValueString($MyFile_Recordset1, "text"));
$Recordset1 = mysql_query($query_Recordset1, $bgf_db) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);
$selectedFile = $row_Recordset1['Image'];
?>
<select name="NameHere">
<option value="<?php echo $selectedFile?>" 
<?php if (!(strcmp($selectedFile, $row_Recordset1['Image']))) {echo "selected=\"selected\"";} ?>><?php echo $selectedFile ?>
</option>
<?php
// Now loop through the files, echoing out a new select option for each one
foreach( $files as $fname )
{
   echo "<option value=\"$fname\">{$fname}</option>\n";
}
echo "</select>\n";
?>


 

Howdy,

I'm trying to display text from a table in a database. It's a list of quotes, so I just need to pull out the quote and the author name. However, the quote and name are not fields in the same record; they are separate records. Example data:

Code: [Select]
quoteid    name        value
1            content You guys are great! Thanks for being awesome.
1            author John Jackson
2            content Gosh you're amazing! Always been so darn helpful!
2            author Peter Davis
So, I just need to rip out the data from 'content' and 'author', and then group them together based on the quoteid. This is my code thus far:

$testimonial_resource = mysql_query("SELECT name, value FROM quotes GROUP BY quoteid ORDER BY author ASC") or die(mysql_error());

while ($testimonial = mysql_fetch_assoc($testimonial_resource)) {

echo '<p>'.$testimonial['value']'.<br /><strong>'.$testimonial['value'].'</strong></p>';

}

Any help would be greatly appreciated for this novice.

Cheers.

Hey guys,
I m not yet an experienced coder. SO faced alot of problems. Here, I'm trying to generate a UNIQUE RANDOM NUMBER set between two numbers. Repetative occurance must be avoided by all means as I want every number so generated bears a unique value. In other words, every values that made entry into the field of my database should be differant from each others.


<?php
$conN=mysql_connect("localhost","root","");
if(!$conN)
{
die('error'.mysql_error());
}
mysql_select_db("freebie_allusers",$conN);
$UIN=mt_rand(1,5);
$locateUIN="SELECT UIN FROM user_info WHERE UIN='".$UIN."'";
$fetchUIN=mysql_query($locateUIN);
$resultUIN=mysql_num_rows($fetchUIN);
if($resultUIN>0)
{



WHAT CODE IS REQUIRED HERE SO AS TO GENERATE UNIQUE RANDOM NUMBER?






}

else 
echo $UIN;
?>

Thanx in advance

Hi all.

The title pretty much says it all.  I really have no idea if this is even possible or where to start.  A recent dilemma caused me to have to change a field name with some spaces in it.  I have a database with ~65 tables, a majority of which were converted from Access.  So I'm curious as to how many others have bad field names.  Is there a way to use PHP to go through all the field names in a database and remove spaces?

i hav dis checkstatus.php page where the user enters a random number sms to him..but after tht he shud be able to see d reply given by admin ..d problem over here is d user only views half of the field called 'anything'
(my database is named "proj")
heres my code
jst cant understand why dis is happening...pls help

Dear all,
I have a problem that I need to be helped with using one form and storing the images in database. I would like to upload multiple images with same title from "title" field.
Can somebody help me with this?

<?php
$db_host = 'localhost'; // don't forget to change 
$db_user = 'mysql-user'; 
$db_pwd = 'mysql-password';

$database = 'test';
$table = 'ae_gallery';
// use the same name as SQL table

$password = '123';
// simple upload restriction,
// to disallow uploading to everyone


if (!mysql_connect($db_host, $db_user, $db_pwd))
    die("Can't connect to database");

if (!mysql_select_db($database))
    die("Can't select database");

// This function makes usage of
// $_GET, $_POST, etc... variables
// completly safe in SQL queries
function sql_safe($s)
{
    if (get_magic_quotes_gpc())
        $s = stripslashes($s);

    return mysql_real_escape_string($s);
}

// If user pressed submit in one of the forms
if ($_SERVER['REQUEST_METHOD'] == 'POST')
{
    // cleaning title field
    $title = trim(sql_safe($_POST['title']));

    if ($title == '') // if title is not set
        $title = '(empty title)';// use (empty title) string

    if ($_POST['password'] != $password)  // cheking passwors
        $msg = 'Error: wrong upload password';
    else
    {
        if (isset($_FILES['photo']))
        {
            @list(, , $imtype, ) = getimagesize($_FILES['photo']['tmp_name']);
            // Get image type.
            // We use @ to omit errors

            if ($imtype == 3) // cheking image type
                $ext="png";   // to use it later in HTTP headers
            elseif ($imtype == 2)
                $ext="jpeg";
            elseif ($imtype == 1)
                $ext="gif";
            else
                $msg = 'Error: unknown file format';

            if (!isset($msg)) // If there was no error
            {
                $data = file_get_contents($_FILES['photo']['tmp_name']);
                $data = mysql_real_escape_string($data);
                // Preparing data to be used in MySQL query

                mysql_query("INSERT INTO {$table}
                                SET ext='$ext', title='$title',
                                    data='$data'");

                $msg = 'Success: image uploaded';
            }
        }
        elseif (isset($_GET['title']))      // isset(..title) needed
            $msg = 'Error: file not loaded';// to make sure we've using
                                            // upload form, not form
                                            // for deletion


        if (isset($_POST['del'])) // If used selected some photo to delete
        {                         // in 'uploaded images form';
            $id = intval($_POST['del']);
            mysql_query("DELETE FROM {$table} WHERE id=$id");
            $msg = 'Photo deleted';
        }
    }
}
elseif (isset($_GET['show']))
{
    $id = intval($_GET['show']);

    $result = mysql_query("SELECT ext, UNIX_TIMESTAMP(image_time), data
                             FROM {$table}
                            WHERE id=$id LIMIT 1");

    if (mysql_num_rows($result) == 0)
        die('no image');

    list($ext, $image_time, $data) = mysql_fetch_row($result);

    $send_304 = false;
    if (php_sapi_name() == 'apache') {
        // if our web server is apache
        // we get check HTTP
        // If-Modified-Since header
        // and do not send image
        // if there is a cached version

        $ar = apache_request_headers();
        if (isset($ar['If-Modified-Since']) && // If-Modified-Since should exists
            ($ar['If-Modified-Since'] != '') && // not empty
            (strtotime($ar['If-Modified-Since']) >= $image_time)) // and grater than
            $send_304 = true;                                     // image_time
    }


    if ($send_304)
    {
        // Sending 304 response to browser
        // "Browser, your cached version of image is OK
        // we're not sending anything new to you"
        header('Last-Modified: '.gmdate('D, d M Y H:i:s', $ts).' GMT', true, 304);

        exit(); // bye-bye
    }

    // outputing Last-Modified header
    header('Last-Modified: '.gmdate('D, d M Y H:i:s', $image_time).' GMT',
            true, 200);

    // Set expiration time +1 year
    // We do not have any photo re-uploading
    // so, browser may cache this photo for quite a long time
    header('Expires: '.gmdate('D, d M Y H:i:s',  $image_time + 86400*365).' GMT',
            true, 200);

    // outputing HTTP headers
    header('Content-Length: '.strlen($data));
    header("Content-type: image/{$ext}");

    // outputing image
    echo $data;
    exit();
}
?>
<html><head>
<title>MySQL Blob Image Gallery Example</title>
</head>
<body>
<?php
if (isset($msg)) // this is special section for
                 // outputing message
{
?>
<p style="font-weight: bold;"><?=$msg?>
<br>
<a href="<?=$PHP_SELF?>">reload page</a>
<!-- I've added reloading link, because
     refreshing POST queries is not good idea -->
</p>
<?php
}
?>
<h1>Blob image gallery</h1>
<h2>Uploaded images:</h2>
<form action="<?=$PHP_SELF?>" method="post">
<!-- This form is used for image deletion -->

<?php
$result = mysql_query("SELECT id, image_time, title FROM {$table} ORDER BY id DESC");
if (mysql_num_rows($result) == 0) // table is empty
    echo '<ul><li>No images loaded</li></ul>';
else
{
    echo '<ul>';
    while(list($id, $image_time, $title) = mysql_fetch_row($result))
    {
        // outputing list
        echo "<li><input type='radio' name='del' value='{$id}'>";
        echo "<a href='{$PHP_SELF}?show={$id}'>{$title}</a> &ndash; ";
        echo "<small>{$image_time}</small></li>";
    }

    echo '</ul>';

    echo '<label for="password">Password:</label><br>';
    echo '<input type="password" name="password" id="password"><br><br>';

    echo '<input type="submit" value="Delete selected">';
}
?>

</form>
<h2>Upload new image:</h2>
<form action="<?=$PHP_SELF?>" method="POST" enctype="multipart/form-data">
<label for="title">Title:</label><br>
<input type="text" name="title" id="title" size="64"><br><br>

<label for="photo">Photo:</label><br>
<input type="file" name="photo" id="photo"><br><br>

<label for="password">Password:</label><br>
<input type="password" name="password" id="password"><br><br>

<input type="submit" value="upload">
</form>
</body>
</html>

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=318312.0

I have a PHP Form where users have to enter and upload a file but when they click upload everything is working except that it is only sending first character of text field to Database!

Can any one help??

Thanks

File upload.php form
Code: [Select]
<form action="add_file.php" method="post" enctype="multipart/form-data">
           <p><br>
      JobID:
        <input name="JobID" type="text" value="<?php echo $row_Recordset1['JobID']; ?>" readonly="readonly" />
    </p>
    <p>Company ID:
      <input name="CompanyID" type="text" value="<?php echo $row_Recordset1['CompanyID']; ?>" readonly="readonly" />
  </p>
    <p>UserID:
      <input name="UsersID" type="text" />
    </p>
    <p>Select File to upload:
      <input type="file" name="uploaded_file" />
  </p>
    <p>Make Sure All Details Are Correct before Submitting!</p>
    <p>
      <input type="submit" value="Submit Application" />
    </p>
        <form>   

add_file.php

// Check if a file has been uploaded
if(isset($_FILES['uploaded_file'])) {
    // Make sure the file was sent without errors
    if($_FILES['uploaded_file']['error'] == 0) {
        // Connect to the database
        $dbLink = new mysqli('localhost', 'user', 'password', 'phpsite');
        if(mysqli_connect_errno()) {
            die("MySQL connection failed: ". mysqli_connect_error());
        }
 
        // Gather all required data
        $JobID = $dbLink->real_escape_string($_POST['JobID']['JobID']);
      $CompanyID = $dbLink->real_escape_string($_POST['CompanyID']['CompanyID']);
      $UsersID = $dbLink->real_escape_string($_POST['UsersID']['UsersID']);
      $name = $dbLink->real_escape_string($_FILES['uploaded_file']['name']);
        $mime = $dbLink->real_escape_string($_FILES['uploaded_file']['type']);
        $data = $dbLink->real_escape_string(file_get_contents($_FILES  ['uploaded_file']['tmp_name']));
       $size = intval($_FILES['uploaded_file']['size']);
 
        // Create the SQL query
        $query = "
            INSERT INTO `application` (
                `JobID`, `CompanyID`, `UsersID`, `name`, `mime`, `size`, `data`, `N_DateAndTime`
            )
            VALUES (
                '{$JobID}', '{$CompanyID}', '{$UsersID}', '{$name}', '{$mime}', {$size}, '{$data}', NOW()
            )";
 
 
        // Execute the query
        $result = $dbLink->query($query);
 
        // Check if it was successfull
        if($result) {
            echo 'Success! Your job application was successfully sent!';
        }
        else {
            echo 'Error! Failed to insert the file, please try again'
               . "<pre>{$dbLink->error}</pre>";
        }
    }
    else {
        echo 'An error accured while the file was being uploaded, please try again. '
           . 'Error code: '. intval($_FILES['uploaded_file']['error']);
    }
 
    // Close the mysql connection
    $dbLink->close();
}
else {
    echo 'Error! A file was not sent! Please try again!';
}
 
// Echo a link back to the main page
echo '<p>Click <a href="year1index.php">here</a> to go back</p>';
?>

MOD EDIT: [code] . . . [/code] BBCode tags added.

Hi Good day Can someone pls help me regarding with this matter.
I have a page for inserting records. And I have a combobox that is fetched through the specific field in database. I also have a table of records, for each row i have an href edit which throws the values to its respective texbox and combobox. 
THE PROBLEM is that when i click the href edit, the value from a selected row was not displayed from the combobox

hiii all, i want to show the database field (alise name) as the url after index.php when page or site loads first time. how i can achieve it? please help.... thanks in advance

Or something like that... I am not sure how to put this..

Anyway, I'll just get started with explaining my problem.

I have an admin-page in which you can delete the comments given on blogs, using checkboxes and clicking on a button with the value 'verwijderenSubmit'. The deletion part works just fine, nothing wrong.

However, I also want to be able to EDIT the comments with an other button called 'bewerkenSubmit', using the same checkboxes that I use for deletion.
Selecting the right CID (CommentID) is no problem, because that works the same as the deletion-part, but selecting the right textarea to update into the database is the problem...

I uploaded a file here with the whole code: http://dhost.info/ddfs/myproblem.html
I escaped the textarea within with square brackets, because otherwise the whole textarea would screw up..
I also added <!-- RELEVANT CODE --> to select the parts that I need to change.

Well, I hope you understand my problem and can help.