PHP - My Form Wont Load For Me To Edit

hi,

I would like to know how do i modify this code if i Have more than 1 option in my dropdown list. Can someone help..Thankz


<select name="mediaList" id="mediaList">
  <option <?php echo ($rows['media_type'] =='Physical Only')?'selected="Selected"':'';?> >Physical Only</option>
</select>

Hello Freaks,

I am currently working on a status report web application.  A user inputs what they are planning on accomplishing that week.  Afterwords realizes that they made a mistake and need to correct it.  The way I want to go about this is sql the database and return * where the username and report date are current.

With the results of this query I insert them into the form as the value.  This yields a populated form.  From here I want the user to be able to change any line and submit the whole form.  However, the post array is only populated with one row of data.  I thought this might have to do with the names of the inputs all being the same so i tried to index those through the while loop with no luck.

Any help would be appreciated!

here's the code


Code: [Select]
<?
// renaming sess var
$UAID = $_SESSION['UAID'];
$WEND = $_SESSION['WEEK_ENDING2'];

//have we submitted our changes?
if($EDIT == TRUE){

//how many task are there this week
$results = mysql_query("Select COUNT(UAID) FROM `WPFTW` where UAID ='$UAID' and Rend = '$WEND'");
$row = mysql_fetch_row($results);

//clean $_POST
print_r($_POST);
unset($_POST['Report']);
echo "<br>";
print_r($_POST);

}
//what do we plan on editing
$results = mysql_query("Select * FROM `WPFTW` where UAID ='$UAID' and Rend = '$WEND'");

//while there are results
while($row = mysql_fetch_array($results)){

//increment for input name
$k = 1;

//date formatting
list($year,$month,$day) = explode("-",$row['Target']);
$row['Target'] = $month."-".$day."-".$year;

//empty string to uncheck checkboxes
$checked = "";

//populate checkboxes
if($row['MS'] == "on"){
$checked = "yes";
}

//select the correct drop down menu option
if(isset($row['Status'])){
        if($row['Status'] == " "){
        $selected = "selected";
        $selected1 = "";
        $selected2 = "";
        $selected3 = "";
        }

        elseif($row['Status'] == "P"){

        $selected = "";
        $selected1 = "selected";
        $selected2 = "";
        $selected3 = "";
        }

        elseif($row['Status'] == "C"){

        $selected = "";
        $selected1 = "";
  $selected2 = "selected";
        $selected3 = "";
        }

        elseif($row['Status'] == "CL"){

        $selected = "";
        $selected1 = "";
        $selected2 = "";
        $selected3 = "selected";
        }
}

//select the correct drop down menu option
if(isset($row['OT'])){
        if($row['OT'] == "Y"){
        $selected4 = "selected";
        $selected5 = "";
        }

        elseif($row['OT'] == "N"){

        $selected4 = "";
        $selected5 = "selected";
        }
}

//var to send form to self
$cani = htmlentities($_SERVER['PHP_SELF']);

//concatenate html & php for form
$test =  "<form id =\"2bigform\" action=$cani method = \"post\">" . "<tr>" .
        "<td><input id=\"hidden\" name=\"UAID$k\" type=\"hidden\" value=" . $_SESSION['UAID'] ."></td>".
        "<td width=\"7%\"> <input id=\"SR\#$k\" name=\"SR#\" type=\"text\" size=\"10%\" maxlength=\"10\" value=" . $row['SRNUM'] . "></td>" .
        "<td width=\"63%\"> <input id=\"Task$k\" name=\"Task\" type=\"textarea\" size=\"63%\"  value=" . $row['Task'] . "></td>" .
        "<td width=\"7%\"> <select id=\"Status$k\" name=\"Status\">".
                                                        " <option value=\" \" $selected></option>".
                                                        " <option value=\"P\" $selected1>P</option>".
                                                        " <option value=\"C\" $selected2>C</option>".
                                                        " <option value=\"CL\" $selected3>CL</option>".
                                                        "</select></td>".

        "<td width=\"7%\"> <input id=\"MS$k\" name=\"MS\" type=\"checkbox\" size=\"7%\"  value=" . $row['MS'] . " checked = $checked></td>".
        "<td width=\"7%\"> <select id=\"OT$k\" name=\"OT\">".
                                                        " <option value=\"Y\" $selected4>Y</option>".
                                                        " <option value=\"N\" $selected5>N</option>".
                                                        "</select></td>".

     "<td width=\"9%\"> <input id=\"Target$k\" name=\"Target\" type=\"text\" size=\"9%\"  value=" . $row['Target'] . "></td>".
        "<input id=\"hidden\" name=\"RDate$k\" type=\"hidden\" value=" . $_SESSION['WEEK_ENDING'] ."></td></tr>";
echo $test;
$k++;
echo $k;
}
echo "</table>".
     "<input id=\"Sub\" type= \"Submit\" name=\"Report\" value= \"Submit\"/>".
     "</form>".
     "</body>".
     "</html>";

//<td width="7%"> <input id="SR#" name="SR#" type="text" size="10%" maxlength="10"/></td>
//<td width="63%"> <input id="Task" name="Task"  type="textarea" size= "63%"/></td>






}
?>

Hello, freaks. I need a little help with the best way to go about this. I have a field in mysql that contains a comma separated string of values. What I need to do is get those values out of the db and list them into editable text fields so the user can update the values in a html form. So far this is my code, I left out all the in betweens - validation, results, etc - for this post...but its there .

SELECT servs FROM services WHERE id = $a_id

$services = $row['servs'];
$srvs = explode(',', $services);

Then for the form:

<?php
foreach ($srvs as $service) { 
?>

<input type="text" name="servs[]" value="<?php echo $service  ?>" />

  <?php } ?>

This gives me a text field for each value of servs, great! But the problem is, the user can enter up to 15 servs. Some currently have less than that saved in the db. I want to give them the ability to 1. edit any of the current servs value and 2. add more values if they have less than 15.

My question is, how do I echo out 15 editable text fields for servs[] no matter the number of values they currently have? And when I go to implode the values, will blank text fields screw everything up?

I can't find anything in the forums that relate to this. I believe it probably will involve $i, but I am not too familiar with counting in php yet, so I could really use some help. Thanks!

Oh - and I am not stuck on using text fields, but I couldn't figure how to make this work using a select list or checkboxes. Any suggestions either way would help. 

I have the code echoing into the next page where it says your mail was sent successfully but it wont email me all the variables here is the code:
<!DOCTYPE HTML>
<html>
<head>
<meta charset="utf-8">
<title>Submitted Form</title>
</head>

<body>
 <?php
 session_start();
  echo "<html>
        <head><title>Posted Variables</title></head>
        <body>";
 $to = "artdept@calmktg.com";
 $subject = "SCE SONGS ORDER FORM";
 $email = $_REQUEST['email'];
  foreach ($_POST as $field => $value)
  {
     echo "$field = $value<br>";
  }
 $headers = "From: $email";
 $sent = mail($to, $subject, $message, $headers) ;
 if($sent)
 {print "Your mail was sent successfully"; }
 else
 {print "We encountered an error sending your mail"; }
 ?>
</body>
</html>

please help me out if you need anything else just ask....my email is artdept@calmktg.com

Each time i submit this form it wont send to mysql, can anyone help?

<?php
$hostname="mysql6.000webhost.com"; //local server name default localhost
$username="a5347792_meto";  //mysql username default is root.
$password="";       //blank if no password is set for mysql.
$database="a5347792_login";  //database name which you created
$con=mysql_connect($hostname,$username,$password);
if(! $con)
{
die('Connection Failed'.mysql_error());
}
 
mysql_select_db($database,$con);
 
//include connect.php page for database connection
include('connect.php');
//if submit is not blanked i.e. it is clicked.
if($_SERVER['POST_METHOD'] == 'POST')
{
if($_POST['name']=='' || $_POST['email']=='' || $_POST['password']==''|| $_POST['repassword']=='')
{
echo "please fill the empty field.";
}
else
{
$sql="insert into student(name,email,password,repassword) values('".$_REQUEST['name']."', '".$_REQUEST['email']."', '".$_REQUEST['password']."', '".$_REQUEST['repassword']."')";
$res=mysql_query($sql);
if($res)
{
echo "Record successfully inserted";
}
else
{
echo "There is some problem in inserting record";
}
 
}
}
?>

I have a table called "colors".  It has 2 columns, id and color.

All I'm trying to do is pull the data into the form, then edit the colors (for example - misspelled, etc.)
Then submit it back to the table.

Here is the code that pulls the data into a form so I can edit it:
Code: [Select]
<form action="adminupdatecolors.php" method="post">

<table width="500" border="1" cellpadding="10">

<tr>
  <td>Color Options:</td>
    <td>
    <?
$result = mysql_query("SELECT * FROM colors");
while ($row = mysql_fetch_assoc($result)) {

echo '<input type="text" name="color[]" value="' . $row['color'] . '"/>' . $row['id'] . '<br />';
}

?>
   </td>
</tr>

   <tr>
  <td>&nbsp;</td>
  <td><input type="Submit" value="Update Colors"></td>
  </tr>
</form>
</table>

Here is my update script:
Code: [Select]
<?php

include("config.php");
include("db.php");

$id=$_POST['id'];
$color=$_POST['color'];

$result = mysql_query("SELECT id FROM colors ORDER BY id DESC LIMIT 0,1");

if ($row = mysql_fetch_assoc($result)) {
$id = $row['id']; 
}
$sql = "DELETE FROM colors WHERE id='$id'";
mysql_query($sql) or die("Error: ".mysql_error());

foreach ($color as $colorvalue) {
$sql2 = "INSERT INTO colors (id,color) VALUES ($id,'$color')";
mysql_query($sql2) or die("Error: ".mysql_error());

}

header("Location: " . $config_basedir . "adminhome.php");

?>

I have a simple form that edits student information.  I have several fields that I have a drop down list pulled as an array from a separate table.  I would like the information that is already entered on the student to be the default but still have the drop down list available if the item was entered in error. For example:  Drop down list of all the states of America, OH was entered for the student, when the edit form comes up for State in Address I would like OH to be the default but still have the list to choose if OH was entered in error. I have tried selected = $selected but then the entire drop down changed to OH.  I am very, very much a beginner in this

<?php
   // list of possible modules comes from database
         
          $state_array=get_states();
       
       
          //display as a drop down list for end user to choose
        if (!is_array($state_array))
      {
         echo "<p>No States are currently available<p>";
         return;
      }
    
             
       
          foreach ($state_array as $row)
        {
               echo "<option value=\"".($row['stateid'])."\"";
               // if existing state, put in current catgory
               echo "<strong>  ".($row['stateid'])."</strong><br />";       
               echo "<option select=\"selected\">OH";
          }
         
   ?>

Need help ASAP. i need the following form to right to mysql table "table1" and columns "name" "email" and "comment". <form action="sign.php" method="post" class="pure-form pure-form-stacked">
<fieldset>

<label for="name">Your Name</label>
<input id="name" type="text" placeholder="Your Name" name="name">


<label for="email">Your Email</label>
<input id="email" type="email" placeholder="Your Email" name="email">

<label for="comment">Your Comments</label>
<input id="comment" type="text" placeholder="Your Comments" name="comment">

<a href="thanks.html" button type="submit" class="pure-button">SUBMIT</button>
</a>

</fieldset>
</form> My PHP File looks like this: <html>
<body>

<?php
$myUser = "dbuser";
$myPassd = "********";
$myDB = "dbname";
$myserver = "192.168.1.80:3306";
$name = $_POST['name'];
$email = $_POST['email'];
$comment = $_POST['comment'];

// Create connection
$dbhandle = mysql_connect($myserver,$myUser,$myPassd)
or die("Unable to connect to MySQL");
echo "Connected to MySQL<br>";

//select a database to work with
$selected = mysql_select_db($myDB, $dbhandle)
or die("Couldn't open database $myDB");
//if (mysqli_connect_errno()) {
//echo "Failed to connect to MySQL: " . mysqli_connect_error();
//}
//echo "connected";

$sql = "INSERT INTO signatures (name, comment) VALUES ( $name, $comment)";

if (!mysql_query($sql))
{
die('Error: ' . mysql_error());
}
echo "record added";

mysql_close($dbhandle);
?>
</body>
</html> I also need this to redirect to another html page after it run. Like i said any help would be great because i cant get it to work!

Can anyone explain... (note this is first time using PHP , dont have a clue, only using it as FLash CS5 wont send emails forms directly) so add all the details and explain it all please.)

why my yahoo email is not detecting this at all. Also I managed some success earlier with earlier attempts but eveythign was surrounded with html tags even though I set them to off in Flash on the property tabs on the input text.

In the Flash file . swf  the send button has this code

Code: [Select]
on(release){

form.loadVariables("email_send.php", "POST");
}
and the movie containing the input text fields has this code

Code: [Select]
onClipEvent(data){
_root.nextFrame();
}
the next page is a thank you screen

The variables in the input text boxes in the Flash file are

name, dept, phone, email and message




Code: [Select]
<?php

$sendTo = "chris.bruneluni@yahoo.co.uk";
$subject = "Message from chriscreativity.com";

$headers = "From: ". $_POST("name");

$headers = "Postion: ". $_POST["dept"];

$headers =  "<". $_POST["email"] .">". "\r\n";

$headers =  "Reply-To: " . $_POST["email"] . "\r\n";

$headers =  "Return-Path: ". $_POST["email"] ;

$headers =  $_POST["phone"];

$message = $_POST["message"];

$message = wordwrap($message, 70);

mail($sendTo, $subject, $headers, $message);

?><\Strong>


Please can you help to


A) get it to email my yahoo email

B) make an email that looks like


Name: Micky
Dept : Disney
Emial : Mick @mouse.com

etc not

<TEXTFORMAT LEADING="2"><P ALIGN="LEFT"><FONT FACE="Comic Sans MS" SIZE="16" COLOR="#000000" LETTERSPACING="0" KERNING="0">Micky</FONT></P></TEXTFORMAT><TEXTFORMAT LEADING="2"><P ALIGN="LEFT"><FONT FACE="Comic Sans MS" SIZE="16" COLOR="#000000" LETTERSPACING="0" KERNING="0">


please

[attachment deleted by admin]

ok..ive done this a million times..i have a working example here and i copied it and amended it for this new project but for some reason i cant get a form to post data to another page.

this is the error message i get

Notice: Undefined index: username in C:\wamp\www\uni\fyp\site\mobile\login.php on line 16

Notice: Undefined index: password in C:\wamp\www\uni\fyp\site\mobile\login.php on line 17

here is my form code:

                    <form method="post" action="login.php">
                        <table align="center" cellpadding="0" cellspacing="0">
                        <tr>
                            <td style="vertical-align:top;">Username:&nbsp; </td><td><input type="text" name="username" value="" /></td>
                        </tr>
                        <tr>
                            <td style="vertical-align:top;">Password:&nbsp; </td><td><input type="password" name="password" value="" /><br /><input type="submit" id="submit" value="Login" /></td>
                        </tr>
                        </table>
                    </form>


and here is the code within the login.php where the form should post to

$username = $_POST['username'];

$password = $_POST['password'];

// Help protect against MySQL injection

$username = stripslashes($username);

$password = stripslashes($password);

$username = mysql_real_escape_string($username);

$password = mysql_real_escape_string($password);

// Selecting data from database where correct username and password are found

$sql="SELECT * FROM customer WHERE username='$username' and password='$password'";

$result=mysql_query($sql) or die(mysql_error());



i cant see anything wrong..been looking for hours...please please help me

In the past I always combine my PHP Form Processing Code with my HTML Forms in one script, and when the User clicks "Submit", the Form/Page reloads itself.

Is it possible to load an Upload Form to itself?

Something like this...
Code: [Select]
<form enctype="multipart/form-data" action="" method="post">


Debbie

I have 10 tables in my database.
I want load these 10 table names into a drop-down list.
If a user selects a value form the drop-down list, a form should be displayed with related to the table. (form fields should be appropriate table's column headers)
When user fill the form and submitted, data should be saved in the selected table.

Give me a code example to do this.

I understand that is might be something that is already answered and I apologize if it is, I could not find it. What I need to do is build a simple form that has two options, they will be dropdown options. Dropdown A and Dropdown B then a Submit button. This part I understand in HTML, although it may be easier in php or javascript. Then I need it to take the two options and create a "if/then" statement that loads a specific pdf that matches the two options selected. Example. If someone selects Option 1 from Dropdown A and Option 2 From Dropdown B then it loads 12.pdf
If someone selects Option 5 from Dropdown A and Option 3 From Dropdown B then it loads 53.pdf
If someone selects Option 2 from Dropdown A and Option 1 From Dropdown B then it loads 21.pdf and so on... It does not have to be the exact thing just some way to take both inputs and have it equal a specific pdf. Here is the form I built but I don't know what to put in the form_action.php file in order to make it work

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<center>
<h1> Get Directions</h1>
<form action="form_action.php" method="get" name="directions" target="_new">

<select name="startpoint" size="1">
<option value="north">North Tower Entrance</option>
  <option value="south">South Tower Entrance</option>
  <option value="moba">MOB A Entrance</option></select>
  
  ----->

<select name="endpoint" size="1">
<option value="onco">Oncology</option>
  <option value="radio">Radiology</option>
  <option value="pulm">Pulmanary</option></select>
<br /><br />
<input type="submit" value="Submit" />
</form>
</center>

</body>
</html>

Any help is appreciated, thanks.

How can i edit just one image at on time with a multiple image upload form? I have the images being stored in a folder and the path being stored in MySQL. I also have the files being uploaded with a unique id. My issue is that I want to be able to pass the values of what is already in $name2 $name3 $name4 if I only want to edit $name1. I don't want to have to manually update the 4 images. Here is the PHP:
Code: [Select]
<?php

require_once('storescripts/connect.php');
mysql_select_db($database_phpimage,$phpimage);
$uploadDir = 'upload/';
if(isset($_POST['upload']))
{


foreach ($_FILES as $file)
{

    $fileName = $file['name'];
    $tmpName = $file['tmp_name'];
    $fileSize = $file['size'];
    $fileType = $file['type'];

    if ($fileName != ""){
        $filePath = $uploadDir;
        $fileName = str_replace(" ", "_", $fileName);

        //Split the name into the base name and extension
        $pathInfo = pathinfo($fileName);
        $fileName_base = $pathInfo['fileName'];
        $fileName_ext = $pathInfo['extension'];

        //now we re-assemble the file name, sticking the output of uniqid into it
        //and keep doing this in a loop until we generate a name that 
        //does not already exist (most likely we will get that first try)
        do {
           $fileName = $fileName_base . uniqid() . '.' . $fileName_ext;
        } while (file_exists($filePath.$fileName));

$file_names [] = $fileName;

        $result = move_uploaded_file($tmpName, $filePath.$fileName);
    }


if(!get_magic_quotes_gpc())
{
$fileName = addslashes($fileName);
$filePath = addslashes($filePath);
}
$fileinsert[] = $filePath;
}
}

$mid   = mysql_real_escape_string(trim($_POST['mid']));
$cat   = mysql_real_escape_string(trim($_POST['cat']));
$item  = mysql_real_escape_string(trim($_POST['item']));
$price = mysql_real_escape_string(trim($_POST['price']));
$about = mysql_real_escape_string(trim($_POST['about']));

$fields = array();
$values = array();
$updateVals = array();
for($i = 0; $i < 4; $i++)
{
    $values[$i] = isset($file_names[$i]) ? mysql_real_escape_string($file_names[$i]) : '';

    if($values[$i] != '') {
        $updateVals[] = 'name' . ($i + 1) . " = '{$values[$i]}'"; 
    }
}
$updateNames = '';
if(count($updateVals))
{
    $updateNames = ", " . implode(', ', $updateVals);
}

$update = "INSERT INTO image
               (mid, cid, item, price, about, name1, name2, name3, name4)
           VALUES
               ('$mid', '$cat', '$item', '$price', '$about', '$values[0]', '$values[1]', '$values[2]', '$values[3]')
           ON DUPLICATE KEY UPDATE
                cid = '$cat', item = '$item', price = '$price', about = '$about' $updateNames";
$result = mysql_query($update) or die (mysql_error());

Hi,   In reference to my first attached image, I have a form which displays two SELECT/drop-down fields (labeled "Store Name" and "Item Description".....and both of which pull-in values from two separate lookup/master tables, in addition to providing an additional option each for "NEW STORE" and "NEW ITEM").   Now, when first-run, and/or if "NEW STORE" and "NEW ITEM" are not selected from the drop-down's then the two fields in green ("New Store Name" and "New Item Name" are hidden, by means of the following code:  

<div class="new-store-container" id="new-store-container" name="new-store-container" style="display:none;">
    <div class="control-group">
        <div class="other-store" id="new_store_name">
            <?php echo standardInputField('New Store Name', 'new_store_name', '', $errors); ?>
        </div>
    </div>
</div>

  Conversely, if "NEW STORE" and/or "NEW ITEM" are selected from the two drop-down's then one (or both) of the "New Name" fields are unhidden by means of the following two pieces of code, one PHP and the second JS:    

<select class="store-name" name="store_id" id="store_id" onclick="toggle_visibility('store_id','new-store-container')">
    <?php echo $store_options; ?>
                            
    <?php
        if($values['store_id'] == "OTH")
            {
                echo "<option value='OTH' selected> <<<--- NEW STORE --->>> </option>";
            }
        else
            {
                echo '<OPTION VALUE="OTH"> <<<--- NEW STORE --->>> </OPTION>';
            }
    ?>
</select>

 

function toggle_visibility(fieldName, containerName)
{
    var e = document.getElementById(fieldName);
    var g = document.getElementById(containerName);
            
    if (e.value == 'OTH')
        {
            if(g.style.display == 'none')
                g.style.display = 'block';
            else
                g.style.display = 'none';
        }
}

  All  of that is working just fine. The problem I'm having is that when I click the "Create" button, after having left any one of the form fields blank, the two "New Name" fields are hidden again, which I don't want to happen i.e. I want them to remain visible (since the values of "store_id" and/or "item_id" are "OTH"), so that the user can enter values into one or both of them, without havng to click on the drop-down a second time in order to execute the "on-click" code.   The second attached image shows how the fields are hidden, after clicking "Create".   How can I achieve that? It would be greate if someone could cobble-up the required code and provide it to me, since I'm relatively new to this.   Thanks much.    Snap1.png   26.14KB   0 downloads    Snap2.png   149.47KB   0 downloads