PHP - Html Form To Phpmyadmin Error

I've decided to rewrite one a couple of my forms, to make them more user friendly.

But I am having problems returning the error. I've created an array ($e) which has the field name ($e['field_name']) for the error, but it isn't displaying any errors.

Here's the code:
case "contact":
$e = array();
if(isset($_POST['sb'])){
$fullname = $sql->real_escape_string($_POST['fullname']);
$email_a = $sql->real_escape_string($_POST['email']);
$email_r = $sql->real_escape_string($_POST['email_r']);
$orderid = $sql->real_escape_string($_POST['orderid']);
$subject = $sql->real_escape_string($_POST['subject']);
$problem = $sql->real_escape_string($_POST['problem']);
# Error Message
function show_error($error,$evar){
$err = ' <span style="color: #ff0000"><strong>'.$error.'</strong></span>';
return $err;
}
# Check for isset
if(!isset($_POST['fullname'])){
$e['fullname'] = show_error('Please enter your full name',$e['fullname']);
}
if(!isset($_POST['email_a'])){
$e['email_a'] = show_error('Please enter your email address.',$e['email_a']);
}
if(!isset($_POST['email_r'])){
$e['email_r'] = show_error('Please re-enter your email address.',$e['email_r']);
}
if(!isset($_POST['subject'])){
$e['subject'] = show_error('Select a subject.',$e['subject']);
}
if(!isset($_POST['problem'])){
$e['problem'] = show_error('Describe your problem.',$e['problem']);
}
if(isset($_POST['email']) AND isset($_POST['email_r']) AND $email_a != $email_r){
$e['email_r'] = show_error('Your email addresses do not match.',$e['email_r']);
}
}
$content = '<h2>
Customer Support - Contact</h2>
<p>
For technical support, sales and product related questions: <a href="'.$tech_support.'" target="_blank">Technical Support</a>.</p>
<p>
For downloading, ordering, refund or other store related questions, please fill out the form below:</p>
<form method="post" name="csupport">
<p>
Please enter your full name:<br />
<input name="fullname" type="text" value="'.$fullname.'" />'.$e['fullname'].'</p>
<p>
Please enter your email address:<br />
<input name="email_a" type="text" value="'.$email_a.'" />'.$e['email_a'].'<br />
Please re-enter your email address:<br />
<input name="email_r" type="text" value="'.$email_r.'" />'.$e['email_r'].'</p>
<p>
Please enter your order number:<br />
<input name="orderid" type="text" value="'.$orderid.'" /></p>
<p>
Please select a subject for your enquiry:<br />
<select name="subject"><option selected="selected" value="">--Select--</option><option value="Billing Question">Billing Question</option><option value="Cancel / Refund Order">Cancel / Refund Order</option><option value="Download Question">Download Question</option><option value="Password Issues">Password Issues</option><option value="Order Question">Order Question</option><option value="Shipping Question">Shipping Question</option><option value="Other">Other</option></select>'.$e['subject'].'</p>
<p>
Describe your problem as best you can:<br />
<textarea cols="45" name="problem" rows="4">'.$problem.'</textarea>'.$e['problem'].'</p>
<p>
<input name="sb" type="hidden" value="sb" /><input name="submit" type="submit" value="Submit" /></p>
</form>
';
$title = 'Contact Support';
break;

Can anyone see what the problem is? Thanks

Hello, I have coded a contact form in PHP and I want to know, if according to you, it is secure! I am new in PHP, so I want some feedback from you.

Moreover, I have also two problems based on the contact form. It is a bit complicated to explain, thus, I will break each of my problem one by one.

FIRST:The first thing I want to know, is if my contact form secure according to you:

The HTML with the PHP codes:

Code: [Select]
<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST') {

//Assigning variables to elements
$first = htmlentities($_POST['first']);
$last = htmlentities($_POST['last']);
$sub = htmlentities($_POST['subject']);
$email = htmlentities($_POST['email']);
$web = htmlentities($_POST['website']);
$heard = htmlentities($_POST['heard']);
$comment = htmlentities($_POST['message']);
$cap = htmlentities($_POST['captcha']);

//Declaring the email address with body content
$to = 'alithebestofall2010@gmail.com';
$body ="First name: '$first' \n\n Last name: '$last' \n\n Subject: '$sub' \n\n Email: '$email' \n\n Website: '$web' \n\n Heard from us: '$heard' \n\n Comments: '$comment'";

//Validate the forms
if (empty($first) || empty($last) || empty($sub) || empty($email) || empty($comment) || empty($cap)) {
echo '<p class="error">Required fields must be filled!</p>';
header ('refresh= 3; url= index.php');
return false;

} 
elseif (filter_var($first, FILTER_VALIDATE_INT) || filter_var($last, FILTER_VALIDATE_INT)) {
echo '<p class="error">You cannot enter a number as either the first or last name!</p>';
return false;
} 
elseif (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
echo '<p class="error">Incorrect email address!</p>';
return false;
}
elseif (!($cap === '12')){
echo '<p class="error">Invalid captcha, try again!</p>';
return false;
}
else {
mail ($to, $sub, $body);
echo '<p class="success">Thank you for contacting us!</p>';
}
}
?>
<form action="<?php echo $_SERVER['PHP_SELF'];?>" method="post">
<p>Your first name: <span class="required">*</span></p>
<p><input type="text" name="first" size="40" placeholder="Ex: Paul"/></p>

<p>Your last name: <span class="required">*</span></p>
<p><input type="text" name="last" size="40" placeholder="Ex: Smith"/></p>

<p>Subject: <span class="required">*</span></p>
<p><input type="text" name="subject" size="40" placeholder="Ex: Contact"/></p>

<p>Your email address: <span class="required">*</span></p>
<p><input type="text" name="email" size="40" placeholder="Ex: example@xxx.com"/></p>

<p>Website:</p>
<p><input type="text" name="website" size="40" placeholder="Ex: http//:google.com"/></p>

<p>Where you have heard us?: <span class="required">*</span></p>
<p><select name="heard">
<option>Internet</option>
<option>Newspapers</option>
<option>Friends or relatives</option>
<option>Others</option>
</select></p>

<p>Your message: <span class="required">*</span></p>
<p><textarea cols="75" rows="20" name="message"></textarea></p>

<p>Are you human? Sum this please: 5 + 7 = ?: <span class="required">*</span></p></p>
<p><input type="text" name="captcha" size="10"/></p>

<p><input type="submit" name="submit" value="Send" class="button"/>
<input type="reset" value="Reset" class="button"/></p>
</form>

SECOND PROBLEM:If a user has made a mistake, he gets the error message so that he can correct! However, when a mistake in the form occurs, all the data the user has entered are disappeared! I want the data to keep appearing so that the user does not start over again to fill the form.

THIRD: When the erro message is displayed to notify the user that he made a mistake when submitting the form, the message is displaying on the top of the page. I want it to appear below each respective field. How to do that? In JQuery it is simple, but in PHP, I am confusing!

i have been working with a website but only locally. i am now trying to put it online which is ok. when i was working with it locally i have been able to create multiple users which have different privileges to each page. for example i said one of my users can only read and the other can read write.
the problem is when i have went to create these users online i can't seem to find the create user option in phpmyadmin.

i am missing the obvious or will i have to code them in and if so can you direct me to a tutorial which shows me how.

thanks in advance     

Hello, first time poster..  I've looked the web over for a long time and can't figure this one out.

- Below is basic code that successfully checks MySQL for a match and displays result.  I was debugging and forced the "height" and "width" to be 24 and 36 to make sure that wasn't the problem.  That's good.. 
- I'd like to give the user ability to select width and height from a form.. and have it do an onchange this.form.submit so the form can be changing as fields are altered (thus the onchange interaction)
- In a normal coding environment I've done this numerous times with no "Page cannot be displayed" problems.  It would simply change one select-option value at a time til they get down the form and click submit...  but in WordPress I'm having trouble making even ONE single onchange work!
- I've implemented the plugins they offer which allows you to "copy+paste" your php code directly into their wysiwyg editor.  That works with basic tests like my first bullet point above.
- I've copied and pasted the wordpress url (including the little ?page_id=123) into the form "action" url... that didn't work...  tried forcing it into an <option value=""> tag.. didn't work.  I'm just not sure. 

I've obviously put xx's in place of private info..

Why does this form give me Page Cannot Be Displayed in WordPress every time?  It won't do anything no matter how simple.. using onchange..

Code..

$con = mysql_connect("xxxx.xxxxxxx.com","xxxxxx","xxxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("xxxxxx", $con);
$myprodwidth=24;
$myprodheight=36;
$result = mysql_query("SELECT * FROM product_sizes WHERE prodwidth='$myprodwidth' and prodheight='$myprodheight'");
while($row = mysql_fetch_array($result))
  {
  echo $row['prodprice'];
  }
mysql_close($con);


<form method="post" action="">
<select name="myheight" onchange="this.form.submit();">
<option selected="selected" value="">select height</option>
<option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">36</option>
<option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">48</option>
</select>

Hi all,

I have an html form that sends formdata to a php script called example.php with the POST method.

example.php then has the following structu

Code: [Select]
<?php
if(isset($_POST['var1'])) {$var1 = $_POST['var1'];} 
else {$var1 = 'English';}
if(isset($_POST['var2'])) {$var2 = $_POST['var2'];} 
else {$var2 = 'French';}

print"
<html>
<body>
<form>
<select id='select1'>
<option>var1</option>
<option>var2</option>
<option>etc...</option>
</select>
<input type='text' id='input1'>
ETC. ETC.
</form>
</body>
</html>
";
?>
So example.php has an embedded html form. How do I specify that some elements of the form should take their values from the variables retrieved at the start of the script via the POST command?

For example, I want select1 to have 'var1' pre-selected and I want input1 to display var2. var1 & var2 get sent through to example.php fine, I just don't know how to get the embedded html form to use them.

Please help! Thanks a lot.

Hi guys

I have two questions:

Question 1: phpmyadmin
If my host does not provide phpmyadmin and I want to install it myself, can I just download phpmyadmin and copy it to my public_html folder or is there another way to install it on my site??

Question 2: Email Script
I have a script that sends an email to a system that sends out multiple sms's. The system works like this, I send an email to an address with an attachment called cellnumbers.txt with a list of all the cellnumbers. I must add a specific subject. When I send the email to my own email address, it comes correctly through. But when it is sent to the address of the server, no sms are received. I attached my script. Please take a look. Your help will be greatly appretiated.

Thank you

I have a form where it ask the user to select the student ID, course ID, enter the grades and comments.
The Student ID and CourseID is selected from a drop down menu, but when the data is sent to PHPMYADMIN it enters a 0 into the SID and CID

How to I get it save the numbers which has been selected

Hi,

I am trying to create a drop down list in php and I want the data to come from a table that I have created in phpmyadmin. The code that I have created allows me to select values from the drop down list and insert the rest of the data. However when I check the the table the SID and Cid are set to 0 and the grade field is empty and the comments field contains the grade. The SID and Cid are both composite keys.


<?php
        $sql = "SELECT Cid FROM course";
        $db1 = new DBStudent_Course();
        $db1->openDB();
        $result = $db1->getResult($sql);

        echo"<select name = Cid>";
        while ($row = mysql_fetch_object($result)) {
            echo "<option value = '" . $row->Cid . "'>$row->Cid</option>";
        }

        echo"</select>";
        echo "</p>";
        ?>

        <?php
        $sql = "SELECT SID FROM student";
        $db1 = new DBStudent_Course();
        $db1->openDB();
        $result = $db1->getResult($sql);

        echo"<select name = SID>";
        while ($row = mysql_fetch_object($result)) {
            echo "<option value = '" . $row->SID . "'>$row->SID</option>";
        }

        echo"</select>";
        echo "</p>";
        ?>
        <?php
        if (!$_POST) { //page loads for the first time
        ?>
            <form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post">
                grade:<input type="text" name="grade"/><br/>
                comments:<input type="text" name="comments" /><br />
                <input type="submit" value="Save" />
            </form>
<?php
        } else {
            $Cid = $_POST["Cid"];
            $SID = $_POST["SID"];
            $grade = $_POST["grade"];
            $comments = $_POST["comments"];
            $db1 = new DBStudent_Course();
            $db1->openDB();
            $numofrows = $db1->insert_student_course("", $SID, $Cid, $grade, $comments);
            echo "Success. Number of rows affected:
<strong>{$numofrows}<strong>";
            $db1->closeDB();
        }
?>


 

Hi, I cannot find where to add foreign key constraints in phpmyadmin!

Any help is appreciated!

Hi

How can I make the data in phpmyadmin accept Arabic Language. Now Arabic language shows as question markss ?????????????.

Is there anyway to do this via php?

I have a db.php and inside it I filled out all the info needed:

<?php
define('DB_HOST', 'example.com');
define('DB_NAME', 'database_name');
define('DB_USERNAME', 'user_name');
define('DB_PASSWORD', '*******');

$odb = new PDO('mysql:host=' . DB_HOST . ';dbname=' . DB_NAME, DB_USERNAME, DB_PASSWORD);
?>

My problem is with the DB_HOST, is that my direct URL to the site? for example, google.com or is it like an IP?  

Hi. I having trouble with counting rows in phpmyadmin.

It works fine this way:
Code: [Select]
$result_rows = mysql_query("SELECT * FROM events");

But when i modify the code to this it doesnt work at all.
Code: [Select]
$result_rows = mysql_query("SELECT * FROM events WHERE category = 'adults' ORDER BY 'date'");

Any idea what is wrong?

Using phpMyAdmin I loaded 6 test records with the id set to auto_increment and it loaded all the data correctly with id # 1-6.

Then from somewhere it got the number 333353  and auto_increments it as the value for the id. So now I have id's 1-6 and 333353, 333354, ect. For every record I add it increments it.

I deleted all but records 1-6 and tried again but it has the last value of 3333xx stored somewhere and increments it. Deleted them again, closed the program, came back and it still does it.

Hi guys, I would like to seek help on inserting data whenever the switch is on or off to my sensor mySQL database in phpMyAdmin from my control.php. I'm using Raspberry PI as my hardware and follow a few tutorials to create my own Web Control Interface, it works perfectly without insert method. After I implemented insert method to my control.php and execute it, it cannot works and cannot store.

Sorry for the caps, but this is relatively time sensitive.  I am trying to make a register form, but when I click the submit button, nothing happens.  It doesn't add to the table, it doesn't bring me home, doesn't even display the errors if the PWD's don't match or the fields are blank.  Here's my code, thanks guys !  PS:  The DB name is phptest, and the table is called users. 
Code: [Select]
<?php
error_reporting(0);

require_once('connector.php');

$errors = array();

if ($_POST["submit"]) {

if (empty($_POST['username'])) { array_push($errors, 'You did not submit a username.');}
if (empty($_POST['email'])) { array_push($errors, 'You did not submit a email.');}
if (empty($_POST['password1'])) { array_push($errors, 'You did not submit a password.');}

$old_usn = mysql_query("SELECT id FROM users WHERE name = '".$_POST['username']."' LIMIT 1") or die (mysql_error());
if (mysql_num_rows($old_usn) > 0) { array_push($errors, 'This username is already registered.');}

$old_email = mysql_query("SELECT id FROM users WHERE email = '".$_POST['email']."' LIMIT 1") or die (mysql_error());
if (mysql_num_rows($old_email) > 0) { array_push($errors, 'This email is already registered.');}

if ($_POST['password1'] != $_POST['password2']) { array_push($errors,'You entered two different passwords');}

if(sizeof($errors) == 0) { 


$username = $_POST['username'];
$email = $_POST['email'];
$password = sha1 ($_POST['password1']);


mysql_query("INSERT INTO users (name, hashed_psw, email, joined)
VALUES ('{$username}', '{$password1}', '{$email}', NOW());") or die (mysql_error());

 

header ('Location: index.php?msg=1');

}
}
?>
<html>
<head>
<title>register</title>
</head>

<body>
<?php
foreach($errors as $e) {
echo $e;
echo "<br/>\n";
}


?>

<form action="register.php" method="post">
<h4>
Username:
<br />
<input name="username" type="text" value="" size="10" maxlength="16" />
<br />
<br />
Email:
<br />
<input name="email" type="text" value="" size="10" maxlength="100" />
<br />
<br />
Password:
<br />
<input name="password1" type="password" value="" size="10" maxlength="16" />
<br />
<br />
Confirm Password:
<br />
<input name="password2" type="password" value="" size="10" maxlength="16" />
<br />
<br />
<input name="submit" type="button" value="Register" />
</h4>
</form>

</body>
</html>

And heres the connector.php script:

Code: [Select]
<?php

mysql_connect("localhost", "***", "***") or die (mysql_error());
mysql_select_db("phptest") or die (mysql_error());

?>(yes, the asterisks have the name and pw, just put them just in caseys!