PHP - Adding Php Field Check With Output

I have a code which I made just up
  Code: [Select]
<?php if ($product->title == '01'); print "Earring"; ?>
<?php if ($product->title == '02'); print "Necklace"; ?>
<?php if ($product->title == '03'); print "Necklace"; ?>
<?php if ($product->title == '04'); print "Ring"; ?>I just want to check the product titles first 2 characters and print according to that.
Example: If $product->title is "01DSE32" print Earring. This checking must be strgn  or something like that which I don't remember. I would appreciate some feedback.

Ok so i have registration page on my webs but i didn't work on empty fields so how could i check all fields?
Code: [Select]
//if no name
if(!$name){
$error = "Username missing.";
}
//if no pass
if($pass == false){
$error = "$error,Password missing";
}
//if no pass conf
if($pass_conf == false) {
$error = "$error,Pass conf missing.";
}
//if no email
if($email == false){
$error = "$error,Email missing.";
}
//pass conf
if($pass != $pass_conf){
$error = "$error,Passwords do not match.";
    }
    echo "<script>alert('$error');</script>;";
//header("Location: register.php");
 My script gives same results every time (even if i fill any of fields)...also if i un-comment header command it wont show alert..it just reloads the page

So can anyone help me with this?
Thanks..

I have this script running:


<?php
$query = mysql_query("SELECT * FROM jobs WHERE event = 'Yes' ORDER BY title");
while ($row = mysql_fetch_array($query)) {
?>


How do I check if a field is empty and NOT display it...?

For instance it has a 'applied' field, if that is empty I dont want it to display, however I still  need the event = 'Yes' part.

Hey All, first off Merry X-Mas

ok so I'm trying to insert something into the batabse in one of three areas' let me explain

there are three fields in the users database bounceone, bouncetwo and bouncethree now what I want to do is when the user clicks add to bounce list insert a value into the databse in one of these three hole, if bounceone is empty put it there, if not check bouncetwo if it's empty put it there, if not check the final one and if it's empty put it there else echo sorry please either remove a bounce

any ideas on how this can be done.

Hello everybody, i try to check if the unique field inventar exists or not before i insert new record. if the inventar exits the user should get error message. but i dont know how to do this.   thank you very much for your help. Rafal   here is my code

<?php
include("123.php");
$inventar = $link1->real_escape_string($_POST["inp_inventar"]);
$product = $link1->real_escape_string($_POST["inp_product"]);
$price = $link1->real_escape_string($_POST["inp_price"]);

$link1 = new mysqli("$hoster", "$nameuser", "$password", "$basedata") or die ("Connection Error!"); error_reporting (0);
$check_inventar = mysqli_query($link1, "SELECT inventar FROM products WHERE inventar='$inventar'");
$check_inventar_num = mysqli_num_rows($check_inventar);
if (isset($_POST['inp_submit']) AND $check_inventar_num = 0)
{
$query1 = "INSERT INTO products
(inventar, product, price)
VALUES ('$inventar', '$product', '$price')";
$result1 = $link1->query($query1) or die("Database Error!");
if ($result1 == true)
{
header ( 'Location:insert.php' );
}
}
else { echo "there is already product with same inventar id!"; }
mysqli_close($link1);
?>

I am building a system where people are able to choose a food menu, they can modify each dish if needed and then say how many of each item they need for the order.

So when they pick the breakfast menu, it selects all starters, mains and desserts.  Should something need to be edited (sausages changed to vegitarian sausages) the user should be able to overtype what has been populated from the select statement)

When the form is submitted, the order needs to be linked to the current job which is in a field ssm_job.job_id but the changes to the menu items need to be stored somewhere for the current order only.  This obviously needs to be accessile later so further changes can be made if needed.  I am not sure of the best way to go about this as i am trying not to save every item in another database (just trying to be tidy)

Any advise on this would be greatly appreciated.

I have had help on this project in relation to an sql and the background can be found below, the help has been amazing but as this is a very different question i thought i would start a new thread.

 

Kind Regards

Hi there once again, sry for previous wrong section, i guess this time i'm on the right place ;p

Well i need following:

I have few forms (input fields) with requested informations, also i have other .php file which i'm calling to send a form. Now i need to know how can i do a check for those fields if they are empty or not. If they are, i need to STOP submiting the form (calling the other .php).

Codes:

Form:
Code: [Select]
<form name="f1" form method="post" action="posalji.php"/>Field:
Code: [Select]
<input type="text" name="hp1" size="20"/>Submit:
Code: [Select]
<input name="f1" type="image" src="registruj.png" onclick="if(f1.hp1.value.length==0) alert('Error: field can't be empty'); else f1.submit(); if(f1.hp2.value.length==0) alert('Error: field can't be empty'); else f1.submit();"/>There are more than one field, that's why you can see more errors in submit. Well all i need is "die" or "stop" calling the "posalji.php" from action above.

Sry 4 poor english.

so I have this code:
Code: [Select]
<input type='checkbox' name='title8' value='$title8' /></td><td><strong>$title8</strong> <br /> $date8 <br />$ $price8<input type='hidden' name='price8' value='$price8'> how do I make is so that the hidden field is only going to submit if the checkbox is checked........

Pehaps this is kidstuff but i'm a kid at heart.

I have only 1 form field on a form
User enters a special code in the form.
Once submitted the form is queried against a mysql database to see if the code exists.
If code does not exist it must be reentered.
If code exists then redirected.

basically its like "get to the next page if you have the code"

thanks

I have a form with a gender control made up of two radio buttons.

If the user doesn't fill out this form, nothing gets submitted (i.e name='gender') in the $_POST array.

So how ae you supposed to verify that the form field was left blank, and use that information to say display an error message on the form?

Edited February 16 by SaranacLake

Now this one is a bit of a shit head. I am using a really comprehensive class i found on on the web for upload / re-size / crop but i want to be able to override it's error handling and i am not good enough to start modifying the code in the class with out fucking it up.

Firstly i will show you how i have invoked the class.
Code: [Select]
if( isset($_FILES['image'] ) ) {

//Class includes
include("../scrips/php/cms/database.insert.class.php");
include ("../scrips/php/cms/img.upload.resize.crop.class.php.php");


//----------|Start: upload, resize & save
$your_image = new _image;

//----------| Upload orginal image
$your_image->uploadTo = 'uploads/';
$upload = $your_image->upload($_FILES['image']);


//----------| Resize and upload small thumbnail
$your_image->newPath = 'thums/';
$your_image->newWidth = 50;
$your_image->newHeight = 50;
$resized = $your_image->resize();



//----------| Resize and upload medium thumbnail
$your_image->newPath = 'thums2/';
$your_image->newWidth = 100;
$your_image->newHeight = 100;
$resized = $your_image->resize();
//----------| Getting the image name to insert into the database futher on in the code
$name = $resized;
$img_str = explode("/",$name);
$final_img_name = $img_str['1'];

echo "Article has been added!";

//----------------|end

//----------------| Start database insert (class manipulation)
$table = "blog_posts";
$title = "'".$_POST['ArticleTitle']."',";
$img = "'".$final_img_name."',";
$post = "'".$_POST['ArticleBody']."',";
$aurthor_id = "'1',";
$category_id = "'".$_POST['Category']."',";
$date_posted = "NOW()";


$values = array("$title","$img","$post","$aurthor_id","$category_id","$date_posted");
$fields = array('title,','img,','post,','aurthor_id,','category_id,','date_posted');
$obj= new DatabaseInsert;
$obj->DatabaseConnectionRequire();
$obj->ArticleInsert($values,$fields,$table);
//----------------|end
}


As you can see it's fairly basic. What i want to do is before it runs this script and starts including/invoking the class etc, i want to be able to check to see if there is a value been posted from a FILE FORM OBJECT and if there is to proceed with this script, alternatively i want it to execute another code which will handle it in regards to the concept of my page. A simple javascript alert with be ok providing it reloaded the page to it's default state. If any one can help me here i would appreciate it a lot. Thanks
 

Hello,

I'm currently using this code to check for blank fields:

if(isset($_POST['submit']))
{
$emailconfirm = $_POST['emailconfirm'];
$user_message = $_POST['message'];
$name = $_POST['name'];
$visitor_email = $_POST['email'];
if(empty($name)|| empty($visitor_email)|| empty($user_message))
{
$errors .= "<div id='contact_form_errorloc' class='err1'>Some of the above fields have not been filled in.</div><br>";
}

This however, does not check for valid emails, I've found this code:

function is_valid_email($email) {
  $result = TRUE;
  if(!eregi("^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,4})$", $email)) {
    $result = FALSE;
  }
  return $result;
}

My question was if the above is possible aswell as incorporating the blank field checks.

Many thanks!

I have a register script, and I am wanting to make it so that if the username field contains, lets say "mod", "ass", and more, then it'll return an error and wont let them register.

Following from my previous post, i have a questionnaire and within each question it has the options to pick from raddio button (strongly agree, agree, disagree, N/A). For each value of each question i want to pass it into the relevant Field in my database but rather than overwriting the current value i want to be added on to it. (For example if the current value is 2, once the new value (1) has added it will make the new value of 3.).

Hiya pplz.

I am trying to add: class="keyboardInput" to activate a javascript script to an input box.

This is the line im trying to add it to:

<td><?php echo form_input(array('name'=>'quantity','value'=>$item['quantity'],'size'=>'2'));?></td>

I have tried:

<td><?php echo form_input(array('name'=>'quantity','value'=>$item['quantity'],'size'=>'2'));? class="keyboardInput"></td>

didn't work.

Any help would be great.

Ian