Javascript Confirm + MySQL query, help!

Trent_Hatred

I have a text field which users can enter in order numbers. If they enter an order number which already exists in the DB, the information is displayed and all is well.

If, however, they enter an order number that is not in the DB, I would like to have a Javascript popup appear asking if they wish to add this new # or not.

Here a bit of the code so you can understand what I mean:

$checknum = "SELECT number FROM table WHERE number = '$number'"; 
$query = mysql_query($checknum); 

if ( mysql_num_rows($query) >= 1 ) {
#Displays information - Working fine
} 
else { 
#Need a popup box here, that allows OK/Cancel.  If OK, continue on somehow to inserting the new information.
}

From what I understand, javascript within PHP does not stop the PHP from completing. So, would I have to stop the code there, and then setup the javascript confirm box to pass a variable back to PHP somehow eg.

if ($yes){
$query = "INSERT INTO ...."

If that's the case, does/can anyone supply me with a small example? I've found a ton of javascript confirm boxes, but am struggling with getting it to work with PHP.

Thanks much,
B

deep

try this:

$checknum = "SELECT number FROM table WHERE number = '$number'"; 
$query = mysql_query($checknum); 
$row = mysql_fetch_array($query);

if(!mysql_num_row($query)){
   echo '<script type="text/javascript">
   if(confirm("add new info")){
      document.location="your_site/index.php&add='.$row['number'].'";
   }
   </script>';
}else{
   //Display info
}

if(isset($_GET['add'])){
   //add the info to the DB.
}

I havnt tested this code but it should work. There are probably better wayt to write it but this is what i could come up whit at the moment. 😉

sneakyimp

you have two options

1) print your prompt in PHP...maybe use red letters or something

2) just write a javascript function in your PHP code that executes when the page loads. this should prompt the user and redirect the browser to the appropriate page.

either way, your page that prompts the user is going to have to load another page and pass the NUMBER to a page which will be responsible for creating the new item.

for option 1, your code might look like this:

$checknum = "SELECT number FROM table WHERE number = '$number'"; 
$query = mysql_query($checknum); 

if ( mysql_num_rows($query) >= 1 ) {
#Displays information - Working fine
} 
else { 
   echo "<FONT color=#ff0000>that record number doesn't exist, would you like to create a new one?</font>";
   echo "<A HREF=\"create.php?number=$number\">YES</a>";
   echo " | ";
   echo "<A HREF=\"do_not_create.php\">NO</a>";
}