How to create a function Javascript pop up box with php code?

Question

In a php page, I want to make a remove button to delete data in my database. This button would be able to show a pop up box to ask 'yes' or 'no' before you confirm to delete.

"<input type=submit value=Remove onclick='confirm(\"Are you sure to remove Title " . $row['AlbumName'] . " ?\");'>"

This code just can call the pop up box but it can't check the user either click 'yes' or 'no'.

I have tried to echo script with confirm function remove() in , but the function is not work when I set onclick=remove() .

Answer 1

Try this

<input type=submit value=Remove onclick='confirmDelete(<?=$row['AlbumName']?>);'>
function confirmDelete(AlbumName)
{
var agree=confirm("Are you sure to remove Title "+AlbumName+"?");
if (agree)
 return true ;
else
 return false ;
}

Answer 2

function doublecheckDelete() {
 if (confirm("really delete") == true) {
 document.deleteform.submit();
 } else {
 alert ("ok, not deleting.");
 }
}
<form name="deleteform" action="delete.php">
... whatever else you need ...
<input type=button onClick="doublecheckDelete()">
</form>