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() .

Zahidul Hossein Ripon · Accepted Answer · 2013-12-08 07:55:25Z

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 ;
}

answered Dec 8, 2013 at 7:55

Zahidul Hossein Ripon

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Guntram Blohm · Accepted Answer · 2013-12-08 07:56:20Z

0

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>

answered Dec 8, 2013 at 7:56

Guntram Blohm

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