my Javascript if statement not functioning

[blumonde]

Hello,

 

The if statement didn't run and the confirm box didn't popup. and I don't know how to fix this after hours of debugging.

 

Below is my code and it is simple: (when I debugging it, the variable $check does have a value of 1 in it)

<?php
$check = 1;
?>

echo "function checkB4SaveForm()";
echo "{";
echo "  var checkacct=".$check.";";
echo "  if (checkacct==1)";
echo "  {";   
echo "    var show = confirm(\"This account exists! Would you like to view it?\");";
echo "    if (show==true)";
echo "    {";
echo "      searchAndDeleteForm();";
echo "      return false;"; // show the record and cancel save transaction
echo "    }";
echo "    else";
echo "    {";
echo "      return false;"; // cancel save transaction
echo "    }";
echo " }";
echo "}"; // end function

Thanks in advance.

 

blumonde

Edited February 24, 2014 by Zane

[Ch0cu3r]

How and when is the checkB4SaveForm() JavaScript function called?

 

If you are trying to call that function from PHP then you cant. PHP and Javascript are two complete different languages, you cannot read each others variables/functions.

[blumonde]

Hello Ch0cu3r,

 

Thank you for your response. I call it via the submit button. The function did get called but the if statement was ignored. The reason I know this is that because I added an alert message to the function for testing and it did pop up.

 

Regards,

 

blumonde

[Zane]

<?php
$check = 1;
echo <<<JSCRIIPT
function checkB4SaveForm() {
     var checkacct={$check};
     if (checkacct==1) { 
          var show = confirm("This account exists! Would you like to view it?");
          if (show) {
              searchAndDeleteForm();
              return false; // show the record and cancel save transaction
          } else { 
              return false;"; // cancel save transaction
          }
     }
} // end function
JSCRIPT;
?>

Try this out, for the sake of syntax and good reabability

[blumonde]

Hello Zane,

 

Thank you for your response. It's still no effect on the conditional statement. Maybe "var checkacct={$check};" didn't transfer the value?

 

 

blumonde

[Zane]

While that could be possible, bluemonde. You would be the only one that can know.  I didn't rewrite you entire script or anything.  I simply made it more human readable.  So any and all syntax errors, variable declarations, server settings, php version, apache version, operating system, etcetera... are on you.

 

 

If you open the source, using view source (or Chrome's development pane), you should be able to see for sure whether or not $check went through.

Edited February 24, 2014 by Zane

[blumonde]

Yes, $check php variable does have a value of 1 in it. I saw it while using Codelobster debugger. It is "checkacct" js variable that I don't know if it gets that same value or not. Too bad there is no debugger that can debug javascript at run time.

[Zane]

Maybe I'm confused, but..

 

Chrome comes pre-equipped with a debugger.....

https://developers.google.com/chrome-developer-tools/

 

Using the console tab, you can easily type alert(checkacct); and press enter, and it will work.

You can even call your function.

Edited February 25, 2014 by Zane

[blumonde]

Zane,

 

My session variable is not working correctly in Drupal. I need to upgrade Drupal to the latest version.

 

Thank you and sorry for my late response.

 

Regards,

 

Dominic