How to debug dynamic javascript functions in FireBug

Question

i have some thing javascript functions like this

function onSelectRow_${itemid}(){
   something;
}

this it is appearing like this in firebug script tab

function onSelectRow_87878(){
     something;
}

I put multiple break points(it has more than 20 lines , i put one ine for example.) in firebug-> script tab.

But the problem is, firebug not able to do debug these methods, ie. it is not stoping execution it executing as usual. I tried multiple times.

this is my actual code and use:

function onSelectRow_${escapedId }(rowId){
}


<jqgrid:grid onSelectRow="onSelectDeviceRow_${escapedId }"

DO you have any solution for it please

How are you generating these functions? And why? – elclanrs Jan 23 '14 at 5:29 — elclanrs, Jan 23 '14 at 5:29

Somnath Kharat · Accepted Answer · 2014-01-24 06:45:51Z

up vote 4 down vote accepted

You can use debugger

function onSelectRow_87878()
{
     debugger; //add here
     something;
}

When you open Firebug, enable Script,and it will automatically go to the debugger point

answered Jan 23 '14 at 5:31

Somnath Kharat
2,13811134

thanks my dear friend SBK, you solved... – Ravi Feb 26 '14 at 5:07

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