为什么我的简单函数没有返回 false 工作？

Question

<form id="form1" onsubmit="return hi(this);">

#1

function hi(bla){
    console.log("abc");
return false
}

#2

function hi(bla){
    console.log(abc);
return false
}

issue is, in #1 the function works perfectly while in #2 the page gets refreshed and i can't see where the issue is. i'd like to log a variable in the console in that function.

thank you for your time.

Answer 1

You need to set the variable abc first. Put this before your existing code:

 var abc;

Answer 2

As @mediaguru said the current code is missing.

var abc;

in #1

function hi(bla){
    console.log("abc");
return false
}

calling

hi();

will console log "abc"

I believe you want to change #2 to

function hi(bla){
console.log(bla);
return false
}

and then you want to call

hi("abc") 

to display "abc" in your console.

or more full example

function hi(bla){
   console.log("abc");
}

var abc = "hi"
hi(abc)
// "hi"

You can go on to debug the value of this from there.

<form id="form1" onsubmit="return hi(this);">