How to Exit If Statement and Run Another IF Statement? - php

Question

$var1="exit";
$var2="run";
$var3="go";
if($var1 != '' && $var2 != '' && $var3 != '' ){
//first condtion
if($var1 == 'clear'){
echo 'clear';
}
else {
exit();
}
//2nd condtion
if($var2 == 'run'){
echo 'run';
}
//3rd condtion
if($var3 == 'go'){
echo 'go';
}
}
**I try to Exit() First If Condition, continue to next IF Condition, But Cannot Continue next IF condition totally Exit Overall Please Fix My Problem Thankyou **

Answer 1

No need to use else condition if you want to run second if statement
if($var1 != '' && $var2 != '' && $var3 != '' ){
//first condtion
if($var1 == 'clear'){
echo 'clear';
}
//2nd condtion
if($var2 == 'run'){
echo 'run';
}
//3rd condtion
if($var3 == 'go'){
echo 'go';
}
}

Answer 2

What About the idea of this one.
<?php
$var1="exit";
$var2="run";
$var3="go";
if($var1 != '' && $var2 != '' && $var3 != '' )
{
if($var2 == 'run'){
echo 'run';
}
elseif($var3 == 'go'){
echo 'go';
}
elseif($var1 == 'clear'){
echo 'clear';
}
else {
exit();
}
}

Answer 3

just use only if.. use return
$var1="exit";
$var2="run";
$var3="go";
if($var1 != '' && $var2 != '' && $var3 != '' ){
//first condtion
if($var1 == 'clear'){
echo 'clear';
}else {
return;
}
//2nd condtion
if($var2 == 'run'){
echo 'run';
}else {
return;
}
//3rd condtion
if($var3 == 'go'){
echo 'go';
}else {
return;
}
}