Regex test to see the content of a array element

1 
  var string_array=[];
 var con =new RegExp(/[bcdfghjklmnpqrstvwxyz.]/);
 string_array = str.split('');

   for(var i=0;i<string_array;i++){
      if(con.test(string_array[i])){
      string_array.shift(string_array[i]);
      console.log(shifted_array);
     }

what would be wrong would this somehow the test method on the array element doesn’t seem to work,even though the array element is a string.
Could someone please help?

2

You’re using console.log(shifted_array) but shifted_array is never declared or defined before then. That’s at least one part of the problem.

4

Sorry this is the whole code.I was just trying to concentrate on that small part of it as I don’t think there is anything wrong
with the rest.


function translatePigLatin(str) {
  var bunch_of_constanents =[];
  var string_array=[];
  var vowel = new RegExp(/[aioue.]/);
  var con =new RegExp(/[bcdfghjklmnpqrstvwxyz.]/);
  var suffix = "way";
  var shifted_array =[];
  var how_Long = str.length;
  
  if(vowel.test(str.charAt(0))){
    var str2 = str+suffix;
    return str2;
    
  }
 
  string_array = str.split('');
  
   for(var i=0;i<string_array;i++){
      if(con.test(string_array[i])){
      string_array.shift(string_array[i]);
      console.log(shifted_array);
     }
     
      else{
           break;
     }
      
   } 
 

}
 
translatePigLatin("california");
5

This is the whole code.I was just trying to concentrate on a small area of it in the previous post.


function translatePigLatin(str) {
  var bunch_of_constanents =[];
  var string_array=[];
  var vowel = new RegExp(/[aioue.]/);
  var con =new RegExp(/[bcdfghjklmnpqrstvwxyz.]/);
  var suffix = "way";
  var shifted_array =[];
  var how_Long = str.length;
  
  if(vowel.test(str.charAt(0))){
    var str2 = str+suffix;
    return str2;
    
  }
 
  string_array = str.split('');
  
   for(var i=0;i<string_array;i++){
      if(con.test(string_array[i])){
      string_array.shift(string_array[i]);
      console.log(shifted_array);
     }
     
      else{
           break;
     }
      
   } 
 

}
 
translatePigLatin("california");
7

I see what you mean.

for(var i=0;i<string_array.length;i++)

I missed out adding the length method.
And then

I was gonna do this with it.

for(var i=0;i<string_array;i++){
      if(con.test(string_array[i])){
      shifted_array  =string_array.shift(string_array[i]);
     string_array.push(shifted_array);
      console.log(shifted_array);
     }

basically using shift and push method to take it from the front of the array and put it on the back.I will try it out later thanks for your help.

8

Any regex with /[{a load of other characters}.]/ will match any character other than line breaks, because that’s what the dot . signifies. So vowel and con are functionally the same as each other.

9

Right managed to do it after a good few hours but its a bit long what do you think?
What gave me a problem was that the shift method was changing the order of the array.So I simply changed this because the shift function kinda iterates the array for you.

for(var i=0;i<string_array.length;i++)

to this

for(var i=0;i<string_array.length;)


function translatePigLatin(str) {
  
  var string_array=[];
  var vowel = /[aioue]/;
  var con = /[bcdfghjklmnpqrstvwxyz]/;
  var suffix = "way";
  var consonant_suffix ="ay";
  var shifted_array =[];
  var consonant=" ";
  
  
  if(vowel.test(str.charAt(0))){
    var str2 = str+suffix;
    return str2;
    
   
  }
  
   string_array = str.split('');
  
  
   for(var i=0;i<string_array.length;){
     if(con.test(string_array[i])){
      console.log(consonant);
      consonant = string_array.shift();
      shifted_array.push(consonant);
       
      }
     
     else{
      var new_string_array = string_array.join('');
      var new_shifted_array = shifted_array.join('');
      var str3 = new_string_array.concat(new_shifted_array+consonant_suffix);
      return str3;
     }
   }
  
     
      
}
 
translatePigLatin("glove");
10

Thanks changed it now been a long time since I did a regex pattern.