CODEWITHSUNDEEP
javascriptjquery
Jonathan
Jonathan

Reputation: 2880

Find string without comma

I have this code where I find keys where the string values are separated by comma 

var db = {
    	"name": "Nkosana",
    	"middle": "Baryy",
    	"surname": "walked",
    	"batch_number": "test,b",
    	"temp": ",,67,6,87,86,5,67865,876,67",
    	"integrity": ",,,,,,,,,,,,,,,,,,,,,,,",
    	"weight": "760,765,755,758,759,758,758,769,758,762,759,7",
    	"comment": "oh la la",
    };

for (var key in db) {
  if (db.hasOwnProperty(key)) {
  	if(db[key].indexOf(',')>-1){
       console.log(key+' | ' + db[key]);
    }
  }
}

As you can see I can retrieve these values

batch_number | test,b
temp | ,,67,6,87,86,5,67865,876,67
integrity | ,,,,,,,,,,,,,,,,,,,,,,,
weight | 760,765,755,758,759,758,758,769,758,762,759,7

Now I would like to know how if there is a way I can retrieve only the keys of values that do not have commas

Upvotes: 1

Views: 107

Answers (4)

Sagar's pet
Sagar's pet

Reputation: 41

Change indexOf(',') > -1 to indexOf(',') == -1:

var db = {
    	"name": "Nkosana",
    	"middle": "Baryy",
    	"surname": "walked",
    	"batch_number": "test,b",
    	"temp": ",,67,6,87,86,5,67865,876,67",
    	"integrity": ",,,,,,,,,,,,,,,,,,,,,,,",
    	"weight": "760,765,755,758,759,758,758,769,758,762,759,7",
    	"comment": "oh la la",
    };

for (var key in db) {
  if (db.hasOwnProperty(key)) {
  	if(db[key].indexOf(',')==-1){
       console.log(key+' | ' + db[key]);
    }
  }
}

Upvotes: 3

CumminUp07
CumminUp07

Reputation: 1978

Simply change if(db[key].indexOf(',')>-1) to if(db[key].indexOf(',') == -1).

Here's your updated code:

var db = {
        "name": "Nkosana",
        "middle": "Baryy",
        "surname": "walked",
        "batch_number": "test,b",
        "temp": ",,67,6,87,86,5,67865,876,67",
        "integrity": ",,,,,,,,,,,,,,,,,,,,,,,",
        "weight": "760,765,755,758,759,758,758,769,758,762,759,7",
        "comment": "oh la la",
    };

for (var key in db) {
  if (db.hasOwnProperty(key)) {
    if(db[key].indexOf(',') == -1){
       console.log(key+' | ' + db[key]);
    }
  }
}

Upvotes: 2

Bhargav Chudasama
Bhargav Chudasama

Reputation: 7161

Try this:

var db = {
    	"name": "Nkosana",
    	"middle": "Baryy",
    	"surname": "walked",
    	"batch_number": "test,b",
    	"temp": ",,67,6,87,86,5,67865,876,67",
    	"integrity": ",,,,,,,,,,,,,,,,,,,,,,,",
    	"weight": "760,765,755,758,759,758,758,769,758,762,759,7",
    	"comment": "oh la la",
    };

for (var key in db) {
  if (db.hasOwnProperty(key)) {
  	if(db[key].indexOf(',')==-1){
       console.log(key+' | ' + db[key]);
    }
  }
}

Upvotes: 2

cs95
cs95

Reputation: 402553

As I mentioned in my comment, the fix is simple. You just need to change the if condition a bit:

if(db[key].indexOf(',')==-1){ ... }

This is because indexOf returns -1 when the search string is not present in the string to be searched.

The complete code listing is:

var db = {
    	"name": "Nkosana",
    	"middle": "Baryy",
    	"surname": "walked",
    	"batch_number": "test,b",
    	"temp": ",,67,6,87,86,5,67865,876,67",
    	"integrity": ",,,,,,,,,,,,,,,,,,,,,,,",
    	"weight": "760,765,755,758,759,758,758,769,758,762,759,7",
    	"comment": "oh la la",
    };

for (var key in db) {
  if (db.hasOwnProperty(key)) {
  	if(db[key].indexOf(',') == -1){
       console.log(key+' | ' + db[key]);
    }
  }
}

Upvotes: 1

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