I have set of documents that contain an array of search terms, e.g.
[ "apples", "oranges", "bananas" ]
The user will enter a search string of keyword prefixes, and I'd like to match all the documents that contain each term in the array. So, for example, "app oranges" will match the list above, but "applet oranges" wouldn't.
It would be fairly trivial to construct a $and query that checked that each term matched one of the items in the array as a prefix using $regex, however that doesn't go far enough...
Each keyword should have a unique match within the set, such that searching "apples app" will not match the list above because the "app" term can't match against "apple" since "apple" has already been matched. This constraint leads to a more subtle problem. Take this set as an example:
[ "france", "fred", "freddy" ]
If the user taps "fr france" then this should match. It's important that the match for "fr" doesn't remove "france" from the possible list of terms for the remaining keywords, otherwise the test for the term "france" that follows would fail.
I need to implement this as a Mongo query. I'm quite new to Mongo and I have't a clue where to start, or even of this is possible. Can it be done? If so, how?
To start with, you can use the $regex operator to match text patterns:
var searchTerms = "app oranges".split(" ");
var arr = [];
searchTerms.forEach(function(i){
var reg = new RegExp("^"+i);
arr.push({"names":{$regex:reg}});
})
db.collection.find({$and:arr});
Would give you the documents with array names containing values starting with app and containing oranges.
Each keyword should have a unique match within the set, such that searching "apples app" will not match the list above because the "app"
term can't match against "apple" since "apple" has already been
matched. This constraint leads to a more subtle problem. Take this set
as an example:
This logic should be carried out in the application server before/after firing the query. If the user enters a string that is a substring of another former input, then the query is bound to fail since it would have already matched the fromer.
Related
I have a string, say "string", and a field in a collection, say "name".
I would like to find all the documents whose name is a substring of "string" (it would return for example documents whose name is "str").
I have looked on internet, and I only find the reciprocal, that is for example documents whose name is "string2".
Would someone know please ?
KR
Zlotz
There is no arbitrary substring match operator in MongDB that I know of, but you could try expressing your condition with $where or $regex.
Recently I wanted to achieve the same in my project query below worked for me
db.find({name:{ $regex: new RegExp("^" + "your string pattern".toLowerCase(), "i") } })
Let me know if you find it useful.
I am learning MongoDB. Getting confused on usage of "$"
I have collection as below schema:
{
_id: 1,
"name": "test",
"city": "gr",
"sector": "IT",
"salary":1000
}
I find below output on executing below query:
Query Result
db.user.find({salary:2000}); Works
db.user.find({$salary:2000}); does not work(unknown top level operator: $salary)
db.user.aggregate({$group:{_id:null,avg:{$avg:"$salary"}}}); Works
db.user.aggregate({$group:{_id:null,avg:{$avg:$salary}}}); does not work($salary is not defined)
db.user.aggregate({$group:{_id:null,avg:{$avg:"salary"}}}); gives wrong output.
Can anyone please explain,what is the syntactical significance of "" and $ in mongoDB.
Hi lets look at these queries
1- db.user.find({salary:2000});
2- db.user.find({$salary:2000});
Take a look at this for find.
According to this find takes {field: value}, your first query works because salary is valid field.
Your second query doesn't work becuase there is no field $salary
3- db.user.aggregate({$group:{_id:null,avg:{$avg:"$salary"}}});
4- db.user.aggregate({$group:{_id:null,avg:{$avg:$salary}}});
5- db.user.aggregate({$group:{_id:null,avg:{$avg:"salary"}}});
For aggregation, lets take a look at this $avg.
Here it says that $avg takes {$avg: expression}. So you are actually keeping expression over there not a field.
Now take a look at this for expression.
Expression can be field paths and system variables, literals, expression objects, and expression operators.
Query numbers 3,4,5 aren't expression objects or expression operators. So lets eliminate these options.
Now lets take a look at $literal.
It states that literals can be of any type, however MongoDB parses literals that start with a dollar sign as a path to a field.
Finally take a look at Field Path and System variables.
It states "To specify a field path, use a string that prefixes with a dollar sign $ ... For example, "$user" to specify the field path for the user field or "$user.name" to specify the field path to "user.name" field."
That means you are specifying $salary as path to the field in $avg:"$salary" and query number 3 works.
Query number 4 doesn't work because $salary is an invalid expression.
This should explain the significance of ""
Query number 5 is not working because again it doesn't find any field to average on. Though it works because its a valid query it simply returns null.
You could have had
db.user.aggregate({$group:{_id:null,avg:{$avg:"some_non_existent_field"}}});
And the query will still run fine but you will get null for your results.
I hope this helps, this was a lot of fun to gather.
How to find partial search?
Now Im trying to find
db.content.find({$text: {$search: "Customer london"}})
It finds all records matching customer, and all records matching london.
If I am searching for a part of a word for example lond or custom
db.content.find({$text: {$search: "lond"}})
It returns an empty result. How can I modify the query to get the same result like when I am searching for london?
You can use regex to get around with it (https://docs.mongodb.com/manual/reference/operator/query/regex/). However, it will work for following :
if you have word Cooking, following queries may give you result
cooking(exact matching)
coo(part of the word)
cooked(The word containing the english root of the document word, where cook is the root word from which cooking or cooked are derived)
If you would like to go one step further and get a result document containing cooking when you type vooking (missplled V instead of C), go for elasticsearch.
Elasticsearch is easy to setup, has extremely powerful edge-ngram analyzer which converts each words into smaller weightage words. Hence when you misspell, you will still get a document based on score elasticsearch gives to that document.
You can read about it here : https://www.elastic.co/guide/en/elasticsearch/reference/current/analysis-edgengram-tokenizer.html
it will always return the empty array for partial words like when you are searching for lond to get this type of text london..
Because it take full words and search same as that they are ..
Not achive same results like :-
LO LON LOND LONDO LONDON
Here you may get help from ELASTIC-SEARCH . It is quite good for full text search when implement with mongoDB.
Refrence : ElasticSearch
Thanks
The find all is to an Array
clientDB.collection('details').find({}).toArray().then((docs) =>
I now used the str.StartWith in a for loop to pick out my record.
if (docs[i].name.startsWith('U', 0)) {
return console.log(docs[i].name);
} else {
console.log('Record not found!!!')
};
This may not be efficient, but it works for now
Is it possible to query documents where a specific field is contained in a given string?
for example if I have these documents:
{a: 'test blabla'},
{a: 'test not'}
I would like to find all documents that field a is fully included in the string "test blabla test", so only the first document would be returned.
I know I can do it with aggregation using $indexOfCP and it is also possible with $where and mapReduce. I was wandering if it's possible to do it in find query using the standard MongoDB operators (e.g., $gt, $in).
thanks.
I can think of 2 ways you could do this:
Option 1
Using $where:
db.someCol.find( { $where: function() { return "test blabla test".indexOf(this.a) > -1; } }
Explained: Find all documents whose value of field "a" is found WITHIN some given string.
This approach is universal, as you can run any code you like, but less recommended from a performance perspective. For instance, it cannot take advantage of indexes. Read full $where considerations here: https://docs.mongodb.com/manual/reference/operator/query/where/#considerations
Option 2
Using regex matching trickery, ONLY under certain circumstances; below is an example that only works with matching that the field value is found as a starting substring of the given string:
db.someCol.find( { a : /^(t(e(s(t( (b(l(a(b(l(a( (t(e(s(t)?)?)?)?)?)?)?)?)?)?)?)?)?)?)?)?$/ } )
Explained: Break up the components of your "should-be-contained-within" string and match against all sub-possibilities of that with regex.
For your case, this option is pretty much insane, but it's worth noting as there may be specific cases (such as limited namespace matching), where you would not have to break up each letter, but some very finite set of predetermined parts. And in that case, this option could still make use of indexes, and not suffer the $where performance pentalties (as long as the complexity of the regex doesn't outweigh that benefit :)
You can use regex to search .
db.company.findOne({"companyname" : {$regex : ".*hello.*"}});
If you are using Mongo v3.6+, you can use $expr.
As you mentioned $indexOfCP can be used to get index, here it will be
{
"$expr": {
{$ne : [{$indexOfCP: ["test blabla test", "$a"]}, -1]}
}
}
The field name should be prefixed with a dollar sign ($), as $expr allows filters from aggregation pipeline.
I have a Mongo search set up that goes through my entries based on numerous criteria.
Currently the easiest way (I know it's not performance-friendly due to using wildcards, but I can't figure out a better way to do this due to case insensitivity and users not putting in whole words) is to use regex wildcards in the search. The search ends up looking like this:
{ gender: /Womens/i, designer: /Voodoo Girl/i } // Should return ~200 results
{ gender: /Mens/i, designer: /Voodoo Girl/i } // Should return 0 results
In the example above, both searches are returning ~200 results ("Voodoo Girl" is a womenswear label and all corresponding entries have a gender: "Womens" field.). Bizarrely, when I do other searches, like:
{ designer: /Voodoo Girl/i, store: /Store XYZ/i } // should return 0 results
I get the correct number of results (0). Is this an order thing? How can I ensure that my search only returns results that match all of my wildcarded queries?
For reference, the queries are being made in nodeJS through a simple db.products.find({criteria}) lookup.
To answer the aside real fast, something like ElasticSearch is a wonderful way to get more powerful, performant searching capabilities in your app.
Now, the reason that your searches are returning results is that "mens" is a substring of "womens"! You probably want either /^Mens/i and /^Womens/i (if Mens starts the gender field), or /\bMens\b/ if it can appear in the middle of the field. The first form will only match the given field from the beginning of the string, while the second form looks for the given word surrounded by word boundaries (that is, not as a substring of another word).
If you can use the /^Mens/ form (note the lack of the /i), it's advisable, as anchored case-sensitive regex queries can use indexes, while other regex forms cannot.
$regex can only use an index efficiently when the regular expression has an anchor for the beginning (i.e. ^) of a string and is a case-sensitive match.