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In my MongoDB, I have stored value as below,

 
 "_id" : ObjectId("5bed3f5019f0431be000412b"), 
 "reference" : "SL2PR01MB2745E4160158C08C4B7A367285C30@SL2PR01MB2745.apcprd01.prod.exchangelabs.com,SL2PR01MB274333160158C08C4B7A367285C30@SL2PR01MB2745.apcprd01.prod.exchangelabs.com", 
 "email_thread_id" : NumberInt(5)

 
 "_id" : ObjectId("5bed3f5019f0431be000412b"), 
 "reference" : "SL2PR01MB2745E4160158C08C4B7A364444C30@SL2PR01MB2745.apcprd01.prod.exchangelabs.com", 
 "email_thread_id" : NumberInt(6)

My search keyword is:

"SL2PR01MB2745E4160158C08C4B7A364444C30@SL2PR01MB2745.apcprd01.prod.exchangelabs.com"

Which is match with second array from above JSON.

This is my query,

$referanceId= "SL2PR01MB2745E4160158C08C4B7A364444C30@SL2PR01MB2745.apcprd01.prod.exchangelabs.com";
$mailExists = DB::connection('mongodb')->collection('email_message')
 ->select('email_thread_id'); 
 if ($referanceId) 
 $mailExists->whereRaw("find_in_set(".$referanceId.", reference)");
 
$query = $mailExists->first();

But it gives me error like below:

ServerException in Find.php line 299:

$or/$and/$nor entries need to be full objects

Please help me to resolve this issue. Thank you.

You can do this easier with the laravel where method, this will do the comparison you want, and also make your code better readable.

You can use the following query to accomplish this:

DB::connection('mongodb')->collection('email_message')
 ->where("reference", $referanceId)->first();

If your value contains multiple reference id's as a string you can use:

$regexQuery = '/.*'.$referanceId.'.*/';
DB::connection('mongodb')->collection('email_message')
 ->where("reference", 'regexp', $regexQuery)->first();

This will match the referenceId in any position in the string.

As @hetalgotel mentioned in the comments another query for this is:

$mailExists->where("reference", 'like', "%$referanceId%")

This one produced the expected result in this scenario