Query returns wrong output result in laravel

Question

I want to compare array of words with a field of database table which contains more than 2000 entries. For example user input a string "this is good" It will explode the string into array and than match the words with the database table , such as good exists in the table then it will return the good keyword. here is my code

$cmnt = $request['comment'];
$parse = explode(' ',$cmnt);
$length = count($parse);
for($i = 0; $i < $length ; $i++)
{
    $keyword = Rate::where('keyword',$parse[$i]);
}
if ($keyword != null)
{
    dd($keyword);
}

Here is my out put

Niklesh Raut · Accepted Answer · 2017-07-01 12:59:37Z

3

You are over writing value of $keyword in each loop iteration do like this

$cmnt = $request['comment'];
$parse = explode(' ',$cmnt);
$parse = array_unique($parse);
$keyword = Rate::whereIn('keyword',$parse);
dd($keyword);

answered Jul 1, 2017 at 12:59

Niklesh Raut

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2 Comments

DevTaabi Over a year ago

But I want to compare all the values of array thats why I am using for loop , how this can work without iterations?

Ruchita Sheth Over a year ago

the above will create a query like select * from rate where keyword in('this','is', 'good').

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