PHP/MYSQL - Generate only one random word inside for loop

Question

I am making a small interraction between a user and the computer in a game. The user is always the first one to make a move, where he inputs a word of his choise. I did a sql query that gets all plays made by the user while he is playing (maximun of 10 attempts). Once the user makes the first move I want the computer to generate a random word, like this:

$sql="Select word from plays where user=1 and id=$id";
$result=mysqli_query($link, $sql);

for($i=0;$i<10 && ($row=mysqli_fetch_assoc($result)) ;$i++){

    // some code here

    $random="select word from words order by Rand() LIMIT 1";
    $result=mysqli_query($link, $random);
    $wrandom=mysqli_fetch_assoc($result);
    $wordR=$wrandom['word']; 
 }

In the first attempt all works correctly, a random word is generated but in the second attempt two random words are generated (instead of just one), in the third attempt, 3 words are generated and so on... What should I do to generate just one random word per attempt?

Are you really using the same $result variable for both queries, or did you change that when copying the code? — Barmar
– Barmar, Commented Jan 3, 2019 at 21:43
I don't see anything here that will generate more words each time. I suspect the problem is in the // some code here. — Barmar
– Barmar, Commented Jan 3, 2019 at 21:48

Barmar · Accepted Answer · 2019-01-03 21:47:29Z

1

You can join your two queries into a single query.

$sql = "SELECT p.word as user_word, w.word as computer_word
        FROM plays AS p
        CROSS JOIN words AS w
        WHERE p.user = 1 AND p.id = $id
        ORDER BY RAND()
        LIMIT 10";
$result = mysqli_query($link, $sql);
while ($row = mysqli_fetch_assoc($result)) {
    $wordU = $row['user_word'];
    $wordR = $row['computer_word'];
    // some code here
}

answered Jan 3, 2019 at 21:47

Barmar

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Comments

Pøziel · Accepted Answer · 2019-01-03 21:38:14Z

0

You should try to change the name of your variable $result in the loop because it creates a conflict with your first $result variable.

answered Jan 3, 2019 at 21:38

Pøziel

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1 Comment

Barmar Over a year ago

That seems to be the solution to a different problem. I don't see how this error would cause him to get more than one word on each iteration.

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