Looping with mysql_fetch_array skipping over first row

Question

I'm having trouble iterating through MySQL rows. This is my current code:

$query = "SELECT * FROM translations
          WHERE iddoc = '$id'
          AND submitted = 1;";
$result= mysqli_query($query);
$numrows = mysqli_num_rows($result);
$row2 = mysql_fetch_row($result);

if ($numrows > 0) {
    while($eachrow = mysqli_fetch_array($result, MYSQLI_NUM)) {
        echo $eachrow[0];
        echo ", ";
        echo $numrows;
   }
}

The result of this is:

6, 2

But if there are 2 rows, why is the while loop ending after only 1 iteration? What am I understanding wrong?

EDIT: It appears to be displaying ONE LESS than the correct amount of rows. I.E. the while loop is running 1 less time than it should.

What is $numtrans? And I'm assuming you left out your select query for a reason... — the_red_baron
– the_red_baron, Commented Aug 22, 2012 at 18:58
You sure there is only one row? did you run it in a mysql broswer? — Naftali
– Naftali, Commented Aug 22, 2012 at 18:59
The code as it is has absolutely no problem! Add more significant code — Razvan
– Razvan, Commented Aug 22, 2012 at 19:00
You're not using mysqli_query correctly here. You should NEVER introduce things like $id directly in the query but should instead use ? placeholders and bind these values before executing the query. Failing to do this could lead to serious consequences. — tadman
– tadman, Commented Aug 22, 2012 at 19:19
@Hubrid What tadman meant was to use the prepare statement, not just putting in the whole query. — lusketeer
– lusketeer, Commented Aug 22, 2012 at 19:22

sir_thursday · Accepted Answer · 2012-08-22 19:37:08Z

3

Found problem. I was fetching the first row with

$row2 = mysql_fetch_row($result);

outside of the while loop, thus causing it to start on the second row and skip over the first.

answered Aug 22, 2012 at 19:37

sir_thursday

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1 Comment

Naftali Over a year ago

@Hubrid... Look it up sometime.

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