PHP while loop omits first entry from MySQL database select

Question

I have four rows in my table. Only three are shown.

$query  = "SELECT * FROM table";
$result = mysql_query($query);
$row    = mysql_fetch_array($result);

while($row = mysql_fetch_array($result)) {
    echo $row['id'];
}

The result is 234, but should be 1234.

What am I doing wrong?

NikiC · Accepted Answer · 2010-10-30 11:17:49Z

10

$row    = mysql_fetch_array($result);

This line already fetches the first entry. Thus in the while loop you fetch the second element.

Correctly it should be:

$query  = "SELECT * FROM table";
$result = mysql_query($query);

while ($row = mysql_fetch_array($result)) {
    echo $row['id'];
}

Alternatively:

$query  = "SELECT * FROM table";
$result = mysql_query($query);
$row    = mysql_fetch_array($result);

do {
    echo $row['id'];
} while ($row = mysql_fetch_array($result));

answered Oct 30, 2010 at 11:17

NikiC

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Comments

Eran Galperin · Accepted Answer · 2010-10-30 11:17:48Z

4

You are already advancing the query buffer one row before the loop by calling mysql_fetch_array() outside of it. Remove that call and it should work as expected

answered Oct 30, 2010 at 11:17

Eran Galperin

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