PHP JSON is dropping first record

Question

I have a line of code to convert data from database to JSON. The only problem is that the code I have seems to be dropping the first record. Any thoughts to why.

$return_arr = Array();

$query_qrySelect = "SELECT * FROM table";
$qrySelect = mysql_query($query_qrySelect, $database) or die(mysql_error());
$row_qrySelect = mysql_fetch_assoc($qrySelect);
$totalRows_qrySelect = mysql_num_rows($qrySelect);

while ($row = mysql_fetch_array($qrySelect, MYSQL_ASSOC)) {
    array_push($return_arr,$row);
}
echo json_encode($return_arr);

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the pink box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — h2ooooooo
– h2ooooooo, Commented Jan 28, 2014 at 14:59

Marc B · Accepted Answer · 2014-01-28 14:58:48Z

2

This line:

$row_qrySelect = mysql_fetch_assoc($qrySelect);

is where you're "losing" your data.

answered Jan 28, 2014 at 14:58

Marc B

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Community · Accepted Answer · 2023-11-17 19:36:10Z

1

Because you call mysql_fetch_assoc and never use the result anywhere ($row_qrySelect is never used).

According to the manual:

Returns an associative array that corresponds to the fetched row and moves the internal data pointer ahead. mysql_fetch_assoc() is equivalent to calling mysql_fetch_array() with MYSQL_ASSOC for the optional second parameter. It only returns an associative array.

edited Nov 17, 2023 at 19:36

CommunityBot

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answered Jan 28, 2014 at 14:58

h2ooooooo

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