This line:
$row_qrySelect = mysql_fetch_assoc($qrySelect);
is where you're "losing" your data.
Domanda
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);
Soluzione
This line:
$row_qrySelect = mysql_fetch_assoc($qrySelect);
is where you're "losing" your data.
Altri suggerimenti
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.