Question

我有以下PHP从MySQL查询创建一些JSON。当有多个元素，即mysql_fetch_assoc＆gt;时，这可以正常工作1但是如果它= 1则没有任何工作和调试，while循环中的任何echo语句都不会被调用！

这是怎么回事？

$byFlight = mysql_query($query_byFlight, $pfArchiveDB) or die(mysql_error());
$row_byFlight = mysql_fetch_assoc($byFlight);
//$totalRows_byFlight = mysql_num_rows($byFlight);
//echo ($query_byFlight);
$flights = array();
if(mysql_num_rows($byFlight)) {
    while($flight = mysql_fetch_assoc($byFlight)) {
        $flights[] = array('flight'=>$flight);
    }
}
header('Content-type: application/json');
echo json_encode(array('flights'=>$flights));

Answer 1

你正在进行两次获取，这可能会消耗你的第一行。这段代码可以正常工作：

$byFlight = mysql_query($query_byFlight, $pfArchiveDB) or die(mysql_error());

 //$row_byFlight = mysql_fetch_assoc($byFlight); // <---- COMMENTED THIS FETCH

//$totalRows_byFlight = mysql_num_rows($byFlight);
//echo ($query_byFlight);
$flights = array();
if(mysql_num_rows($byFlight)) {
    while($flight = mysql_fetch_assoc($byFlight)) {
        $flights[] = array('flight'=>$flight);
    }
}
header('Content-type: application/json');
echo json_encode(array('flights'=>$flights));

MYSQL虽然在= 1时没有返回mysql_fetch_assoc

1 个答案: