php - 当数组来自 mysql_fetch_array 时，在 php 中显示数组差异

Question

我正在设计一个 PHP 脚本，它显示给定日期某个旅行社的可用巴士。这是 php 代码片段。

$query1 = mysql_query("SELECT bus_id FROM schedule WHERE travel_agency_id = '".$travel_agency_id."' AND scheduleddate ='".$busDate."'");
$query2 = mysql_query("SELECT bus_id FROM pc WHERE travel_agency_id = '".$travel_agency_id."' ");

$array1 = array(@mysql_fetch_array($query1));
$array2 = array(@mysql_fetch_array($query2));
$diff = array_diff($array2, $array1);
print_r ($diff);

我认为上述查询将在给定日期 "$busDate"打印旅行社 ID "$travel_agency_id" 的 "bus_id" 。但是出现了如下错误。

Notice: Array to string conversion in.......... Array ( [0] => )

谁能解释一下为什么会发生这个错误..？如何使用 "echo" 显示可用总线数组 (id) ？

Answer 1

试试这个..

    $query1 = mysql_query("SELECT bus_id FROM schedule WHERE travel_agency_id = '".$travel_agency_id."' AND scheduleddate ='".$busDate."'");
    $query2 = mysql_query("SELECT bus_id FROM pc WHERE travel_agency_id = '".$travel_agency_id."' ");
    if(mysql_num_rows($query1) > 0 && mysql_num_rows($query2) > 0 ){
    $array1 = mysql_fetch_array($query1);
    $array2 = mysql_fetch_array($query2);
    $diff = array_diff($array2, $array1);
    print_r ($diff);
}