JS代碼:如何在Ajax調用中獲得兩個數組?
$.ajax({
url: 'assignavailtrainers.php',
data: {action:'test'},
type: 'post',
success: function(data) {
}
});
PHP代碼:
<?php
$username = "trainerapp";
$password = "password";
$hostname = "localhost";
$link = @mysql_connect($hostname, $username, $password);
if(@mysql_select_db("trainer_registration"))
{
$select_query_num = @mysql_query("select program_id,facilitator_id,availability_status from program_facilitator where availability_status in (1,2)");
$select_query_name = @mysql_query("select facilitator_id,firstname,lastname,email_id from facilitator_details");
$num_rows = @mysql_num_rows($select_query_num);
$trainerdetails = [];
$traineravaildetails = [];
$i = 0;
$j = 0;
while($row = @mysql_fetch_assoc($select_query_num))
{
$trainerdetails[$i]['pgidi'] = $row['program_id'];
$trainerdetails[$i]['facilitatorid'] = $row['facilitator_id'];
$trainerdetails[$i]['avail_status'] = $row['availability_status'];
$trainerdetails[$i]['idi'] = $row['facilitator_id'];
$i++;
}
while($row1 [email protected]_fetch_assoc($select_query_name))
{
$traineravaildetails[$j]['facilitatorid'] = $row1['facilitator_id'];
$traineravaildetails[$j]['firstname'] = $row1['firstname'];
$traineravaildetails[$j]['lastname'] = $row1['lastname'];
$traineravaildetails[$j]['emailidvalue'] = $row1['email_id'];
$j++;
}
echo json_encode(array('result1'=>$trainerdetails,'result2'=>$traineravaildetails));
}
?>
請幫我在阿賈克斯成功功能區的代碼。我試過使用initChart2,但我得到一個錯誤,說initChart2沒有定義。我似乎並不瞭解如何從ajax中獲取兩個PHP數組,因爲我是一個新手ajax。如果有人能幫我解釋代碼,那會很棒。而且我還需要知道如何區分從PHP發送的ajax中的輸出。
不要使用'mysql_ *'函數,因爲它們已被棄用了很長時間,並且將在幾個月內被PHP7刪除。改用PDO或mysql ** i **。 – TiMESPLiNTER
檢查響應console.log(data.toSource());也爲ajax設置數據類型 –
我試過console.log(data.toSource());之前..但沒有工作..任何其他方法? –