我想问一下如何显示多行数据而不是一行.以下代码仅显示一行记录而不是多条记录.我在这里使用
mysqli_prepare语句.或问题是我的
android studio编码?我的应用程序使用登录功能和编码实现如下.
<?php $host="DB_HOST"; $user="DB_USER"; $password="DB_PASSWORD"; $db="DB_NAME"; $con = mysqli_connect($host,$user,$password,$db); $parentic=$_POST["ParentIC"]; $password=$_POST["Password"]; $selectquery = mysqli_prepare($con, "SELECT Parents_Data.Name, Student_List.StudName, Student_List.StudIC, Student_List.Form,Student_List.Class,discipline_record.Date,discipline_record.RulesCode, discipline_record.TypesofMistakes,discipline_record.Punishment FROM discipline_record LEFT JOIN Student_List ON discipline_record.StudIC = Student_List.StudIC LEFT JOIN Parents_Data ON Student_List.ParentIC = Parents_Data.ParentIC WHERE Parents_Data.ParentIC = ? AND Parents_Data.Password = ? "); mysqli_stmt_bind_param ($selectquery, "ss", $parentic, $password); mysqli_stmt_execute($selectquery); mysqli_stmt_store_result($selectquery); mysqli_stmt_bind_result($selectquery,$name,$studname,$studic,$form,$classs,$ddate,$code,$mistakes,$punishment); $user = array(); while(mysqli_stmt_fetch($selectquery)) { $user[name]=$name; $user[studname]=$studname; $user[studic] = $studic; $user[form]=$form; $user[classs]=$classs; $user[ddate]=$ddate; $user
最佳答案 我会选择这样的东西:=$code;
$user[mistakes]=$mistakes;
$user[punishment]=$punishment;}
echo json_encode($user);
mysqli_stmt_close($selectquery);
mysqli_close($con);
?>
$userGroup = array(); $user = array(); while(mysqli_stmt_fetch($selectquery)) { $user[name]=$name; $user[studname]=$studname; $user[studic] = $studic; $user[form]=$form; $user[classs]=$classs; $user[ddate]=$ddate; $user
=$code;
$user[mistakes]=$mistakes;
$user[punishment]=$punishment;array_push($userGroup,$user);
}
echo json_encode($userGroup);