php脚本只在android studio中显示一行数据而不是很多

2023年2月15日 166次阅读

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;
    $user
=$code;
 $user[mistakes]=$mistakes;
 $user[punishment]=$punishment;
}
 echo json_encode($user);
 mysqli_stmt_close($selectquery);
mysqli_close($con);
?>
  最佳答案   我会选择这样的东西：
$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);
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