我将把数组发送到java以填充ListView，如何获取以下代码来输出数组？
<?php
$user = 'root';
$pass = '';
$db = 'uopuser';
$con=mysqli_connect('localhost', $user, $pass, $db) or die('Unable to connect');
$statement = mysqli_prepare($con, 'SELECT * FROM society');
mysqli_stmt_execute($statement);
mysqli_stmt_store_result($statement);
mysqli_stmt_bind_result($statement, $society_id, $name, $email, $description);
$society = array();
while(mysqli_stmt_fetch($statement)){
$society['society_id'] = $society_id;
$society['name'] = $name;
$society['email'] = $email;
$society['description'] = $description;
echo json_encode($society);
}
echo json_encode($society);
mysqli_stmt_close($statement);
mysqli_close($con);
?>
而不是：
{"society_id":1,"name":"TestName1","email":"Test@email1","description":"TestDes1"}
{"society_id":2,"name":"TestName2","email":"Test@email2","description":"TestDes2"}
{"society_id":3,"name":"TestName3","email":"Test@email3","description":"TestDes3"}
在发布此内容之前，我已经查看了互联网，但感到非常困惑！提前感谢任何人。
如果你
想要一个数组，你就走得太远了，也错过了一步。您正在 while 函数中创建一个数组，但您不需要使用 json_encode，除非您将其传递给需要 json 格式的内容。要创建一个大数组，只需创建另一个数组：
$society = array();
while(mysqli_stmt_fetch($statement)){
$new_soc = array();
$new_soc['society_id'] = $society_id;
$new_soc['name'] = $name;
$new_soc['email'] = $email;
$new_soc['description'] = $description;
//$new_soc is now a single array. You add it to the larger array next
$society[] = $new_soc;
}
现在，您可以对大型$society数组执行任何需要的操作。
如果您只是想要数组形式的输出，那么为什么要将其编码为 JSON？ 您可以简单地使用，
print_r($society);
附言：如果您正在为 AJAX 调用执行此操作，那么使用此json_encode函数会更好。你可以在 JavaScript 代码中使用 JSON.parse（） 方法来解析收到的响应。