我想将所选结果转换为JSON。这是我的代码：
<?php
include("DbConnect.php");
$connection=new DbConnect();
$sth = mysqli_query($connection->_con,"SELECT * FROM account WHERE ac_id='1'");
if($sth){
$rows = array();
while($row = mysqli_fetch_assoc($sth)){
$users = mysqli_query($connection->_con,"SELECT user.user_id,user.name,user.email,ac_detail.ac_id,ac_detail.amount FROM user,ac_detail WHERE ac_detail.ac_id='1' AND user.user_id=ac_detail.user_id");
$usersArray = array();
while($userRow = mysqli_fetch_assoc($users)){
$usersArray[]=$userRow;
}
$a=array("users"=>$usersArray);
//$row["user"]=$usersArray
array_push($row,$a);
$rows[] = $row;
}
echo json_encode(array('data'=>$rows));
}else{
echo json_encode(array('message'=>'error - 2'));
}
?>
通过执行此代码，它生成类似JSON的：
{"data":[{"ac_id":"1","user_id":"2","title":"Travel","ac_for":"Traveling","required_amount":"50","current_amount":"0","initial_date":"2014-11-11","final_date":"2014-11-14","is_shared":"1","status":"1","0":{"users":[{"user_id":"2","name":"Muhammad Imran","email":"macrotechnolgies@gmail.com","ac_id":"1","amount":"0"},{"user_id":"3","name":"Muhammad Imran","email":"macrotecholgies@gmail.com","ac_id":"1","amount":"0"}]}}]}
但我不想要"0"｛"user:：…｝
应该如何（预期结果）：
{"data":[{"ac_id":"1","user_id":"2","title":"Travel","ac_for":"Traveling","required_amount":"50","current_amount":"0","initial_date":"2014-11-11","final_date":"2014-11-14","is_shared":"1","status":"1","users":[{"user_id":"2","name":"Muhammad Imran","email":"macrotechnolgies@gmail.com","ac_id":"1","amount":"0"},{"user_id":"3","name":"Muhammad Imran","email":"macrotecholgies@gmail.com","ac_id":"1","amount":"0"}]}]}
提前感谢
您正在执行：
while($row = mysqli_fetch_assoc($sth)){
[...snip...]
array_push($row,$a);
while行创建一个数组$row，然后使用它的一部分来创建$a。然后将该$a推回到原始$row阵列上。但是$row已经是一个关联数组，所以按下的$a得到了键0。
由于您现在将一个关联数组（非数字键）与一个数字键数组（推送操作）混合在一起，PHP必须将数字键添加到您的推送项中：数组中不能有没有键的元素。
然后，由于JS不允许实际的JS数组（[]）具有非数字键，因此必须将整个数组转换为对象（{}）。
你可能想要的东西更像：
while($row = ...) {
... build $a ...
array_push($row['users'], $a);
相反。
为什么不代替array_push($row,$a)尝试以下操作：
<?php
include("DbConnect.php");
$connection=new DbConnect();
$sth = mysqli_query($connection->_con,"SELECT * FROM account WHERE ac_id='1'");
if($sth){
$rows = array();
while($row = mysqli_fetch_assoc($sth)){
$users = mysqli_query($connection->_con,"SELECT user.user_id, user.name, user.email, ac_detail.ac_id, ac_detail.amount FROM user,ac_detail WHERE ac_detail.ac_id='1' AND user.user_id=ac_detail.user_id");
$usersArray = array();
while($userRow = mysqli_fetch_assoc($users)){
$usersArray[]=$userRow;
}
// here comes the change
// $a = array("users"=>$usersArray);
// //$row["user"]=$usersArray
// array_push($row,$a);
$row['users'] = $usersArray;
$rows[] = $row;
}
echo json_encode(array('data'=>$rows));
}else{
echo json_encode(array('message'=>'error - 2'));
}
这应该行得通。没有样本数据来测试它。