提问人：Mary Reejo R 提问时间：4/25/2020 最后编辑：Alessio CantarellaMary Reejo R 更新时间：4/25/2020 访问量：145
获取多行并以 JSON 格式显示
fetching multiple rows and display in json
问：
我有一个PHP代码，如下所示：
while ($row2 = mysqli_fetch_assoc($check2)) {
// $leadarray[] = $row2;
$service_id[$i] = $row2['service_id'];
$row1['service_id'][$i] = $service_id[$i];
$service_name[$i] = $row2['service_name'];
$row1['service_name'][$i] = $service_name[$i];
$i++;
}
$leadarray[] = $row1;
$trimmed1['leaddetails'] = str_replace('\r\n', '', $leadarray);
echo json_encode($trimmed1);
获得类似
{
"leaddetails": [
{
"service_id": [
"7",
"2"
],
"service_name": [
"Past Control Services",
"Civil Finishing"
],
}
]
}
我想要这样的输出：
"service_id": [
1: "7",
2: "2"
],
"service_name": [
1: "Past Control Services",
2: "Civil Finishing"
],
或
[
"service_id": "7",
"service_name": "Past Control Services",
"service_id": "2",
"service_name": "Civil Finishing",
]
答：
0赞
Jeroen van der Laan
4/25/2020
#1
$trimmed1在将它传递给（PHP 中的关联数组）时被视为对象，因为您显式地给它一个键（在 JSON 中被视为对象属性）：json_encode
$trimmed1 = [];
$trimmed1['leaddetails'] = 'foobar';
这将导致编码到一个对象中：json_encode$trimmed1
{
"leaddetails": "foobar"
}
要获得所需的结果，请在传递给 时被视为 ARRAY。$trimmed1json_encode
$trimmed1[] = str_replace('\r\n', '', $leadarray); // push instead of assign
0赞
wayan edi
4/25/2020
#2
第二种选择是不可能的
你可以像这样做[{"service_id":"7", "service_name":"Past Control Services"}, {"service_id":"2", "service_name":"Civil Finishing"}]
评论
0赞
wayan edi
4/25/2020
使用类 ex：class Data{ $service_id; $service_name; } $obj = new Array(); while ($row2 = mysqli_fetch_assoc($check2)){ $data = new Data(); $data->service_id = $row2['service_id']; $data->service_name = $row2['service_name']; array_push($obj, $data); } echo json_encode($obj);
0赞
David Buck
4/25/2020
请编辑您的答案以添加格式正确的 bove 代码。
0赞
wayan edi
4/25/2020
#3
[{"service_id":"7", "service_name":"Past Control Services"}, {"service_id":"2", "service_name":"Civil Finishing"}]
像这样使用类
class Data{
$service_id;
$service_name;
}
$obj = new Array();
while ($row2 = mysqli_fetch_assoc($check2)){
$data = new Data();
$data->service_id = $row2['service_id'];
$data->service_name = $row2['service_name'];
array_push($obj, $data);
}
echo json_encode($obj);
评论
0赞
Mary Reejo R
4/29/2020
这段代码是哎呀，你能帮我把这段代码转换为过程编程吗？
0赞
Barmar
4/25/2020
#4
首先创建二维数组。然后在循环期间推送到嵌套数组上。
$result = ["service_id" => [], "service_name" = []];
while ($row2 = mysqli_fetch_assoc($check2)) {
$result["service_id"][] = $row["service_id"];
$result["service_name"][] = $row["service_name"];
}
echo json_encode($result);
评论