Php 将JSON结果分离为变量
我生成了以下JSON结果,并想了解如何将结果分为两个变量: JSON 预期结果:Php 将JSON结果分离为变量,php,json,Php,Json,我生成了以下JSON结果,并想了解如何将结果分为两个变量: JSON 预期结果: { "data": [ { "source": "server1", "host": "pc1", "description": "SSH server is down on {HOSTNAME}", }, {
{
"data": [
{
"source": "server1",
"host": "pc1",
"description": "SSH server is down on {HOSTNAME}",
},
{
"source": "server2",
"host": "pc2",
"description": "webapp down",
}
]
}
及
PHP代码:
<?php
include '../include/db_conn.php';
print to_json(get_all_alert());
$return = get_all_alert();
print to_json($return["data"]);
print to_json($return["error"]);
?>
php代码仍然会打印两次合并的结果。谢谢好吧,你有一条可能不需要的额外线路
print to_json(get_all_alert());
去掉这个,你的问题很可能会消失。好吧,你有一条额外的线,你可能不需要
print to_json(get_all_alert());
去掉这个,你的问题很可能会消失。好吧,你有一条额外的线,你可能不需要
print to_json(get_all_alert());
去掉这个,你的问题很可能会消失。好吧,你有一条额外的线,你可能不需要
print to_json(get_all_alert());
删除该选项,您的问题可能会消失。您的json数据不正确,您可以在此处检查: 从每个阵列部件中删除额外的逗号 试试这个:
<?php
$data1='{
"data": [
[
{
"source": "server1",
"host": "pc1",
"description": "SSH server is down on {HOSTNAME}"
}
],
[
{
"source": "server2",
"host": "pc2",
"description": "webapp down"
}
]
],
"error": {
"server3": "Host is not allowed to connect to this MySQL server",
"server4": "Cant connect to MySQL server"
}
}';
$val_array = json_decode($data1,true);
print_r($val_array['data']);
print_r($val_array['error']);
您的json数据不正确您可以在此处检查:
从每个阵列部件中删除额外的逗号
试试这个:
<?php
$data1='{
"data": [
[
{
"source": "server1",
"host": "pc1",
"description": "SSH server is down on {HOSTNAME}"
}
],
[
{
"source": "server2",
"host": "pc2",
"description": "webapp down"
}
]
],
"error": {
"server3": "Host is not allowed to connect to this MySQL server",
"server4": "Cant connect to MySQL server"
}
}';
$val_array = json_decode($data1,true);
print_r($val_array['data']);
print_r($val_array['error']);
您的json数据不正确您可以在此处检查:
从每个阵列部件中删除额外的逗号
试试这个:
<?php
$data1='{
"data": [
[
{
"source": "server1",
"host": "pc1",
"description": "SSH server is down on {HOSTNAME}"
}
],
[
{
"source": "server2",
"host": "pc2",
"description": "webapp down"
}
]
],
"error": {
"server3": "Host is not allowed to connect to this MySQL server",
"server4": "Cant connect to MySQL server"
}
}';
$val_array = json_decode($data1,true);
print_r($val_array['data']);
print_r($val_array['error']);
您的json数据不正确您可以在此处检查:
从每个阵列部件中删除额外的逗号
试试这个:
<?php
$data1='{
"data": [
[
{
"source": "server1",
"host": "pc1",
"description": "SSH server is down on {HOSTNAME}"
}
],
[
{
"source": "server2",
"host": "pc2",
"description": "webapp down"
}
]
],
"error": {
"server3": "Host is not allowed to connect to this MySQL server",
"server4": "Cant connect to MySQL server"
}
}';
$val_array = json_decode($data1,true);
print_r($val_array['data']);
print_r($val_array['error']);
谢谢,这很有效,但还有一个问题。json格式还不是很好,谢谢,虽然有效,但仍然有一个问题。json格式还不是很好,谢谢,虽然有效,但仍然有一个问题。json格式还不是很好,谢谢,虽然有效,但仍然有一个问题。json格式仍不正确谢谢,我可以使用您的答案更正错误并修复问题。谢谢,我可以使用您的答案更正错误并修复问题。谢谢,我可以使用您的答案更正错误并修复问题。谢谢,我可以使用您的答案更正错误并修复问题。